@@yoprofmatt The maria occupy one third of the visible NEAR side of WHAT IS THE MOON. This IS fundamentally related to the surface gravity of WHAT IS THE MOON. Attention: The rotation of WHAT IS THE MOON matches the revolution. Accordingly, ON BALANCE, the gravity of the Sun upon the Moon IS about TWICE that of the Earth. Therefore, on balance, the Moon's crust is about TWICE as thick on it's far side. SO, we multiply one half times one third in order to obtain the surface gravity on the Moon in comparison with that of what is THE EARTH/ground. We then ALSO determine (ON BALANCE) that the NEAR side of what is the Moon (basically) DOES contain the maria (AND at what is then 33 percent) !! Great. By Frank Martin DiMeglio
I was scared to open the video since the thumbnail consists of different formulas and what not. Well can't believe I finished the video and actually understood what he was explaining.. Nice.
I do love the way he's always being honest of the fact that we might not always immediately know the answers to certain problems. He uses the words: 'might', 'we don't know '... 🙏🏼👏🏼
I love the way this guy teaches, I never finished college, but learn so much from these lectures, thanks for posting them. There was such a great but missed opportunity at 5:50. Could have said “to infinity, and beyond!!!”
A Quick and (Somewhat) Dirty Way to Calculate Escape Velocity - I was doing some calculations using the escape velocities from Earth, Moon and Mars. Then by chance I calculated the velocities attained when an object was "dropped" from a height of the radius of each of these bodies, ASSUMING THE ACCELERATION DUE TO GRAVITY REMAINED CONSTANT DURING THE FALL, AND WAS EQUAL TO G AT EARTH'S SURFACE. Escape velocity is the minimum velocity needed to escape a gravitational field. For example, escape velocity on earth, Vesc = 40,270 km/h (given). Using the simple formula V2 = (2ar)^0.5, where V2 = velocity after falling distance r; a = acceleration due to gravity; r = radius of earth (initial velocity = 0), and a = 9.80665 m/s^2 r = 6,378,000 m V2 = (2*9.80665*6378000)^0.5 = 11,184.53 m/s = 40,264.29 km/h This value of 40,264 km/h compares very closely to the escape velocity of 40,270 km/h I thought perhaps this was a coincidence, so I calculated V2 for the Moon and Mars and compared them to the known escape velocities for each body. Moon: Escape velocity = Vesc = 8,533.6 km/h a = 1.62 m/s^2 r = 1,737,150 m V2 = (2*1.62*1737150)^0.5 = 2,372.418 m/s = 8,540.7 km/h Again, a very close fit. Mars: Escape velocity = Vesc = 18,108 km/h a = 3.72761 m/s^2 r = 3,389,500 m V2 = (2*3.72761*3389500)^0.5 = 5026.875 m/s = 18,097 km/h Another very close approximation. I haven't done the calculations for the other planets, moons, or the sun, but I expect I would get similar results. From these results I speculate that Vesc = V2. This seems to be a quick and dirty formula for calculating escape velocity, and might be handy for a stranded astronaut. But am I really just reiterating Vesc = (2Gm/r)^0.5? (G = gravitational constant; m = mass, e.g., of earth; r = radius, e.g., of earth). I any case, using this method avoids a lot of the very intricate calculations necessary for determining Vesc the standard way, as long as the acceleration due to gravity and radius are known. Sources: www.livescience.com/50312-how-long-to-fall-through-earth.html keisan.casio.com/exec/system/1360310353 www.wolframalpha.com/input/?i=escape+velocity+Mars
@Ari Parker Basically Gravitational force 'F' is directly proportional to the mass of the planet (the larger mass) 'M', the smaller mass, 'm', and inversely proportional to r,^2 the distance between the centres of the two masses. Meaning F is proportional to Mm/r^2. G, Newton's universal gravitational constant is the constant of proportionality between them. Meaning that F=GMm/r^2
@Ari Parker G is the constant of proportionality in the universal law of gravitation. It was first measured by Henry Cavendish, who sought to measure the mass of Earth in the interest of geology. He measured it by "weighing" lead balls in each others' gravitational fields, on a torsion balance to increase its sensitivity. By measuring the period without the stationary balls, and with the stationary balls, he could determine the impact of the gravity between objects whose mass he could know in advance, and back-out the universal constant of gravitation.
To the OP, it turns out that you get the correct result, but your reasoning is inconsistent with what physical principles are actually at play. The reason is that gravitational fields are not uniform from the radius of a planet out to twice the radius of the planet, and a body isn't in deep space when it is twice the radius of the planet. It is a mathematical coincidence that this method works. The reason it works: v = sqrt(2*g*h) vesc = sqrt(2*G*M/r) Gravitational field: g = G*M/r^2 Assume h = r, and v = vesc, and combine: vesc = sqrt(2*(G*M/r^2) * r) = sqrt(2*G*M/r)
Escape velocity is obtained considering the equality: Kinetic energy of the object with this escape velocity=work needed to bring the object to infinity (definite integral from R to infinity of F ) which is GMm/R.
Sir I have seen your many videos since the time I have entered to class 11 and from the time I found your RUclips channel . I am soo greatful and blessed to find your RUclips channel. Sir I know it may sound like something impossible but I really have a wish to have a live class with you when I would be one of your students whome you would be teaching offline and to listen you. Sir thank you for all your such great explaintion.
you was talking about a rocket in the beginning.But for a rocket which has a fuel,there is no meaning in escape velocity.it can escape the earth with any velocity it wants. escape velocity makes sense only for projectiles with no fuel
unfortunately, this gets way more complicated to decipher and grasp as you start think about multiple gravitational fields (e.g. first you escape the earth but then you're stuck in the solar system). Note also that this all depends on where you first start. For example, it's easier to escape the solar system from the earth than it is to do so from the surface of the sun
BUT...rockets are not "projectiles" like cannon balls or bullets. They carry a fuel source which can continue to 'burn' and provide acceleration. An object could 'escape' Earth at a sustained velocity as slow as only 1 mile per hour, correct? The term "escape velocity" should really be something like "muzzle velocity to escape" or "initial blast velocity to escape"....right???
6:39 yes we launched voyager and many other satellites into orbit or into space. But every launch (I think) first went into orbit, going further and further into high-orbit to eventually swing away with very little energy needed. This is correct right?
Arman, Heard that before. My kids agree. Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A
(Just saw previous post... had same basic idea) --- escaping earth gravity by going at any speed (even very slow speed) , just have steady thrust until you arrive at a Lagrangian point (or a point where gravity attraction from another object such as the Moon or another planet takes over, thus canceling out Earth's gravitational pull). If your spacecraft was very light, you may even be able to hitch a ride on a nearby asteroid or satellite of some kind. The ability of light spacecraft to have efficient steady thrust, would allow for low cost space visits, especially when the return to earth can be a glide and you are able to re-use the craft. I suppose the main problem here is getting the small spacecraft (or just a spacesuit) light enough and still have ability to carry some kind of reliable steady thrust / power system).
I have exactly same question. Escape velocity talks about projectiles but what about an object that's rocket powered that keeps it powered until it crosses the limit of Earth's gravitational pull. Does it need to have velocity equal to Escape velocity or not? I think it does not need to be going at ecape velocity if it has additional power source.
The rocket example everyone gives when explaining escape velocity is very misleading, since rockets are usually percieved as having their own means of acceleration, so they can escape the Earth at basically any speed as long as said acceleration is greater than or equal to 1g (assuming its moving directly away from the planet) . Escape velocity, however, is about initial speed of a body without any forces applied to it other than gravity, just like you said in the video
Without any excess final velocity, how can we consider rfinal as infinity because since there is no excess final velocity the object will not keep moving after it escaping the earth's gravitation but should exactly escape and stop in that space while not being under earth's influence anymore..
