The skater in your example accelerates opposite to you only as long as you're pushing the rock with your hand. How does the rocket push on the outward flowing gas so that the rocket accelerates in the opposite direction? There is no hand or piston to push the gas. If the rocket doesn't push on the gas the gas can't push back on the rocket. So, how does Newton's third law come into play?
I absolutely loved these examples. I am trying to teach myself physics and this is the most fun I have had learning it without being overwhelmed with symbols and concepts I haven't yet mastered. Thank you, as always.
But the velocity of fuel he mentioned is in the respect of rocket but the actual velocity is measured with the respect of earth?!! can we use the same for earth as well??
Tip, keep your suicide burn TWR as close to 1 as possible for the whole burn. Means a longer burn but it is easier to calculate and you won't have the problem of over thrusting and bobbing up and down.
How do i calculate exhaust velocity? If i know the : Throat Nozzle size at the end Propellant nature Flow rate Engine geometry pressure of the propellant
I'm a young man trying to be a engineer at space x and trying to understand how rockets work and this.. was way more easier to understand then the stuff you learn in middle or high school
@@wyattb3138 There are limited options on creating or designing a rocket engine as pretty much all of them has a thin throat and a big nozzle to convert Sub Sonic speeds into super sonic.
Grimm_m, start with subsonic then run CFD software to optimize the nozzle design and do supersonic later. You have to design your rocket engine off your chemical reactants.
Major flow: who says that the delta m is the mass of the fuel?! by definition it is the change in mass of the expelled gases; there is a correlation for sure but delta m in the equation is not the mass of the fuel that need to be burned, it is much more complex than that.
Why didn’t you use t in question 1 and from what I’ve heard is that if you have even 1 thing missing like t (time ) or a (acceleration) you can’t have the correct and proper answer
So the wind blowing out of the LEM's engine nozzle, as it landed on the moon, would need to be more than the mass of the vehicle to account for deceleration? Do I have that right?
@@PsychoMuffinSDM Thanks. So can you help me get an idea of the speed or velocity of the gas. I know that in throat it cannot exceed local mach and I have no idea what that would be except that it speeds up from there.
GoneBamboo - You can create the same "Force" by pushing away a very large mass SLOWLY or a very small mass VERY FAST. You need to generate MOMENTUM which is defined as MASS times VELOCITY. So you can create a large MOMENTUM with high velocity and low mass (like a rocket engine) or low velocity and high mass (like a catapult throwing rocks) I have a 44 magnum desert eagle that fires very large bullets and a relatively low muzzle velocity. Don't you agree that my smaller .223 rifle could create the SAME RECOIL as the desert eagle IF it can get the smaller bullet to much higher velocity than the 44 magnum?
Couldn't it be like, that we could differentiate momentum w.r.t time and get m.dv/dt & v.dm/dt and consdering velocity constant it would become F=v.dm/dt. Also what if the rocket is still leaving eath shouldn't mass.g be applied
can someone help me with this: NASA is launching a rocket into space from Earth. This particular rocket burns its fuel for 5 minutes and then turns off its engines. At this point, the rocket keeps floating along through space at a speed of 10 km/s. a.) What is the rocket’s average acceleration during these first 5 minutes? b.) How far did the rocket travel during the first 5 minutes? c.) The Moon is approximately 3.8 x 10^5km from Earth. At its final speed, how long will it take the rocket to reach the Moon? d.) The star nearest to us, called Alpha Centauri, is 4.1 x 10^13 km away. Could this rocket get a man to Alpha Centauri before he dies of old age? e.) The rocket is half way to the Moon when its pilot suddenly notices he forgot his camera! He radios a second rocket crew on Earth and tells them to bring his camera and meet him on the Moon when he lands. Unfortunately, the second rocket can only accelerate at 3 m/s^2. Can this second rocket make it to the Moon before the first one lands?
