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Variation of potential energy with height

  • #1

Homework Statement


Show that the variation of gravity with height can be accounted for approximately by the following potential function

V = mgz(1+z/re)

in which re is the radius of the Earth. find the force given by the above potential function.

Homework Equations



V = GM/r

The Attempt at a Solution


Let z be a point above the surface of the Earth and R be the radius of the earth

V(z) = -GM/(R+z) = (-GM/R)(1+z/R)-1

Using the binomal expansion of (1+x)n ≈ 1 +nx +n(n-1)x2/2

I get

V = (-Gm/R)(1-z/R + (z/r)2

Expanding gives

V = -GM/R + GMz/R2 - GMz2/R3

but the force is given by

F = GM/R2 = mg

replace GM/R2 with mg to get

V = -mgR + mgz - mgz2/R

= -mgR + mgz(1-z/R)

I'm not sure why I have an extra -mgR term.
 

Answers and Replies

  • #2
andrewkirk
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Gravitational potential energy, like gravitational and electrical potential, has no absolute level, so we can add any constant we like to it.
Gravitational potential energy near the Earth's surface is usually given a gauge (choice of a constant added) such that te value at the Earth's surface is 0. In that gauge, the gravitational PE of an item of mass m that is z metres above the Earth's surface, which is distance R from the centre, is the formula you got, minus the grav PE it would have at the Earth's surface, which is -mMG/R. Taking the difference removes the unwanted term.

Strictly speaking, the formula is for the grav PE of the mass relative to what it would have at the Earth's surface.

By the way, the function given in the OP is of Potential Energy, not potential. To get potential we have to divide by m.
 
  • #3
Okay, so I multiply by m to get the potential energy, but I don't understand what you mean by adding a constant. So my first term in my expansion should just be 0 somehow?
 
  • #4
andrewkirk
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Okay, so I multiply by m to get the potential energy, but I don't understand what you mean by adding a constant. So my first term in my expansion should just be 0 somehow?
Your expansion is fine. But the potential energy your formula gives is relative to what the object's PE would be at the centre of the Earth. The formula you are asked to prove is for PE relative to what it would be at the surface of the Earth. Since the latter is -mMG/R, you subtract that from what you got, to get the PE relative to the Earth's surface.
 
  • #5
I see. Thanks!
 

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