Marc Sleiman he is technically right since as rf gets bigger and bigger (to infinity) that whole term will get smaller and smaller to essentially zero, it's just that he decided to call it "zero". I know it's a late but just in case someone else looks at this comment
We could try it "differently", but the result is obviously the same. Using the Work energy theorem aka that Summa F*s = change in kinetic energy = 0.5 mv(final)^2 - 0.5mv(initial)^2, we could make the problem like this: Let us assume that we launch a projectile with the speed necessery to reach an R distance where gravitational force converges to 0 (basically R= infinity). Let us also assume that only G is the (only) working force here. Now by using that Work (of G) = change of kinetic energy and that work of G is calculated as integrandus r to infinity of the function G(r)=mGM/r^2, we basically get the same result. Of course in my example we did not assume v(final) being 0, which is technically saying "we started with the minimum required amount of kinetic energy". Ofc it all depends on how well versed one is in calculus. If someone' just learned about potential energy with its formula definition, but no calculaus/proof of it, then your example is easier to understand, but people can get confused by you saying "we dont wanna have any excess speed in the end at 3:48". The thing with Newtonian physics is that the math part is actually pretty easy, the tougher part is getting the ideas and implications right.
The correct answer to the question of when it will stop is, it will never stop. It will approach zero velocity with reference to the earth asymptotically (never reaching zero). The assumption that Vf reaches zero in the solution to the problem is contingent on the assumption that Rf is infinity.
Thank you sir you really helped a lot but can you please tell how the potential energy at 2:57 is -GMm/r. In particular I didn't understand why the potential energy is negative. Thanks
Lime Pie, See this one: ruclips.net/video/19wjt_zM1xU/видео.html Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A
Sanidhya Upadhyay, Thanks to all my Fandersons in India! You're awesome. Check out my new website: www.universityphysics.education Coupon for 50% off: RUclipsFANDERSON Let's keep learning physics together! Cheers, Dr. A
The derived equation describes an escape velocity from Earth's gravity, but how would you describe the escape velocity of our solar system? Would it be the escape velocity from earth plus an additional term for the sun's mass and earth's distance from the sun etc?
Not 100% sure, but I think the reason we call it solar system, is because the Sun is the central, "fattest" object. I would say that we only need to calculate this very same equation, but with the Sun's mass and radius probably. Try that and compare it to a wiki/googled solar EV estimate. :D
The sun contains 99% of the mass of the solar system, and the gravity of the other planets is close to insignificant, unless we deliberately take credit for them in a gravitational slingshot. Since the spacecraft doesn't go deep in the gravitational wells of other planets, it is reasonable to neglect their contributions. The way you'd calculate escape velocity for an object starting on Earth, and leaving the solar system, is to first calculate the change in GPE necessary to permanently leave the planet Earth from the Earth's surface. Then calculate the change in GPE necessary to permanently leave the sun's gravitational field, from the position of Earth's obit. Add up these two changes in GPE, and equate to the KE of the object leaving Earth's surface. Then solve for velocity. The equation: 1/2*m*v^2 = -(-G*Me/Re - G*Ms/re) Solved for velocity: v = sqrt(2*G*(Me/Re + Ms/re)) where: Me and Ms are masses of the Earth and sun. Re is the radius of Earth re is the orbital radius of Earth at the instant of launch Plugging in data, assuming re = 1 AU, Re is the volumetric average radius of Earth we get the following, for the escape velocity of the solar system at the position of Earth's surface. v = 43600 m/s
Good explanation. But, doesn't V have to be relative to the center of the planet? The reason I ask is because most rockets accelerate in a near circular trajectory around the earth to get into orbit. (Orbit velocity has similar mechanics). But, in orbit, the velocity creates a centrifugal force to counteract gravity. I guess what I'm asking is - is the math the same for a rocket in an angular direction versus one in a radial direction?
You are correct. To ultimately escape, you have to hit this speed (which is relative to the planet's center). If you're at the equator and already moving at 1000 mph, you've got a head start. Cheers, Dr. A
so i have an issue, this all relies on the term Rf being infinity but surely an infinite distance is somewhat theoretical. so assuming there were no objects for it to hit or be caught in the gravitational field of would an object not always after some time be attracted back to earth? i have a feeling there is an answer lying in the fact v is 0 at the point R is infinity but an explanation would help
Dan Evetts, If objects are going at escape velocity, they have zero velocity when they get very far away (r -> infinity). If they are shot at a faster velocity than escape velocity, they still have some velocity even as r -> infinity. Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A
When you throw a ball straight up in the air, that ball immediately starts slowing down. Eventually, it reaches a point where it's velocity is zero, and then starts falling back down. If a projectile is launched at escape velocity, it will simply never slow down to zero - unless it travels an infinite distance - which would taken an infinite amount of time. Any arbitrarily small amount slower, and you could theoretically calculate when the projectile would reach zero and begin falling back down. Anything AT escape velocity or higher simply would never do it - because it will never reach an infinite distance to reach zero.
Okay but say i had a rocket that did just a constant 10mph heading towards space would it never escape because it would need infinite fuel? thats what i never understood
Hey Prof, I got a question. Escape velocity talks about projectiles but what about an object that's rocket powered that keeps it powered until it crosses the limit of Earth's gravitational pull. Does it need to have velocity equal to Escape velocity or not?
I don't understand escape velocity. The further from the centre of the Earth, the less the Earth's gravitational force therefore the escape velocity should be constantly changing. As your distance gets further and further from Earth, you're required less and less velocity to escape because Earth's gravitational force keeps getting weaker and weaker as you get further and further therefore how can we define 1 specific excape velocity? I've watched 6 escape velocity videos now and this seemingly simple question/confusion/misunderstanding of mine is yet to be answered...
Spinelli, Great question. At 4:42 we derive the equation for escape velocity, and it has ri in the denominator. This ri is the distance from the center of the planet. (Later we set this to R, the radius of the planet.) If you use ri as your starting point, then this is your escape velocity from that point. For example, pretend you start at a distance of 2R from the planet. Set ri = 2R to solve. Since the denominator just got bigger, the escape velocity gets smaller. This is in line with your intuition, so nice job. Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A
Please, somebody help me, why is it -G.M.m/r and not G.M.m/r^2 ? at 3:00 I just found out you have a other video explaining that, but, sorry I don't understand that either, escpecially the integrals thingy. (Sorry for any possible spelling mistakes.)
It s easy really this simbolises the potential energy and the other simbolises the gravitational force exerted on an object on a specific point in space
When you integrate G*M*m/r^2 relative to r, you get -G*M*m/r. That's what this formula calculates, is the integration from a point in space to a point within the gravitational well of an astronomical body. Integrate relative to r, and you get the quantity we call gravitational potential energy. It is used as a shortcut for evaluating work done by a gravitational field, when moving an object from point A to point B, which is path-independent, because gravity is a conservative force.
John Terpack, There must be, since obviously you can't escape a black hole. But for rockets with continuous burn rate, you don't talk about escape velocity in quite the same way. The limit is when the gravity pulling down equals the maximum upward thrust that the rocket can generate. Forces are equal and opposite thus the acceleration is zero. Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A
@@yoprofmatt I'll see if I can find out the max thrust of a chemical rocket. My question was inspired by the frequent appearance in sci-fi of civilizations on high-gravity planets. At some point, gravity would be too strong for artificial structures entirely... maybe even too strong for terrestrial life. But I always thought it would be interesting if there was an advanced race that was limited by the fact they just couldn't launch rockets. And I wondered how much stronger gravity would need to be, compared to earth, to achieve that.
So at the black hole, event horizon, escape velocity is the speed of light... That means, if you shine a light, it goes out a bit, then returns back? But isn't the speed of light constant through space, so deceleration is not allowed? When the light starts to return, it will reach the speed of 0 m/s briefly? Is it allowed in physics? How about we have a very fast spaceship? Even if the escape velocity is the speed of light, if the spaceship keeps accelerating at the rate slightly higher than the surface gravity at the event horizon.... Then the spaceship can escape the black hole from inside the event horizon as well?
