Variation of gravity with height

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The discussion centers on deriving the variation of gravity with height using the potential function V(z) = mgz(1 - z/R). The user begins with the potential energy function V(r) = -GmM/r and modifies it to account for height by defining r as R + z. After expanding the potential energy function around z = 0, they attempt to adjust the reference point to ensure V = 0 when z = 0. The user encounters confusion regarding an extra factor of two in their final expression and realizes they neglected to divide by 2 factorial during the series expansion. The conversation highlights the importance of careful mathematical manipulation in gravitational potential calculations.
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Homework Statement


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

V(z)=mgz(1-z/R)

Where R is the radius of the Earth and z the height above the surface.


Homework Equations


r=R+z
V=-GmM/r
F=GmM/r^2


The Attempt at a Solution



First, I said define z such that r = R + z. We have the potential energy function for the Earth as V(r)=-GmM/r, so V(z)=-GmM/(R+z). I then expanded this around z=0 and took the first two terms in the series to get:

V=-GmM/R + GmMz/R^2

and you can factor this to get

V=GmM/R * [1-z/R]

but the force F given by the potential energy function is GmM/R^2 at the surface, so this is equal to mg, so g = GM/R^2

So the V function is then V = mgR[1-z/R]

And here I am stuck. It appears that I am still taking the centre of the planet as my reference point. Can someone help me? I'm stuck.


Somehow I need to redefine the reference point so that V = 0 for z = 0. How can I do this?
 
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Additional work:

I now have something very close, but not quite.

Since V is unchanged by adding a constant, I added GmM/R to V and swapped r = R+z to get

V(z) = GmM/R - GmM/(R+z), which is 0 at z = 0 as it should be.

Now I expand this around 0 and take the first three terms in the series:

V(0) + zV'(0) + z^2 V''(0)

to get:

0 + zGmM/R^2 -2z^2GmM/R^3

which can be factored and the identity g = GM/R^2 put into get

V = mgz[1-2z/R]

Why do I have an extra factor of two in my answer...?

Oops: forgot to divide by 2 factorial in the expansion...
 
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The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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