Hello, I need help for followinng question. Please! A small research rocket of mass 250kg is launched vertically as part of a weather study. It sends out 50 kg of burnt fuel and exhaust gases with a velocity of 180 ms-1 in a 2 s initial acceleration period. (a) What is the velocity of the rocket after this initial acceleration? (Answer v = 40 ms-1) (b) What upwards force does this apply to the rocket? (Answer F = 4500N) (c) What is the net upwards acceleration acting on the rocket? (Use g = 10 ms-2 if required) (Answer a = 10 ms-2)
Hi In the first part of the question I am obtaining 36 as answer. I guess the answer is wrong. Since in thus we just have to apply the formula MV=mv and mass needn't be included. In the second part you have to apply first equation of motion (v=u+at) but using the downward value and by third law of motion upward and downward values will be equal. I am not sure abt 3rd. I will confirm once and leave reply as soon as possible. BTW it had been 3 months since you posted this question. I think you have got the answer. And if yes please tell where I am making mistake. Thanks
@@arpanarora Lets' compare solution! Could you please tell me your solution for this problem. Here is my solution: (a) Given: m1 = m2 + m3 = 250 kg u1 = 0 ms^-1 m2 = m_rocket = 250 kg - 50 kg = 200 kg v2 = ? m3 = m_fuel gas = 50 kg v3 = 180 ms^-1 t = 2 s (initial acceleration) From m1.u1 = m2.v2 + m3.v3 (250kg x 0 ms^-1) = (200 kg x v2) + (50 kg x 180 ms^-1) ==> v2 = -45 ms^1 = 45 ms^-1 in the direction opposite to the exhausted gas (Hence, answer = 40 ms^-1 is WRONG) (b) m2 = m_rocket = 250 kg - 50 kg = 200 kg u2 = u_rocket = 0 ms^-1 v2 = 45 ms^-1 (as calculated in (a) above) t2 = 2 s a2 = a_apply = ? From a = (v-u)/t ==> a2 = a_apply = (45 ms^-1 - 0 ms^-1) / 2 s = 22.5 ms^-2 ==> F_apply = m.a_apply = 200 kg x 22.5 ms^-2 = 4500 N (c) m2 = m_rocket = 250 kg - 50 kg = 200 kg F_g = m2.g = 200 kg X 10 ms^-2 = 2000 N (only m_rocket as fuel and gas are burning off) F_net = F_appy - F_g and F_appy = 4500 N (found in (b)) ==> F_net = 4500 N - 2000 N = 2500 N From F_net = m.a ==> a = F_net / m = 2500 N / 250 kg = 10 ms^-2 Now your turn to show me your solution PLEASE
@@huyanhle it seems your solution is absolutely right. But this is how I did this question 1) Mass of rocket M= 250kg Velocity of the rocket V=? Mass of the burnt fuel m =50kg escape Velocity v=180 m/s Using conservation of momentum M*V=m*v 250*V= 50*180 V=9000/250 V=36m/s. 2) by applying F=ma F=(m*v)/ time F= 50*180/2 3) I was unable to solve 3rd part but I guess yours is right.
I didn't understand what you meant when you said we can write the change of momentum as the product of a mass and its change in velocity or the product of a changing mass with a constant velocity. How can the mass change? At least in Newtonian mechanics, isn't mass supposed to be constant?
The man feels the force to the opposite direction than the block. In a closed system the force should be taken as negative since all forces in a closed system always add up to 0. But the force that the man feels is not negative from his perspective.
No, but if you would like some numbers to "noodle with", a Space Shuttle main engine ingests 500 kilograms of liquid oxygen and liquid hydrogen PER SECOND, combusts it and expels PURE WATER at 2500 meters per second (at sea level)
And where do you get this Ball in space? A ball has inertia. Expanding rocket gas particles do not have inertia.in a vacuum. because the gas particles are in 'free expansion'.
You are not "shooting out matter" from a rocket in the vacuum of space. In space , the vacuum evacuates the rocket gases at the rate at which they expand due to there being no resistance. A rocket works on the pressure principle. You need to create pressure behind the rocket to create movement. It is impossible to create pressure in a vacuum.
- "It is impossible to create pressure in a vacuum." *This is FALSE* !! Rocket engines can EASILY create nozzle pressures above 3000 PSI *IN HARD VACUUM* . I have SHOWN you videos of rocket propulsion IN A VACUUM, but you just keep dismissing it with your expert opinion of "Nuh-UHHHHHH" Now tell us your level of physics education, or JUST THE FUCK UP ALREADY with your nonsense!! ACTUAL *Rocket Propulsion ENGINEERS* have told you that you are mistaken, but you dont care since you are "self-educated" by YT videos... It pathetic how far we have slipped into UN-educated masses having access to modern technology and still being able to vote.
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The skater in your example accelerates opposite to you only as long as you're pushing the rock with your hand.
How does the rocket push on the outward flowing gas so that the rocket accelerates in the opposite direction? There is no hand or piston to push the gas. If the rocket doesn't push on the gas the gas can't push back on the rocket. So, how does Newton's third law come into play?
I absolutely loved these examples. I am trying to teach myself physics and this is the most fun I have had learning it without being overwhelmed with symbols and concepts I haven't yet mastered. Thank you, as always.