I am certainly not a black hole expert, but I would say this: Inside the event horizon, light (or anything else) cannot escape. Outside the event horizon, light (and possibly other stuff) can escape. There's LOTS of questions about what happens right at the event horizon, including stuff like Hawking radiation. And I would say many of the questions do not have definitive answers yet because black holes are still a relatively new study. Cheers, Dr. A
@@yoprofmatt I found out the answer... from the internet...:) There are two kinds of black holes, Newtonian and General Relativity (GR). Newtonian black holes, light cannot escape, but anything can, as long as they have a powerful enough jetpack. GR black holes, light or anything else cannot escape for some reason related to spacetime thing... But the strange thing is, if the black hole is 1 trillion solar mass (and such black hole is allegedly found) The surface gravity is only 1.5 g, and the density is thinner than Martian air. In such situation, how come we cannot escape from the event horizon, even with GR black hole? For that, I have not found an answer. Thanks!!!
YR NAGARAJU, Currently yes. Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A
I'm still missing something in the equation. What about the mass of the object and the density of both? If there is a neutron star sitting on the surface, does it need the same speed if it's innertia is so large? Yea, and what about applying the velocity horizontally? I need a better vid obviously
As long as the escaping object is insignificant compared to the mass of the planet, and as long as the planet's mass is moderate enough that relativistic effects don't come in to play, this formula is valid. The mass of the escaping object "cancels out" of the equation of energy balance, and it turns out that the escape velocity is independent of the escaping object's mass. It still works if you escape horizontally, instead of vertically, because gravity is a conservative vector field. No matter which direction you launch, you escape the planet's gravitational field, as long as you aren't on a crash trajectory.
Do you really can write backwards??... or is this some new technology? BTW I loved the explanation... I always thought spacecraft attained this speed to reach outer space.. my whole childhood was a lie.
санкет мхаске, Not writing backwards (I'm not that talented). The board is called Learning Glass. You can check it out at www.learning.glass Cheers, Dr. A
in my book, the approach used is the gravitational force acts as a centripetal force (mv^2)/R=GmM/R^2. So, we found v^2=GM/R. Whats wrong with my book?
indra, Nothing wrong with your book, but that is the speed of orbit, not escape speed. Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A
That's the escape velocity derivation that you're looking at which is the minimum velocity an object/satellite should have to orbit the planet given an orbital radius. Interestingly enough it's also related to the escape velocity as the escape velocity is the square root of 2 times the orbital velocity. That means the escape velocity for a given object would be about 41.4% more than it's orbital velocity.
Radius and mass determine the gravitational field at every point surrounding a massive body. Integrating this gravitational field relative to radial position, and we get the GPE at every distance from the body, relative to it being defined as zero, infinitely far away. If we add enough kinetic energy for it to coast the whole way out, then the initial KE equalling the total necessary change in KE, is what we need in order to escape. A black hole is an extreme case, where there exists a special radius at which not even light can escape the gravitational field, because the escape velocity exceeds the speed of light. This is called the Schwarzschild radius, named after a physicist whose name fortuitously means "black shield". This sets the boundary for the point-of-no-return, for an object falling in to the black hole, that we call the event horizon.
But eventually the velocity f will decrease as gravity is acting and we want the minimum possible velocity that will take projectile put of the gravitational influence. So, as we require minimum velocity it will slow down and come at rest till the boundary of gravitational influence
Why is escape velocity important for a powered craft. Can't a hypothetical rocket thrust away from the earth at 30 miles per hour if it has enough fuel. Surely escape velocity is only necessary if I'm fired out of a cannon on Earth's surface and am unpowered from then on. Also with black holes and the event horizon, can light cross out of the event horizon and go very far but must dive back in, like a dolphin diving popping out of the sea? The event horizon is not an impenetrable wall after all. The light just cannot go infinitely far from the black hole but can reach a finite radius away. Would the event horizon appear to shrink away from you as you approached a black hole? Seeing as the light is strong enough to jump up to your eyes but no further, the black edge would represent the furthest point at which light can jump into your eyes. As your eyes got nearer a black hole, it is easier for light to reach them? Seeing as we have only observed black holes from "infinite" distance such that their gravity is negligible, the schwarzchild radius formula becomes 2GM/c^2
Hello Prof Matt Anderson.... I assume it's Possible to go faster than the calculated Escape velocity as you launch from earth... So, IF the velocity at launch is Greater than the Escape Velocity, then does that Rocket go BEYOND Infinity?... and if so where does that rocket end up?... :D ...
We have a convention of defining GPE to equal zero when the object is infinitely far away from the source of gravity. When we set final GPE equal to zero, and compare it to the GPE at the launch point, we determine what initial kinetic energy is needed to travel from launch point to infinitely far away, if our spacecraft coasts coast the whole way with no additional rocket thrust, and no other forces act on our spacecraft.
2019 JAY MODI, We are just considering one path here escaping from one planet, but since gravity is a conservative force, it is path independent. Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A
It is the inertial system we use. It is subjective, but makes sense once you define your system. In this case the kinetic energy's "direction" is in opposition to the gravitational force which points towards the earth, while the kinetic energy is moving/pointing away from the earth (well, the speed vector is pointing away actually). If we define the kinetic energy as the positive (which we do), then logically potential energy is negative. But in general we always treat the potential energy as negative. Without using an inertial system, positive and negative energy as terms are meaningless, unless you are into some quantum woo magic bs. But when 2 forces (opposing each other) affect an object on D distance, one can be treated as positive work (energy), while the other would be negative work (or energy). Hope this helps.
Hi, As a simple 'laymen' I have a question that has been bugging me for years regarding escape velocity. Am I correct in understanding that the 11.186 km/s figure is if a projectile/rocket is launched like a bullet or slingshot from the surface in one almighty 'shot' so to speak, with no additional prolonged input such as a rocket. Because hypothetically, i have always thought that surely if a rocket had an almost limitless, or super efficient fuel-source, then theoretically it could be gaining altitude at far less speed IF it could keep accelerating, even if at a much slower speed, until it eventually escaped the earth's gravity. Hopefully it's not too silly a question.
Hi Mark, You are indeed correct. Escape velocity refers to the single shot from sea level. That's why the velocity is so high. Per your example, if your rocket has enough fuel, then it could increase its altitude at an infinitesimal speed and still ultimately escape earth's gravity (of course it would take a long time). Thanks for the comment. Cheers, Dr. A
confusing ... surely a rocket starts at 0km/h and accelerates to to 40 000 km/h to break the pull of earth's gravity ... learning something new everyday sometimes confuses historical understandings ... also confusing me now, after watching your brilliant video , is the launch of Apollo from the moon which does not seem realistically possible to reach a speed of 8568 km/h to break the moon's gravitational pull ... hoping you can open my eyes via your simple but superb delivery of such a tricky topic
Ravin, Thanks for the comment. I'm working on a video for proper rocket motion. Stay tuned. Also, remember that the moon has a very different mass and radius than the earth, and therefore has a much lower escape velocity. You should calculate it and see what you get. Cheers, Dr. A
earth = 11.2 km/sec = 672 km/min = 40320 km/hour moon = 2.4 km/sec = 144 km/min = 8640 km/hour we can assume a moon escape velocity of over 8000 km/hour many other factors will play a role but it seems highly unlikely to reach 8000 km/hour considering the effort that went into the earth launch ... limited fuel and a few thrusters ? ... it seems that you do not mince words when it comes to math ... the real world appreciates such !
Are you not using the equation wrong, and introducing an infinity? Also this is the Newtonian model, that is wrong. That equation uses m1 and m2, the mass of the object, if the object has mass, and the earth was the only object in the universe, there is no velocity that is an escape velocity, at some point it will fall back to earth, even if it is 100 billion light years away, if it has mass, m1 and m2 will come back together. So this model fails in a system where there is only an earth, but in our universe you just need to get far enough away to be captured by some other matters gravity. This is still being taught as fact! The universe is relative, not Newtonian..
