The video adds a negative (-) sign at the end in an attempt to make sense of gravitational PE equation as function of distance from center of mass of planet (eg earth). There is a more formal approach where you can see how the negative sign is derived from first principles. Think of gravitational PE (U) of a mass object in terms of the force of the gravitational field (F) on the mass, such that the vector F = - gradient U. Gradient U is defined as change of U over distance -- that is, slope dU/dr in the radial direction, since the vector F for gravitational field has only a radial component. If slope dU/dr > 0 (positive), then you expect the gravitational force to be F < 0 (negative) since F must oppose the increase of gravitational PE of the mass object (the vector is also opposite direction of the increasing direction of PE along the radial direction). So that gives us... dU/dr = - F (for radial component of the gradient U = - vector F) where vector F = - GMm/r^2 (along unit vector r) showing vector F is pointed opposite direction of radial unit vector r (Note: M=planet mass, m= object mass, r = radial distance between them) Now take... integral of (dU/dr) * dr [over limits r to infinity] = integral of - F * dr [over limits r to infinity] = integral of - (- GMm/r^2) * dr [over limits r to infinity] U(infinity) - U(r) = GMm (-1/r) [from r to infinity] = - GMm (1/infinity - 1/r) U(r) - U(infinity) = GMm (1/infinity - 1/r) = GMm (0 - 1/r) ... since 1/infinity = 0 Therefore, PE as a function of radial distance r is U(r) = - GMm/r ... since we define gravitational PE, U --> 0 as r --> infinity (ie, as mass m object gets infinitely far away from mass M)
Thanks a lot. I already did the math myself, but I didn't know you add a minus, so I was confused how potential energy gets lower as you move closer to an object, thanks for solving that mystery for me.
I don't understand why you can just put a minus to make the math work. Seems to me it indicates a flawed logic in PE and gravity. For instance how do I work with negative PE values? Plz help with understanding. So is PE zero at infinity and at zero distance?
@BrinkMan49 Negative sign also indicates that the force acting on m1 by m2 is in opposite direction to that of displacement ( when m1 is moving away from m2 )
Technically r=0 is not possible as two masses will always occupy some volume. Now shells of matter, well they could nest, but let's not go there right now. As r approaches 0, the PE would approach infinity. As for the negative sign, this is actually the result of calculus, and just sticking it in as this video does seems sneaky. Sometimes us physics teachers pull this kind of sneaky move when avoiding higher maths. Not cool, but done sometimes. You can think of the negative being necessary because as you separate masses (lift) this takes a positive amount of energy. You would need to do work on a mass to lift it. But when you lift it all the way to infinity, you add more and more positive energy into the system, until finally it reaches zero. Huh? What kind of number can you add a positive to, and get to zero? Well, a negative number. Any system where things are attracted (opposite electric charges, or magnetic poles), you will find negative potential energy. If stuff repels, then the PE will be positive.
i guess , for example , we know the magnitude of Potential difference is Electric field times distance . BUT we know electric field passes from High potential Low potential so we add a -ve sign to the equation dV = (-) Edr
Gravitational Potential Energy is Gravitation? every mass in solar system or universe have Gravitation to each other much or less? so there should be no reference point? or every mass in earth should have reference point inside center of the earth not earth surface?
You just got to keep rewatching it. I just rewatched the part where he explained why it's negative like 5 or 6 times, drawing it out and I finally got it. The reason why you're even on here has to be rock solid
I think the main concept behind this is we are always bind to earths gravitational field , in other word we have to do some work (+ve energy) on us to escape from earths gravitational field .
When you integrate 1/r^2 with respect to r you get -1/r. If you evalute from height a to height b, you get -1/a -(-1/b) = -1/a +1/b , since a is greater as it is your starting heigh, 1/a is smaller than 1/b , thus -1/a + 1/b is positive.
David think about it this way... on earth, the gravitational potential energy would be 0 at sea level, so when you lift an object up the potential energy increases. But if there were only two objects in space, where would must they be placed so that the potential energy is 0? Theoretically, they have to be placed an infinite distance apart. This is where there is maximum potential energy. So as you try to approach infinity, the amount of potential energy will approach 0. Therefore, universal gravitational potential energy can never be positive
When r tends to infinity the gravitational potential energy becomes zero it means when you increase the radius the gravitational potential energy decreases but we want it to increase so we use a negative in front of it
@3:45 Sal says: "Let's put a negative upfront". Well, if you put a negative in front of one side of the equation, you must do the same for the other side of the equation, which means that the Gravitational potential energy should also be negative. And so we are back to square one. Why can we do this step, and seemingly break basic algebraic rules?
You can derive that you need the negative sign using calculus. It's quite fascinating when you learn calculus and realize, "oh, this equation must be negative." Without calculus it is hard to explain, but Sal is just trying to give an intuitive reason for it without calculus.
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The video adds a negative (-) sign at the end in an attempt to make sense of gravitational PE equation as function of distance from center of mass of planet (eg earth). There is a more formal approach where you can see how the negative sign is derived from first principles.
Think of gravitational PE (U) of a mass object in terms of the force of the gravitational field (F) on the mass, such that the vector F = - gradient U.
Gradient U is defined as change of U over distance -- that is, slope dU/dr in the radial direction, since the vector F for gravitational field has only a radial component.
If slope dU/dr > 0 (positive), then you expect the gravitational force to be F < 0 (negative) since F must oppose the increase of gravitational PE of the mass object (the vector is also opposite direction of the increasing direction of PE along the radial direction).