@@dapperwolf465 same bro... same...
I bought a 10$ rocket on Amazon as a starting point and then learned the basics components of a rocket along with the science and physics behind it
@@loganpace1260 That's cool!
@@dapperwolf465 I’m working on engineering a real rocket that will go into orbit & space, not some toy one off Amazon
@@macysondheim bro what
Your channel has been a lifesaver, wish I found it sooner.
Thank you so much for your help! Your lessons are always clear and the complex examples are broken down in a way it is easy to understand.
Is it just me or do school teachers oversimplify and furnish the equations to mug up instead of explaining the origin of the concept?
I see people make this mistake often. It’s not VELOCITY times mass flow rate ( dm/dt ), it’s EXHAUST VELOCITY.
It's not rate of change of mass?
@@daffavirwandy7694 no.
That’s mass flow
you make every concept so much easier to understand. thank you so much
But the velocity of fuel he mentioned is in the respect of rocket but the actual velocity is measured with the respect of earth?!! can we use the same for earth as well??
this guy really has a solution for all my academic problems.
love u, my hero for clarifying every concept I don't understand in lecture
U kinda inspired me to read my textbook
I'm here from Kerbal Space Program and trying to perform an automated suicide burn SpaceX style
Nice.
Wrong channel. Check Scott Manley. You are welcome!
Tip, keep your suicide burn TWR as close to 1 as possible for the whole burn. Means a longer burn but it is easier to calculate and you won't have the problem of over thrusting and bobbing up and down.
nice and clear explanation.thank you
How do i calculate exhaust velocity? If i know the :
Throat
Nozzle size at the end
Propellant nature
Flow rate
Engine geometry
pressure of the propellant
Google it
At least it’s not rocket science
Can't believe. Man you are the life saver. Thanks for help 🙏
I'm a young man trying to be a engineer at space x and trying to understand how rockets work and this.. was way more easier to understand then the stuff you learn in middle or high school
Good. You should try to design your own rocket engine using math you already know.
@@wyattb3138 There are limited options on creating or designing a rocket engine as pretty much all of them has a thin throat and a big nozzle to convert Sub Sonic speeds into super sonic.
Grimm_m, start with subsonic then run CFD software to optimize the nozzle design and do supersonic later. You have to design your rocket engine off your chemical reactants.
Wow I skipped all my lectures and this just made it super clear
Mashallah tbark allah alhamudillah inshallah better AStgfrallah
@@allahm-ast3mnlywlatstbdlny164 have u just converted to Islam mate
But the mass of the whole rocket decreases every second, so the thrust force will be greater due to mass reducing. Can you do a video on that?
I love you videos! Very understandable!
Thank you Professor From Pakistan 😍
Thank you so much for sharing your knowledge with us for free
Great video, thanks
What about the acceleration???
Thanks sir, it helped me a lot.😇
Thank you so much it’s the only video I’ve found useful and you gained a new sub
Major flow: who says that the delta m is the mass of the fuel?! by definition it is the change in mass of the expelled gases; there is a correlation for sure but delta m in the equation is not the mass of the fuel that need to be burned, it is much more complex than that.
You think that the same video that talks about newtons 3rd law is gonna talk about actual rocket science?
Frogadron hahh seems i had over expectations 😂
@@alekos5916 he doesn't even get N3 right.
Mistake:
4kg*25m/s/5=20 newton. not 100 newton
isn't the division by 5 the mistake ?
Oh my god. Its Spring Break and i'm watching physics videos. Just give me the "A" rn
Why didn’t you use t in question 1 and from what I’ve heard is that if you have even 1 thing missing like t (time ) or a (acceleration) you can’t have the correct and proper answer
You make it really easy... Thanks
So the wind blowing out of the LEM's engine nozzle, as it landed on the moon, would need to be more than the mass of the vehicle to account for deceleration?
Do I have that right?
No, because you are forgetting about speed. The faster the fuel is propelled, the more force, thus, less is needed. Think of it in terms of p=mv.
@@PsychoMuffinSDM
Thanks. So can you help me get an idea of the speed or velocity of the gas. I know that in throat it cannot exceed local mach and I have no idea what that would be except that it speeds up from there.
@@gonebamboo4116 It's probably in here somewhere: www.hq.nasa.gov/alsj/a11/a11fltpln_final_reformat.pdf
It would actually be less, just enough to sustain some of the weight of the vehicle but not enough to propel it on the opposite direction.
GoneBamboo - You can create the same "Force" by pushing away a very large mass SLOWLY or a very small mass VERY FAST. You need to generate MOMENTUM which is defined as MASS times VELOCITY.