He explains with authority but also very academic, theoretical, in the sense of impractical. Ok, so you want to escape earth and travel to infinity, and get launched with the calculated 'escape' velocity, and ignore the (enormous) gravity from sun and galaxy. When you pass the moon, at 300,000 km, 50 times the radius of earth, you already lost 6/7 of the launch speed. Square root of 50 is about 7. When you get past mars, after just 80 million km, 10 thousand earth radii, you slowed down to 1 percent of launch speed, 0.1 km/s. It will take you some 20 years. Now you will further slow down, and it is still a very long trip to the next planet, let alone the last planet, or the next star. Yes, you will eventually reach infinity but it will also take literally infinite amount of time.
What happens if you fly a rocket in the opposite direction That the earth revolves around the sun at 66,000 MPH?...Are you effectively free of earths momentum around the sun?
How can the value of escape velocity of Earth be 11.2 km/s as this relation is obtain this value by substituting radius of Earth i.e. 6400 km. But 'r' is nothing but the distance between the centre of Earth and the radius of orbit. So how can we substitute the value of radius of earth?
a tab off for me. escape velocity for powered flight vs an object launched at surface but no additional force like a artillery shell moving at 25 k mph.
Jr nu mex, You are correct, we are assuming no additional propulsion after initial launch. Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A
i dont get it...........seems like the (magic) could be distilled near the beginning. Clearly the rocket is 'escaping' gravity w thrust.......... why not just keep continuing?
-G*M*m/r is what you get when you integrate G*M*m/r^2 with respect to r. This is the equation of gravitational potential energy. The negative gradient of GPE is the force of gravity. The gravitational potential field is the negative potential function of the vector field of the gravitational force.
I'm used to using gravity of earth (local) as lower case 'g' and the gravitational constant as upper case 'G'. I would assume his 'G' is equal to local gravity of whatever planet you launch from. Can anyone confirm?
Swart Michael Law yes, it is in fact uppercase G referring to the universal gravitational constant (6.67x10^-11 N/m^2/kg^2)... The mass and radius (or rather dist. from center of mass to center of mass) of the objects are used to find little g, or local g, and he included Fgrav=GMm/r^2 in his derivation. I missed that at first!
It states Ganymede is the largest of Jupiters moons. Consider a rocket on the surface of Ganymede, at the point farthest from the planet. Model the rocket as a particle. (a). Does the presence of Ganymede make Jupiter exert a larger, smaller, or same size force on the rocket compared with the force it would exert if Ganymede were not interposed? (b). Determine the escape speed for the rocket from the planet-satellite system. The radius of Ganymede is 2.64*10^6m, and its mass is 1.495*10^23kg. The distance between Jupiter and Ganymede is 1.071*10^9m, and the mass of Jupiter is 1.90*10^7kg. Ignore the motion of Jupiter and Ganymede as they revolve about their center of mass.
Swart Michael Law Swart Michael Law (a) is a bit confusing, I’m not sure I follow. The gravity of Jupiter and the gravity of Ganymede both act upon the rocket separately. Just cause the rocket is on a moon does not mean Jupiters gravity has less/more pull. However, this distance is greater from Jupiter so in that sense, yes, it would have less of an effect. (Note: Fgrav=GMm/distance^2).. for (b) my guess is solve using exactly this formula but with masses added together in the M variable and the distance between moon&jupiter plus radius of moon? The small relative mass and far distance from Jupiter should have negligible effect on a launch from that moon in my opinion. I am no physicist, just a student currently. Might be wrong myself.
@@carultch Fair Point Carl. Also some Friction from the Atmosphere. Also No Launch Assist, hence This Method assumes Launch from Zero Speed on Earth. However it would be Good if Someone got Serious with Space Planes.
Dhan Prasad Pradhan, Thanks for the comment, and keep up with the physics! You might also like my new website: www.universityphysics.education Cheers, Dr. A
At 4:12 your reference to rf =0 isn’t right. At infinity, rf = infinity resulting in Ugf =0. Great videos though. Appreciate your simple and clear explanation of concepts.
Nice, but somewhat inaccurate, escape velocity means you never fall back to Earth, but it does not mean you will escape the galaxy. Escape velocity from Earth is around 11 km/s but to get from Sun orbit out of the galaxy you would need a staggering 300-500 km/s, and to get to infinity who knows..?
One of the best explanations I have ever heard in my life.
Thanks for the kinds words. But keep looking. There must be better explanations out there!
Cheers,
Dr. A
Question, a wheel, big, orbiting earth, also spinning at orbit velocity?
@@yoprofmatt The maria occupy one third of the visible NEAR side of WHAT IS THE MOON. This IS fundamentally related to the surface gravity of WHAT IS THE MOON. Attention: The rotation of WHAT IS THE MOON matches the revolution. Accordingly, ON BALANCE, the gravity of the Sun upon the Moon IS about TWICE that of the Earth. Therefore, on balance, the Moon's crust is about TWICE as thick on it's far side. SO, we multiply one half times one third in order to obtain the surface gravity on the Moon in comparison with that of what is THE EARTH/ground. We then ALSO determine (ON BALANCE) that the NEAR side of what is the Moon (basically) DOES contain the maria (AND at what is then 33 percent) !! Great.
By Frank Martin DiMeglio
Still , Ve is valid for projectiles and not rockets.
Salute the professor. He explained it so easily unlike other people who make things unnecessarily complicated..
I was scared to open the video since the thumbnail consists of different formulas and what not. Well can't believe I finished the video and actually understood what he was explaining.. Nice.
This was also me!😂
He's definitely the Professor he said he was.👌🏼👏🏼
This guy is writing backwards. Respect
I think the video is mirrored
ruclips.net/video/GJKNR_zb0BY/видео.html
@@davidfrancesconi6557 that's so cool
No there either using a software or a mirror.
Pingum0n I didn’t realize that till I saw this comment
I do love the way he's always being honest of the fact that we might not always immediately know the answers to certain problems.
He uses the words: 'might', 'we don't know '...
🙏🏼👏🏼
I love the way this guy teaches, I never finished college, but learn so much from these lectures, thanks for posting them. There was such a great but missed opportunity at 5:50. Could have said “to infinity, and beyond!!!”
Doh! Totally missed that one. Thanks.
Cheers,
Dr. A
Excellent good vibes physics lectures!! Making it easy to comprehend.
Thank you for the content. You are an excellent teacher.
A Quick and (Somewhat) Dirty Way to Calculate Escape Velocity - I was doing some calculations using the escape velocities from Earth, Moon and Mars. Then by chance I calculated the velocities attained when an object was "dropped" from a height of the radius of each of these bodies, ASSUMING THE ACCELERATION DUE TO GRAVITY REMAINED CONSTANT DURING THE FALL, AND WAS EQUAL TO G AT EARTH'S SURFACE. Escape velocity is the minimum velocity needed to escape a gravitational field.
For example, escape velocity on earth, Vesc = 40,270 km/h (given).
Using the simple formula V2 = (2ar)^0.5, where
V2 = velocity after falling distance r; a = acceleration due to gravity; r = radius of earth (initial velocity = 0), and
a = 9.80665 m/s^2
r = 6,378,000 m
V2 = (2*9.80665*6378000)^0.5 = 11,184.53 m/s = 40,264.29 km/h
This value of 40,264 km/h compares very closely to the escape velocity of 40,270 km/h
I thought perhaps this was a coincidence, so I calculated V2 for the Moon and Mars and compared them to the known escape velocities for each body.
Moon:
Escape velocity = Vesc = 8,533.6 km/h
a = 1.62 m/s^2
r = 1,737,150 m
V2 = (2*1.62*1737150)^0.5 = 2,372.418 m/s = 8,540.7 km/h
Again, a very close fit.
Mars:
Escape velocity = Vesc = 18,108 km/h
a = 3.72761 m/s^2
r = 3,389,500 m
V2 = (2*3.72761*3389500)^0.5 = 5026.875 m/s = 18,097 km/h
Another very close approximation.