So that gives us... dU/dr = - F (for radial component of the gradient U = - vector F)
where vector F = - GMm/r^2 (along unit vector r) showing vector F is pointed opposite direction of radial unit vector r
(Note: M=planet mass, m= object mass, r = radial distance between them)
Now take...
integral of (dU/dr) * dr [over limits r to infinity] = integral of - F * dr [over limits r to infinity]
= integral of - (- GMm/r^2) * dr [over limits r to infinity]
U(infinity) - U(r) = GMm (-1/r) [from r to infinity]
= - GMm (1/infinity - 1/r)
U(r) - U(infinity) = GMm (1/infinity - 1/r)
= GMm (0 - 1/r) ... since 1/infinity = 0
Therefore, PE as a function of radial distance r is
U(r) = - GMm/r ... since we define gravitational PE, U --> 0 as r --> infinity (ie, as mass m object gets infinitely far away from mass M)
fam for my tiny highschool mind the added minus sign is enough; he does this for simplicity or we would not understand
U knew what u were about to comment wasn’t simple but u still did it anyway 🙄
Thanks a lot.
I already did the math myself, but I didn't know you add a minus, so I was confused how potential energy gets lower as you move closer to an object, thanks for solving that mystery for me.
Sir I saw your FACE today. And your VOICE has seemed to belong to ANYBODY but the real owner of it.
I don't understand why you can just put a minus to make the math work. Seems to me it indicates a flawed logic in PE and gravity. For instance how do I work with negative PE values? Plz help with understanding. So is PE zero at infinity and at zero distance?
@BrinkMan49 Negative sign also indicates that the force acting on m1 by m2 is in opposite direction to that of displacement ( when m1 is moving away from m2 )
Technically r=0 is not possible as two masses will always occupy some volume. Now shells of matter, well they could nest, but let's not go there right now. As r approaches 0, the PE would approach infinity. As for the negative sign, this is actually the result of calculus, and just sticking it in as this video does seems sneaky. Sometimes us physics teachers pull this kind of sneaky move when avoiding higher maths. Not cool, but done sometimes. You can think of the negative being necessary because as you separate masses (lift) this takes a positive amount of energy. You would need to do work on a mass to lift it. But when you lift it all the way to infinity, you add more and more positive energy into the system, until finally it reaches zero. Huh? What kind of number can you add a positive to, and get to zero? Well, a negative number. Any system where things are attracted (opposite electric charges, or magnetic poles), you will find negative potential energy. If stuff repels, then the PE will be positive.
love to see u back khan!
This guy has taught me more than my teachers
A elegant derivative of potential energy. Well done 👏
So we integrating S (F)dr from r to inf ?
I love Khan Academy 💕🤩
Thank you Thank you Thank you!!!
Thanks a lot.finally understood it.
So let me get this right, we added in a negative sign to fit the equation into our framework?! Is that even mathematically legal ?
i guess , for example , we know the magnitude of Potential difference is Electric field times distance . BUT we know electric field passes from High potential Low potential so we add a -ve sign to the equation dV = (-) Edr
You don't know Hindi
Gravitational Potential Energy is Gravitation?
every mass in solar system or universe have Gravitation to each other much or less?
so there should be no reference point?
or every mass in earth should have reference point inside center of the earth not earth surface?
👍🏼👍🏼👍🏼
👍🏼👍🏼👍🏼
I can tell it's a useful video, but I still can't understand a thing in physics...
Qian Nicole same
You just got to keep rewatching it. I just rewatched the part where he explained why it's negative like 5 or 6 times, drawing it out and I finally got it. The reason why you're even on here has to be rock solid
gravitational PE can never be positive
i literally never comment but seriously this video helped sm, thank you!
Excellent. I finally understood the reason for the negative. Just learnt this chapter for CIE A2 level Physics today!
Well explained!!!
Hey
wow i love the way you explained this
your explanation will not be in vain
Thx! Really helped me with this
60 minutes into 5 minutes
WHAT A VOICE!!
Please make some videos on gauss law
Hi
hi sal.good night.it is india..
Nice trick
Thanks Sal. You're the best teacher in the world
sal try to explain a concept without software by using black board and chalk.
I dont understand why you make it negative
I think the main concept behind this is we are always bind to earths gravitational field , in other word we have to do some work (+ve energy) on us to escape from earths gravitational field .
If the negative wasn't there, when you increase r, your potential energy would decrease while we want the opposite to happen.
When you integrate 1/r^2 with respect to r you get -1/r. If you evalute from height a to height b, you get -1/a -(-1/b) = -1/a +1/b , since a is greater as it is your starting heigh, 1/a is smaller than 1/b , thus -1/a + 1/b is positive.
David think about it this way... on earth, the gravitational potential energy would be 0 at sea level, so when you lift an object up the potential energy increases.
But if there were only two objects in space, where would must they be placed so that the potential energy is 0? Theoretically, they have to be placed an infinite distance apart. This is where there is maximum potential energy. So as you try to approach infinity, the amount of potential energy will approach 0. Therefore, universal gravitational potential energy can never be positive
When r tends to infinity the gravitational potential energy becomes zero it means when you increase the radius the gravitational potential energy decreases but we want it to increase so we use a negative in front of it
Genius!
@3:45 Sal says: "Let's put a negative upfront". Well, if you put a negative in front of one side of the equation, you must do the same for the other side of the equation, which means that the Gravitational potential energy should also be negative.
And so we are back to square one. Why can we do this step, and seemingly break basic algebraic rules?
Because he was interpreting the expected result and modified his result from that
Because you're not using the minus in the actual calculation you are using it to show that the gravitational field is only an attractive force
You can derive that you need the negative sign using calculus. It's quite fascinating when you learn calculus and realize, "oh, this equation must be negative." Without calculus it is hard to explain, but Sal is just trying to give an intuitive reason for it without calculus.
Sir please make all videos of gravitation class 11th in Hindi
Please sir