So you can create a large MOMENTUM with high velocity and low mass (like a rocket engine) or low velocity and high mass (like a catapult throwing rocks)
I have a 44 magnum desert eagle that fires very large bullets and a relatively low muzzle velocity.
Don't you agree that my smaller .223 rifle could create the SAME RECOIL as the desert eagle IF it can get the smaller bullet to much higher velocity than the 44 magnum?
In my book a more complex rocket propulsion equation is given...
By the way the last question you can use F = m x delta(v)/delta(t)
Dude u copied the equation on the cover
.
Simplest explanation
Couldn't it be like, that we could differentiate momentum w.r.t time and get m.dv/dt & v.dm/dt and consdering velocity constant it would become F=v.dm/dt.
Also what if the rocket is still leaving eath shouldn't mass.g be applied
what about this formula f+u(change in m)=ma
So when calculating the thrust force,are we always supposed to calculate for the fluid first?
F = (m/t) x V
F=thrust
M=weight
T=time
V=velocity
Nice but you dont take into account the relative change in mass ejection as the rocket increases speed per unit time.
Its the speed of the exhaust, not the rocket. Exhaust velocity will be constant no matter how much mass is expelled.
can someone help me with this:
NASA is launching a rocket into space from Earth. This particular rocket burns its fuel for 5 minutes and
then turns off its engines. At this point, the rocket keeps floating along through space at a speed of 10
km/s.
a.) What is the rocket’s average acceleration during these
first 5 minutes?
b.) How far did the rocket travel during the first 5
minutes?
c.) The Moon is approximately 3.8 x 10^5km from Earth. At its final speed, how long will it take the rocket to reach the Moon?
d.) The star nearest to us, called Alpha Centauri, is
4.1 x 10^13 km away. Could this rocket get a man to Alpha Centauri before he dies of old age?
e.) The rocket is half way to the Moon when its pilot
suddenly notices he forgot his camera! He radios a second rocket crew on Earth and tells them to bring his camera and meet him on the Moon when he lands. Unfortunately, the second rocket can only accelerate at 3 m/s^2. Can this second rocket make it to the Moon before the first one lands?
Do you still need help?
@@warzonemoments3970 it's okay now 😊👌🏻thank you though
Give me the solution
*While watching the video, an ad about rocket equation appeared.*
Chris Hadfield?
when calculating the force of rocket why we use fuel velocity relative to rocket instead of using the actual velocity of fuel?
Because the relative velocity is the one moving the rocket, not the fuel itself. Moving fuel and moving rocket are two different things.
Bro fuel velocity depends on volume of fuel after burning
Hello,
I need help for followinng question. Please!
A small research rocket of mass 250kg is launched vertically as part of a weather study. It sends out 50 kg of burnt fuel and exhaust gases with a velocity of 180 ms-1 in a 2 s initial acceleration period.
(a) What is the velocity of the rocket after this initial acceleration? (Answer v = 40 ms-1)
(b) What upwards force does this apply to the rocket? (Answer F = 4500N)
(c) What is the net upwards acceleration acting on the rocket? (Use g = 10 ms-2 if required) (Answer a = 10 ms-2)
Hi
In the first part of the question I am obtaining 36 as answer. I guess the answer is wrong. Since in thus we just have to apply the formula MV=mv and mass needn't be included.
In the second part you have to apply first equation of motion (v=u+at) but using the downward value and by third law of motion upward and downward values will be equal.
I am not sure abt 3rd. I will confirm once and leave reply as soon as possible.
BTW it had been 3 months since you posted this question. I think you have got the answer. And if yes please tell where I am making mistake.
Thanks
@@arpanarora Lets' compare solution! Could you please tell me your solution for this problem.
Here is my solution:
(a) Given:
m1 = m2 + m3 = 250 kg u1 = 0 ms^-1
m2 = m_rocket = 250 kg - 50 kg = 200 kg v2 = ?
m3 = m_fuel gas = 50 kg v3 = 180 ms^-1 t = 2 s (initial acceleration)
From m1.u1 = m2.v2 + m3.v3
(250kg x 0 ms^-1) = (200 kg x v2) + (50 kg x 180 ms^-1)
==> v2 = -45 ms^1 = 45 ms^-1 in the direction opposite to the exhausted gas (Hence, answer = 40 ms^-1 is WRONG)
(b) m2 = m_rocket = 250 kg - 50 kg = 200 kg
u2 = u_rocket = 0 ms^-1
v2 = 45 ms^-1 (as calculated in (a) above) t2 = 2 s
a2 = a_apply = ?