I haven't done the calculations for the other planets, moons, or the sun, but I expect I would get similar results. From these results I speculate that Vesc = V2. This seems to be a quick and dirty formula for calculating escape velocity, and might be handy for a stranded astronaut. But am I really just reiterating Vesc = (2Gm/r)^0.5? (G = gravitational constant; m = mass, e.g., of earth; r = radius, e.g., of earth).
I any case, using this method avoids a lot of the very intricate calculations necessary for determining Vesc the standard way, as long as the acceleration due to gravity and radius are known.
Sources:
www.livescience.com/50312-how-long-to-fall-through-earth.html
keisan.casio.com/exec/system/1360310353
www.wolframalpha.com/input/?i=escape+velocity+Mars
@Ari Parker Basically Gravitational force 'F' is directly proportional to the mass of the planet (the larger mass) 'M', the smaller mass, 'm', and inversely proportional to r,^2 the distance between the centres of the two masses. Meaning F is proportional to Mm/r^2. G, Newton's universal gravitational constant is the constant of proportionality between them. Meaning that F=GMm/r^2
@Ari Parker G is the constant of proportionality in the universal law of gravitation. It was first measured by Henry Cavendish, who sought to measure the mass of Earth in the interest of geology. He measured it by "weighing" lead balls in each others' gravitational fields, on a torsion balance to increase its sensitivity. By measuring the period without the stationary balls, and with the stationary balls, he could determine the impact of the gravity between objects whose mass he could know in advance, and back-out the universal constant of gravitation.
To the OP, it turns out that you get the correct result, but your reasoning is inconsistent with what physical principles are actually at play. The reason is that gravitational fields are not uniform from the radius of a planet out to twice the radius of the planet, and a body isn't in deep space when it is twice the radius of the planet. It is a mathematical coincidence that this method works.
The reason it works:
v = sqrt(2*g*h)
vesc = sqrt(2*G*M/r)
Gravitational field:
g = G*M/r^2
Assume h = r, and v = vesc, and combine:
vesc = sqrt(2*(G*M/r^2) * r) = sqrt(2*G*M/r)
Escape velocity is obtained considering the equality:
Kinetic energy of the object with this escape velocity=work needed to bring the object to infinity (definite integral from R to infinity of F ) which is GMm/R.
Your lectures are great, Professor!!
What a detailed explanation❤
Thank you so much sir for ur explanation it's useful Even after 8yrs♥️
Sir I have seen your many videos since the time I have entered to class 11 and from the time I found your RUclips channel . I am soo greatful and blessed to find your RUclips channel. Sir I know it may sound like something impossible but I really have a wish to have a live class with you when I would be one of your students whome you would be teaching offline and to listen you. Sir thank you for all your such great explaintion.
Now that warms my heart. I would love to have you as one of my students.
Cheers,
Dr. A
you was talking about a rocket in the beginning.But for a rocket which has a fuel,there is no meaning in escape velocity.it can escape the earth with any velocity it wants.
escape velocity makes sense only for projectiles with no fuel
bingo
unfortunately, this gets way more complicated to decipher and grasp as you start think about multiple gravitational fields (e.g. first you escape the earth but then you're stuck in the solar system). Note also that this all depends on where you first start. For example, it's easier to escape the solar system from the earth than it is to do so from the surface of the sun
Gustavo,
Great comment, thanks. You're making me think!
You might also like my new website: www.universityphysics.education
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Dr. A
Hats off...💜
That's the best lecture I've ever had....
Thanks! But keep looking, there must be better ones out there.
Cheers,
Dr. A
BUT...rockets are not "projectiles" like cannon balls or bullets. They carry a fuel source which can continue to 'burn' and provide acceleration. An object could 'escape' Earth at a sustained velocity as slow as only 1 mile per hour, correct? The term "escape velocity" should really be something like "muzzle velocity to escape" or "initial blast velocity to escape"....right???
True , Ve is indeed considered for the object "shot "at that speed at the radius R.
It is so clear and simple... thank u sir
ABSOLUTELY AMAZING EXPLANATION
6:39 yes we launched voyager and many other satellites into orbit or into space.
But every launch (I think) first went into orbit, going further and further into high-orbit to eventually swing away with very little energy needed.
This is correct right?
What? A nice physics teacher looking like Dr. Strange.
Arman,
Heard that before. My kids agree.
Thanks for the comment, and keep up with the physics!
You might also like my new website: www.universityphysics.education
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Dr. A
(Just saw previous post... had same basic idea) --- escaping earth gravity by going at any speed (even very slow speed) , just have steady thrust until you arrive at a Lagrangian point (or a point where gravity attraction from another object such as the Moon or another planet takes over, thus canceling out Earth's gravitational pull).
If your spacecraft was very light, you may even be able to hitch a ride on a nearby asteroid or satellite of some kind. The ability of light spacecraft to have efficient steady thrust, would allow for low cost space visits, especially when the return to earth can be a glide and you are able to re-use the craft.
I suppose the main problem here is getting the small spacecraft (or just a spacesuit) light enough and still have ability to carry some kind of reliable steady thrust / power system).
I have exactly same question.
Escape velocity talks about projectiles but what about an object that's rocket powered that keeps it powered until it crosses the limit of Earth's gravitational pull. Does it need to have velocity equal to Escape velocity or not? I think it does not need to be going at ecape velocity if it has additional power source.
Thank you Sir. You explained really well, things got simpler for me and it did help a lot.
Great explanation
Really understood everything
Thank you very much professor matt
The rocket example everyone gives when explaining escape velocity is very misleading, since rockets are usually percieved as having their own means of acceleration, so they can escape the Earth at basically any speed as long as said acceleration is greater than or equal to 1g (assuming its moving directly away from the planet) . Escape velocity, however, is about initial speed of a body without any forces applied to it other than gravity, just like you said in the video
Without any excess final velocity, how can we consider rfinal as infinity because since there is no excess final velocity the object will not keep moving after it escaping the earth's gravitation but should exactly escape and stop in that space while not being under earth's influence anymore..
Thanks! Helped me a lot with my SAT Physics subject test.
My saviour in college,
I just found my self anew physics teacher.thanks prof
Small mistake: at 4:06 you said “this term will drop out because we put rf =0” although you meant infinity.
Or he meant to say "this over rf = 0"
Marc Sleiman he is technically right since as rf gets bigger and bigger (to infinity) that whole term will get smaller and smaller to essentially zero, it's just that he decided to call it "zero". I know it's a late but just in case someone else looks at this comment
Well, yeah... but 1/infinity is 0... its the reciprocal of rf there...
@@music_heals_souls rf is taken as zero bruh
i think he meant rf=♾️, so the value of -Gmm/rf will equal to 0. He misspoke.
Brilliant explanation
Cheers,
Dr. A
We could try it "differently", but the result is obviously the same. Using the Work energy theorem aka that Summa F*s = change in kinetic energy = 0.5 mv(final)^2 - 0.5mv(initial)^2, we could make the problem like this: Let us assume that we launch a projectile with the speed necessery to reach an R distance where gravitational force converges to 0 (basically R= infinity). Let us also assume that only G is the (only) working force here. Now by using that Work (of G) = change of kinetic energy and that work of G is calculated as integrandus r to infinity of the function G(r)=mGM/r^2, we basically get the same result. Of course in my example we did not assume v(final) being 0, which is technically saying "we started with the minimum required amount of kinetic energy". Ofc it all depends on how well versed one is in calculus. If someone' just learned about potential energy with its formula definition, but no calculaus/proof of it, then your example is easier to understand, but people can get confused by you saying "we dont wanna have any excess speed in the end at 3:48". The thing with Newtonian physics is that the math part is actually pretty easy, the tougher part is getting the ideas and implications right.
Sir your explanations are so easy to grasp. Thank you for this. And could you please prepare a video on escape energy. Pleaee professor.
Thanks. Escape energy shouldn't be hard once you know the speed. Just calculate 1/2 mv^2.
Cheers,
Dr. A
I was about to sleep, but that was so cooooool
Awesome. Glad to not wake you up with a major annoyance.
