Gravitational potential energy at large distances | AP Physics 1 | Khan Academy

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Learn how to calculate gravitational potential energy when we can no longer assume the gravitational field is uniform.

AP Physics 1 on Khan Academy: Meet one of our writers for AP¨_ Physics, Sean. A physics teacher for seven years, Sean has taught AP¨_ Physics 1, AP¨_ Physics C, and Conceptual Physics. HeÕs also a former mechanical engineer. Sean is based in Boise, Idaho, and is a Khan Academy physics fellow, creating awesome new exercises and articles for AP¨_ Physics.

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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.

nafrost
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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)

cesarjom
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This guy has taught me more than my teachers

blakeguarnieri
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College Physics student here!
This can be confusing and Very misleading. It's easier to understand if you take the indefinate integral of the force (GMm/R^2) with respect to R (say R=r+h where h is from PE=mgh and r is the radius of the planet). Say that this equation is labeled U(h) (setting it as a function of h).
Now what you need to do is find the Constant of Integration by setting the boundary condition U(h=0)=0 and solve for the constant. You should end up with something like this:

U(h)=(GMm/r)-(GMm/(r+h)) or more suggestively:
U(h)=(GMm/r)*(h/(r+h))

This also mathematically explains why its negative as the indefinate integral of 1/x^2 with respect to x is -1/x. Inputing the values into U(h) and PE(h) for large r and you'll see that the two are approximately equal to each other for relatively low values of h.

This is why it's important to NEVER FORGET THE CONSTANT OF INTEGRATION!

eliirrii
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A elegant derivative of potential energy. Well done 👏

johnholme
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i literally never comment but seriously this video helped sm, thank you!

whothax
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Sir I saw your FACE today. And your VOICE has seemed to belong to ANYBODY but the real owner of it.

ummeyusuf
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So we integrating S (F)dr from r to inf ?

Sneakyne
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wow i love the way you explained this
your explanation will not be in vain

mangakasaide
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Excellent. I finally understood the reason for the negative. Just learnt this chapter for CIE A2 level Physics today!

oneinabillion
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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?

brinkman
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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?

zizerokub
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So let me get this right, we added in a negative sign to fit the equation into our framework?! Is that even mathematically legal ?

davidespejo
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@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?

stavshmueli
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I can tell it's a useful video, but I still can't understand a thing in physics...

nicoleqian
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sal try to explain a concept without software by using black board and chalk.

.ayirp.
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gravitational PE can never be positive

qualquan
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I dont understand why you make it negative

davidsweeney
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Sir please make all videos of gravitation class 11th in Hindi
Please sir

codebros