From a = (v-u)/t
==> a2 = a_apply = (45 ms^-1 - 0 ms^-1) / 2 s = 22.5 ms^-2
==> F_apply = m.a_apply = 200 kg x 22.5 ms^-2 = 4500 N
(c) m2 = m_rocket = 250 kg - 50 kg = 200 kg
F_g = m2.g = 200 kg X 10 ms^-2 = 2000 N (only m_rocket as fuel and gas are burning off)
F_net = F_appy - F_g
and F_appy = 4500 N (found in (b))
==> F_net = 4500 N - 2000 N = 2500 N
From F_net = m.a
==> a = F_net / m = 2500 N / 250 kg = 10 ms^-2
Now your turn to show me your solution PLEASE
@@huyanhle it seems your solution is absolutely right.
But this is how I did this question
1) Mass of rocket M= 250kg Velocity of the rocket V=?
Mass of the burnt fuel m =50kg escape Velocity v=180 m/s
Using conservation of momentum M*V=m*v
250*V= 50*180
V=9000/250
V=36m/s.
2) by applying
F=ma
F=(m*v)/ time
F= 50*180/2
3) I was unable to solve 3rd part but I guess yours is right.
clear explanation man
Istg u made my life easier
I didn't understand what you meant when you said we can write the change of momentum as the product of a mass and its change in velocity or the product of a changing mass with a constant velocity. How can the mass change? At least in Newtonian mechanics, isn't mass supposed to be constant?
Mass changes because fuel is being used and the fuel has mass
WAXOGEN WILL BECOME A ROCKET FUEL IN THE FUTURE
5:16 So the force that man feels should be taken negative?
The man feels the force to the opposite direction than the block. In a closed system the force should be taken as negative since all forces in a closed system always add up to 0. But the force that the man feels is not negative from his perspective.
Keep going
thanks mate
Anybody happen to know the mass & velocity exiting the LEM's engine nozzle as it descended to the moon's surface?
Why? Are you trying to replicate the Apollo program?
No, but if you would like some numbers to "noodle with", a Space Shuttle main engine ingests 500 kilograms of liquid oxygen and liquid hydrogen PER SECOND, combusts it and expels PURE WATER at 2500 meters per second (at sea level)
Sir love from Infia
India
@@siyarana2122 😂
i believe the second exercise is wrong, you do not take into account the force of gravity that has an effect on the spaceship
he took the equation to be happening in space. not near/on earth surface. have a good day
Mass flowrate =Fuel + air ?.
Was the negative sign also indicates the mass is decreasing?
No it only represent the direction
Please teach Chemical Engineering lol!!
And where do you get this Ball in space? A ball has inertia. Expanding rocket gas particles do not have inertia.in a vacuum. because the gas particles are in 'free expansion'.
CLEARLY, your knowledge of physics is very limited.
No doubt, you're one of the deluded flat earth believing fools too.
How we can say "Change in mass"
Mass is constant every where
The rocket loses mass as it ejects the exhaust gas
Please do face cam it make more interactive ☹️☹️
critical gas inertia
Physics is tricky
Van gogh are you?
Pd: Good video :)
So I learnt rocket science
slow down💀😅😅
You are not "shooting out matter" from a rocket in the vacuum of space. In space , the vacuum evacuates the rocket gases at the rate at which they expand due to there being no resistance.
A rocket works on the pressure principle. You need to create pressure behind the rocket to create movement. It is impossible to create pressure in a vacuum.
- "It is impossible to create pressure in a vacuum."
*This is FALSE* !!
Rocket engines can EASILY create nozzle pressures above 3000 PSI *IN HARD VACUUM* .
I have SHOWN you videos of rocket propulsion IN A VACUUM, but you just keep dismissing it with your expert opinion of "Nuh-UHHHHHH"
Now tell us your level of physics education, or JUST THE FUCK UP ALREADY with your nonsense!!
ACTUAL *Rocket Propulsion ENGINEERS* have told you that you are mistaken, but you dont care since you are "self-educated" by YT videos...
It pathetic how far we have slipped into UN-educated masses having access to modern technology and still being able to vote.
bo ptah = ignorant fool.
I dont need to know this Im just high af. And I like learning when Im high XD. I dont get it btw
😊
ترجمه عربي
ترجمه
😁🧒🧒
why are 6th graders doing this
is this rocket science?
Yes.
@@hscc0rp898 no it's pseudoscience.
انته لو كماك عشرة ، كان ما حد أيروح العسكرية . أرفع الكمه عشانك....
nah
The narrator is boring.