Cheers,
Dr. A
The correct answer to the question of when it will stop is, it will never stop. It will approach zero velocity with reference to the earth asymptotically (never reaching zero). The assumption that Vf reaches zero in the solution to the problem is contingent on the assumption that Rf is infinity.
Correct interpretation. If instead you launch the object faster than escape velocity, it will still have velocity out at rf = infinity.
Cheers,
Dr. A
Very nice video sir.. Hats of
Thank you sir you really helped a lot but can you please tell how the potential energy at 2:57 is -GMm/r. In particular I didn't understand why the potential energy is negative. Thanks
Lime Pie,
See this one: ruclips.net/video/19wjt_zM1xU/видео.html
Thanks for the comment, and keep up with the physics!
You might also like my new website: www.universityphysics.education
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Dr. A
But would the satellie keep on going if its final velocity is 0?
which technology this is writing on glass shield
www.learning.glass
Cheers,
Dr. A
Very cool!
Sir I am a 9th grader from India and your channel is beautiful it helped me learn class 11th concept please reply if you can
Sanidhya Upadhyay,
Thanks to all my Fandersons in India! You're awesome.
Check out my new website:
www.universityphysics.education
Coupon for 50% off: RUclipsFANDERSON
Let's keep learning physics together!
Cheers,
Dr. A
The derived equation describes an escape velocity from Earth's gravity, but how would you describe the escape velocity of our solar system? Would it be the escape velocity from earth plus an additional term for the sun's mass and earth's distance from the sun etc?
Not 100% sure, but I think the reason we call it solar system, is because the Sun is the central, "fattest" object. I would say that we only need to calculate this very same equation, but with the Sun's mass and radius probably. Try that and compare it to a wiki/googled solar EV estimate. :D
The sun contains 99% of the mass of the solar system, and the gravity of the other planets is close to insignificant, unless we deliberately take credit for them in a gravitational slingshot. Since the spacecraft doesn't go deep in the gravitational wells of other planets, it is reasonable to neglect their contributions.
The way you'd calculate escape velocity for an object starting on Earth, and leaving the solar system, is to first calculate the change in GPE necessary to permanently leave the planet Earth from the Earth's surface. Then calculate the change in GPE necessary to permanently leave the sun's gravitational field, from the position of Earth's obit. Add up these two changes in GPE, and equate to the KE of the object leaving Earth's surface. Then solve for velocity.
The equation:
1/2*m*v^2 = -(-G*Me/Re - G*Ms/re)
Solved for velocity:
v = sqrt(2*G*(Me/Re + Ms/re))
where:
Me and Ms are masses of the Earth and sun.
Re is the radius of Earth
re is the orbital radius of Earth at the instant of launch
Plugging in data, assuming re = 1 AU, Re is the volumetric average radius of Earth we get the following, for the escape velocity of the solar system at the position of Earth's surface.
v = 43600 m/s
Great explanation sir
Good explanation. But, doesn't V have to be relative to the center of the planet? The reason I ask is because most rockets accelerate in a near circular trajectory around the earth to get into orbit. (Orbit velocity has similar mechanics). But, in orbit, the velocity creates a centrifugal force to counteract gravity. I guess what I'm asking is - is the math the same for a rocket in an angular direction versus one in a radial direction?
You are correct. To ultimately escape, you have to hit this speed (which is relative to the planet's center). If you're at the equator and already moving at 1000 mph, you've got a head start.
Cheers,
Dr. A
so i have an issue, this all relies on the term Rf being infinity but surely an infinite distance is somewhat theoretical. so assuming there were no objects for it to hit or be caught in the gravitational field of would an object not always after some time be attracted back to earth? i have a feeling there is an answer lying in the fact v is 0 at the point R is infinity but an explanation would help
Dan Evetts,
If objects are going at escape velocity, they have zero velocity when they get very far away (r -> infinity). If they are shot at a faster velocity than escape velocity, they still have some velocity even as r -> infinity. Thanks for the comment, and keep up with the physics!
You might also like my new website: www.universityphysics.education
Cheers,
Dr. A
When you throw a ball straight up in the air, that ball immediately starts slowing down. Eventually, it reaches a point where it's velocity is zero, and then starts falling back down. If a projectile is launched at escape velocity, it will simply never slow down to zero - unless it travels an infinite distance - which would taken an infinite amount of time. Any arbitrarily small amount slower, and you could theoretically calculate when the projectile would reach zero and begin falling back down. Anything AT escape velocity or higher simply would never do it - because it will never reach an infinite distance to reach zero.
@@willoughbykrenzteinburg this was very clear, thanks
Okay but say i had a rocket that did just a constant 10mph heading towards space would it never escape because it would need infinite fuel? thats what i never understood
Hey Prof, I got a question.
Escape velocity talks about projectiles but what about an object that's rocket powered that keeps it powered until it crosses the limit of Earth's gravitational pull. Does it need to have velocity equal to Escape velocity or not?
ok so it just how much kinteic energy required to overcome the inital poteial energy when on surface
I don't understand escape velocity. The further from the centre of the Earth, the less the Earth's gravitational force therefore the escape velocity should be constantly changing. As your distance gets further and further from Earth, you're required less and less velocity to escape because Earth's gravitational force keeps getting weaker and weaker as you get further and further therefore how can we define 1 specific excape velocity? I've watched 6 escape velocity videos now and this seemingly simple question/confusion/misunderstanding of mine is yet to be answered...
Spinelli,
Great question. At 4:42 we derive the equation for escape velocity, and it has ri in the denominator. This ri is the distance from the center of the planet. (Later we set this to R, the radius of the planet.)
If you use ri as your starting point, then this is your escape velocity from that point. For example, pretend you start at a distance of 2R from the planet. Set ri = 2R to solve. Since the denominator just got bigger, the escape velocity gets smaller. This is in line with your intuition, so nice job.
Thanks for the comment, and keep up with the physics!
You might also like my new website: www.universityphysics.education
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Dr. A
Please, somebody help me, why is it -G.M.m/r and not G.M.m/r^2 ? at 3:00
I just found out you have a other video explaining that, but, sorry I don't understand that either, escpecially the integrals thingy.
(Sorry for any possible spelling mistakes.)
It s easy really this simbolises the potential energy and the other simbolises the gravitational force exerted on an object on a specific point in space
When you integrate G*M*m/r^2 relative to r, you get -G*M*m/r. That's what this formula calculates, is the integration from a point in space to a point within the gravitational well of an astronomical body. Integrate relative to r, and you get the quantity we call gravitational potential energy. It is used as a shortcut for evaluating work done by a gravitational field, when moving an object from point A to point B, which is path-independent, because gravity is a conservative force.
How strong could gravity get before a chemical rocket couldn't reach escape velocity? There has to be a limit, right?
John Terpack,
There must be, since obviously you can't escape a black hole. But for rockets with continuous burn rate, you don't talk about escape velocity in quite the same way. The limit is when the gravity pulling down equals the maximum upward thrust that the rocket can generate. Forces are equal and opposite thus the acceleration is zero.
Thanks for the comment, and keep up with the physics!
You might also like my new website: www.universityphysics.education
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Dr. A
@@yoprofmatt I'll see if I can find out the max thrust of a chemical rocket.
My question was inspired by the frequent appearance in sci-fi of civilizations on high-gravity planets. At some point, gravity would be too strong for artificial structures entirely... maybe even too strong for terrestrial life. But I always thought it would be interesting if there was an advanced race that was limited by the fact they just couldn't launch rockets. And I wondered how much stronger gravity would need to be, compared to earth, to achieve that.
So at the black hole, event horizon, escape velocity is the speed of light...
That means, if you shine a light, it goes out a bit, then returns back?
But isn't the speed of light constant through space, so deceleration is not allowed?
When the light starts to return, it will reach the speed of 0 m/s briefly?
Is it allowed in physics?
How about we have a very fast spaceship?
