We know that the magnitude of the gravitational force is given by:
Fg | = |
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Use the connection between force and potential energy to determine the general form of gravitational potential energy. U = mgh applies only for a uniform field, so it does not apply here where the field goes as 1/r2.
F | = | - |
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ΔU = - ∫ F dr
This gives: U | = |
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if we define the potential energy to be zero at r = infinity. You are NOT free to define the zero anywhere you want - it is pre-defined to be zero at infinity.
Does it matter that the potential energy is negative everywhere? Not at all - all that matters is how the potential energy changes. If a mass moves from close to an object to further away, the potential energy changes from a larger negative number to a smaller one - this is an increase, as we expect.
This is consistent with the mgh we used for potential energy near the surface of the Earth, which can be proved.
If you move an object up a height h from ground level, the potential energy changes as follows:
ΔU = Uf - Ui = -GmM/(R+h) - -GmM/R.
1/(R+h) = 1/[R(1+h/R)] = (1/R)*(1 - h/R) when h is small compared to R.
Substituting this in gives:
ΔU = -GmM/R + GmMh/R2 + GmM/R = GmMh/R2
We showed previously that g = GM/R2, so:
ΔU = mgh.
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How fast would you have to throw an object so it never came back down? Ignore air resistance. Let's find the escape speed - the minimum speed required to escape from a planet's gravitational pull.
Apply the master energy equation:
Ui + Ki = Uf + Kf
Set both terms on the right to zero. We want the object to reach infinity, where the potential energy is zero, and we want it to just reach infinity...so it can arrive there with negligible speed.
Basically the total initial energy must be at least zero for the object to escape.
For a planet of mass M and radius R, that potential energy of an object of mass m at the planet's surface is:
U | = |
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Therefore: |
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+ Kescape = 0 |
½ mv2 | = |
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vescape | = | ( |
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) | ½ |
For the Earth this works out to 11.2 km/s.