Even if the escape velocity is the speed of light, if the spaceship keeps accelerating at the rate slightly higher than the surface gravity at the event horizon....
Then the spaceship can escape the black hole from inside the event horizon as well?
I am certainly not a black hole expert, but I would say this:
Inside the event horizon, light (or anything else) cannot escape.
Outside the event horizon, light (and possibly other stuff) can escape.
There's LOTS of questions about what happens right at the event horizon, including stuff like Hawking radiation. And I would say many of the questions do not have definitive answers yet because black holes are still a relatively new study.
Cheers,
Dr. A
@@yoprofmatt I found out the answer... from the internet...:)
There are two kinds of black holes, Newtonian and General Relativity (GR).
Newtonian black holes, light cannot escape, but anything can, as long as they have a powerful enough jetpack.
GR black holes, light or anything else cannot escape for some reason related to spacetime thing...
But the strange thing is, if the black hole is 1 trillion solar mass (and such black hole is allegedly found)
The surface gravity is only 1.5 g, and the density is thinner than Martian air.
In such situation, how come we cannot escape from the event horizon, even with GR black hole?
For that, I have not found an answer.
Thanks!!!
@@yoprofmatt The lecture was great BTW, the best regarding escape velocity I've ever seen. Thanks!
Hey do you vedios on only physics ?? 🙄🤔
YR NAGARAJU,
Currently yes.
Thanks for the comment, and keep up with the physics!
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Dr. A
realllly helpful sir
You r d saviour sir..........
I'm still missing something in the equation. What about the mass of the object and the density of both? If there is a neutron star sitting on the surface, does it need the same speed if it's innertia is so large? Yea, and what about applying the velocity horizontally? I need a better vid obviously
As long as the escaping object is insignificant compared to the mass of the planet, and as long as the planet's mass is moderate enough that relativistic effects don't come in to play, this formula is valid. The mass of the escaping object "cancels out" of the equation of energy balance, and it turns out that the escape velocity is independent of the escaping object's mass.
It still works if you escape horizontally, instead of vertically, because gravity is a conservative vector field. No matter which direction you launch, you escape the planet's gravitational field, as long as you aren't on a crash trajectory.
Does it technically mean that the object will have zero velocity when it reaches infinity?
Yes.
Do you really can write backwards??... or is this some new technology?
BTW I loved the explanation... I always thought spacecraft attained this speed to reach outer space.. my whole childhood was a lie.
санкет мхаске,
Not writing backwards (I'm not that talented). The board is called Learning Glass. You can check it out at www.learning.glass
Cheers,
Dr. A
in my book, the approach used is the gravitational force acts as a centripetal force (mv^2)/R=GmM/R^2. So, we found v^2=GM/R. Whats wrong with my book?
indra,
Nothing wrong with your book, but that is the speed of orbit, not escape speed.
Thanks for the comment, and keep up with the physics!
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Dr. A
That's the escape velocity derivation that you're looking at which is the minimum velocity an object/satellite should have to orbit the planet given an orbital radius. Interestingly enough it's also related to the escape velocity as the escape velocity is the square root of 2 times the orbital velocity. That means the escape velocity for a given object would be about 41.4% more than it's orbital velocity.
good explanation sir. but i can't understand your writing.. but everything is cool. take respect from Bangladesh..
His writing is perfectly legible . Maybe, you should try attending some English classes first before you take on Physics.
How does the radius and mass of body i.e black hole responsible for its escape velocity???
Radius and mass determine the gravitational field at every point surrounding a massive body. Integrating this gravitational field relative to radial position, and we get the GPE at every distance from the body, relative to it being defined as zero, infinitely far away. If we add enough kinetic energy for it to coast the whole way out, then the initial KE equalling the total necessary change in KE, is what we need in order to escape.
A black hole is an extreme case, where there exists a special radius at which not even light can escape the gravitational field, because the escape velocity exceeds the speed of light. This is called the Schwarzschild radius, named after a physicist whose name fortuitously means "black shield". This sets the boundary for the point-of-no-return, for an object falling in to the black hole, that we call the event horizon.
How?vf =0 when vf is the velocity make the object still moving in the space?
But eventually the velocity f will decrease as gravity is acting and we want the minimum possible velocity that will take projectile put of the gravitational influence. So, as we require minimum velocity it will slow down and come at rest till the boundary of gravitational influence
thanks sir for good explanation
Why is escape velocity important for a powered craft. Can't a hypothetical rocket thrust away from the earth at 30 miles per hour if it has enough fuel. Surely escape velocity is only necessary if I'm fired out of a cannon on Earth's surface and am unpowered from then on.
Also with black holes and the event horizon, can light cross out of the event horizon and go very far but must dive back in, like a dolphin diving popping out of the sea? The event horizon is not an impenetrable wall after all. The light just cannot go infinitely far from the black hole but can reach a finite radius away. Would the event horizon appear to shrink away from you as you approached a black hole? Seeing as the light is strong enough to jump up to your eyes but no further, the black edge would represent the furthest point at which light can jump into your eyes. As your eyes got nearer a black hole, it is easier for light to reach them? Seeing as we have only observed black holes from "infinite" distance such that their gravity is negligible, the schwarzchild radius formula becomes 2GM/c^2
Sir please make a video on derivation of acceleration due to gravity.
Sky King,
Working on new videos. Stay tuned.
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Cheers,
Dr. A
Well done You help me alottttt
Great to hear. Keep up with the physics.
Cheers,
Dr. A
Hello Prof Matt Anderson.... I assume it's Possible to go faster than the calculated Escape velocity as you launch from earth... So, IF the velocity at launch is Greater than the Escape Velocity, then does that Rocket go BEYOND Infinity?... and if so where does that rocket end up?... :D ...
very good explanation
Cheers
Student N.
Back at ya!
Cheers,
Student A
What is the reason that the velocity of any object is heard by going to infinity?
any point far away enough for the gravitational force to stop effecting is referred to as infinity
How can we consider
Consider .... what? Don't leave me hanging.
Cheers,
Dr. A
anyway,thanks for the video
Thanks Sir
It really help me and I love the way you write on
Hey, write on!
Cheers,
Dr. A
Why we have putten rf=O in final potential energy?? Couldn't understand please help!!!
We have a convention of defining GPE to equal zero when the object is infinitely far away from the source of gravity. When we set final GPE equal to zero, and compare it to the GPE at the launch point, we determine what initial kinetic energy is needed to travel from launch point to infinitely far away, if our spacecraft coasts coast the whole way with no additional rocket thrust, and no other forces act on our spacecraft.
how many paths of an escape velocity?
2019 JAY MODI,
We are just considering one path here escaping from one planet, but since gravity is a conservative force, it is path independent.
Thanks for the comment, and keep up with the physics!
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Dr. A
can someone explain why is the potentiol energy negative?
It is the inertial system we use. It is subjective, but makes sense once you define your system. In this case the kinetic energy's "direction" is in opposition to the gravitational force which points towards the earth, while the kinetic energy is moving/pointing away from the earth (well, the speed vector is pointing away actually). If we define the kinetic energy as the positive (which we do), then logically potential energy is negative. But in general we always treat the potential energy as negative. Without using an inertial system, positive and negative energy as terms are meaningless, unless you are into some quantum woo magic bs. But when 2 forces (opposing each other) affect an object on D distance, one can be treated as positive work (energy), while the other would be negative work (or energy). Hope this helps.
@@ktaikamatika1397 omg i finally understand thanks for explaining
@@ktaikamatika1397 I think you mean reference point, not "inertial system". This has nothing to do with inertial reference frames.
Plz make a new vdo of velocity
Noob Gamer Boy,
About what aspect?
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Cheers,
Dr. A
Hi, As a simple 'laymen' I have a question that has been bugging me for years regarding escape velocity. Am I correct in understanding that the 11.186 km/s figure is if a projectile/rocket is launched like a bullet or slingshot from the surface in one almighty 'shot' so to speak, with no additional prolonged input such as a rocket. Because hypothetically, i have always thought that surely if a rocket had an almost limitless, or super efficient fuel-source, then theoretically it could be gaining altitude at far less speed IF it could keep accelerating, even if at a much slower speed, until it eventually escaped the earth's gravity. Hopefully it's not too silly a question.
Hi Mark,
You are indeed correct. Escape velocity refers to the single shot from sea level. That's why the velocity is so high.
Per your example, if your rocket has enough fuel, then it could increase its altitude at an infinitesimal speed and still ultimately escape earth's gravity (of course it would take a long time).
Thanks for the comment.
Cheers,
Dr. A
confusing ... surely a rocket starts at 0km/h and accelerates to to 40 000 km/h to break the pull of earth's gravity ... learning something new everyday sometimes confuses historical understandings ... also confusing me now, after watching your brilliant video , is the launch of Apollo from the moon which does not seem realistically possible to reach a speed of 8568 km/h to break the moon's gravitational pull ... hoping you can open my eyes via your simple but superb delivery of such a tricky topic
Ravin,
Thanks for the comment. I'm working on a video for proper rocket motion. Stay tuned.
Also, remember that the moon has a very different mass and radius than the earth, and therefore has a much lower escape velocity. You should calculate it and see what you get.
Cheers,
Dr. A
earth = 11.2 km/sec = 672 km/min = 40320 km/hour
moon = 2.4 km/sec = 144 km/min = 8640 km/hour
we can assume a moon escape velocity of over 8000 km/hour
many other factors will play a role but it seems highly unlikely to reach 8000 km/hour considering the effort that went into the earth launch ... limited fuel and a few thrusters ? ...
it seems that you do not mince words when it comes to math ... the real world appreciates such !
Awesome. Thanks for the update.
Cheers,
Dr. A
Are you not using the equation wrong, and introducing an infinity? Also this is the Newtonian model, that is wrong.
That equation uses m1 and m2, the mass of the object, if the object has mass, and the earth was the only object in the universe, there is no velocity that is an escape velocity, at some point it will fall back to earth, even if it is 100 billion light years away, if it has mass, m1 and m2 will come back together.
So this model fails in a system where there is only an earth, but in our universe you just need to get far enough away to be captured by some other matters gravity.
This is still being taught as fact!
The universe is relative, not Newtonian..
He explains with authority but also very academic, theoretical, in the sense of impractical. Ok, so you want to escape earth and travel to infinity, and get launched with the calculated 'escape' velocity, and ignore the (enormous) gravity from sun and galaxy. When you pass the moon, at 300,000 km, 50 times the radius of earth, you already lost 6/7 of the launch speed. Square root of 50 is about 7. When you get past mars, after just 80 million km, 10 thousand earth radii, you slowed down to 1 percent of launch speed, 0.1 km/s. It will take you some 20 years. Now you will further slow down, and it is still a very long trip to the next planet, let alone the last planet, or the next star. Yes, you will eventually reach infinity but it will also take literally infinite amount of time.
my man is actually writing backwards. im speechless
Your man, maybe. But this man is definitely not. See www.learning.glass
Cheers,
Dr. A
What happens if you fly a rocket in the opposite direction That the earth revolves around the sun at 66,000 MPH?...Are you effectively free of earths momentum around the sun?
What does the G stand for?
The gravitational constant. (6,67428 ± 0,00067) . 10^-11 Nm^2 kg^-2.
How can the value of escape velocity of Earth be 11.2 km/s as this relation is obtain this value by substituting radius of Earth i.e. 6400 km. But 'r' is nothing but the distance between the centre of Earth and the radius of orbit. So how can we substitute the value of radius of earth?
The radius of the earth is where we started. Infinity is where we ended. This is how to calculate escape velocity.
Cheers,
Dr. A
Very nice
Thank you.
Cheers,
Dr. A
a tab off for me. escape velocity for powered flight vs an object launched at surface but no additional force like a artillery shell moving at 25 k mph.
Jr nu mex,
You are correct, we are assuming no additional propulsion after initial launch.
Thanks for the comment, and keep up with the physics!
You might also like my new website: www.universityphysics.education
Cheers,
Dr. A
i dont get it...........seems like the (magic) could be distilled near the beginning. Clearly the rocket is 'escaping' gravity w thrust.......... why not just keep continuing?
4:05 correction: rf is equal to infinity not zero. the "term" equals zero as rf approaches infinity
Newton's law of gravitation is Gmm/r^2, not Gmm/r, so how come you used Gmm/r.
-G*M*m/r is what you get when you integrate G*M*m/r^2 with respect to r. This is the equation of gravitational potential energy. The negative gradient of GPE is the force of gravity. The gravitational potential field is the negative potential function of the vector field of the gravitational force.
I'm used to using gravity of earth (local) as lower case 'g' and the gravitational constant as upper case 'G'. I would assume his 'G' is equal to local gravity of whatever planet you launch from. Can anyone confirm?
Hi did you get an answer for this? its an interesting question
Swart Michael Law yes, it is in fact uppercase G referring to the universal gravitational constant (6.67x10^-11 N/m^2/kg^2)... The mass and radius (or rather dist. from center of mass to center of mass) of the objects are used to find little g, or local g, and he included Fgrav=GMm/r^2 in his derivation. I missed that at first!
Okay cool. i have this other problem here. Can you help me sort it out?
It states
Ganymede is the largest of Jupiters moons. Consider a rocket on the surface of Ganymede, at the point farthest from the planet. Model the rocket as a particle.
(a). Does the presence of Ganymede make Jupiter exert a larger, smaller, or same size force on the rocket compared with the force it would exert if Ganymede were not interposed?
(b). Determine the escape speed for the rocket from the planet-satellite system. The radius of Ganymede is 2.64*10^6m, and its mass is 1.495*10^23kg. The distance between Jupiter and Ganymede is 1.071*10^9m, and the mass of Jupiter is 1.90*10^7kg. Ignore the motion of Jupiter and Ganymede as they revolve about their center of mass.
Swart Michael Law Swart Michael Law (a) is a bit confusing, I’m not sure I follow. The gravity of Jupiter and the gravity of Ganymede both act upon the rocket separately. Just cause the rocket is on a moon does not mean Jupiters gravity has less/more pull. However, this distance is greater from Jupiter so in that sense, yes, it would have less of an effect. (Note: Fgrav=GMm/distance^2).. for (b) my guess is solve using exactly this formula but with masses added together in the M variable and the distance between moon&jupiter plus radius of moon? The small relative mass and far distance from Jupiter should have negligible effect on a launch from that moon in my opinion. I am no physicist, just a student currently. Might be wrong myself.
hello! sir can you reply us?
2019 JAY MODI,
Sure, can you repeat the question?
Cheers,
Dr. A
This Method assumes No Lift from Earth's Atmosphere such as Planes Achieve?.
This method assumes no forces other than gravity act on the spacecraft from the instant of launch to when it escapes.
@@carultch Fair Point Carl.
Also some Friction from the Atmosphere.
Also No Launch Assist, hence This Method assumes Launch from Zero Speed on Earth.
However it would be Good if Someone got Serious with Space Planes.
Rf=0? You can divide by zero? You mean Rf=infinity?
So u just consider the vf = 0 ?
What if i want it to launch a rocket to mars ? I can't consider it 0
thats when all the kinetic energy has turned to potential energy to offset the gravity
it already left sir...2020
Dhan Prasad Pradhan,
Thanks for the comment, and keep up with the physics!
You might also like my new website: www.universityphysics.education
Cheers,
Dr. A
Is he writing backwards?
Cure to insomnia.
At 4:12 your reference to rf =0 isn’t right. At infinity, rf = infinity resulting in Ugf =0. Great videos though. Appreciate your simple and clear explanation of concepts.
Nice, but somewhat inaccurate, escape velocity means you never fall back to Earth, but it does not mean you will escape the galaxy. Escape velocity from Earth is around 11 km/s but to get from Sun orbit out of the galaxy you would need a staggering 300-500 km/s, and to get to infinity who knows..?