Show That the Variation of Gravity With Height Can Be Accounted for Approximately
Variation of gravity with height
- Thread starter KBriggs
- Outset date
Homework Argument
Show that the variation of gravity with top 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 tiptop above the surface.
Homework Equations
r=R+z
V=-GmM/r
F=GmM/r^ii
The Attempt at a Solution
Kickoff, 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 and so expanded this around z=0 and took the get-go two terms in the series to get:
Five=-GmM/R + GmMz/R^ii
and you tin factor this to get
Five=GmM/R * [1-z/R]
but the strength F given past the potential energy function is GmM/R^ii at the surface, so this is equal to mg, then g = GM/R^2
Then the V function is then 5 = mgR[1-z/R]
And here I am stuck. It appears that I am yet taking the centre of the planet equally my reference point. Tin someone help me? I'thou stuck.
Somehow I need to redefine the reference betoken and then that V = 0 for z = 0. How tin can I do this?
Answers and Replies
I now have something very close, but not quite.
Since V is unchanged past calculation a constant, I added GmM/R to V and swapped r = R+z to become
V(z) = GmM/R - GmM/(R+z), which is 0 at z = 0 every bit it should be.
At present I expand this around 0 and take the first three terms in the series:
V(0) + zV'(0) + z^ii V''(0)
to get:
0 + zGmM/R^two -2z^2GmM/R^iii
which can be factored and the identity g = GM/R^2 put in to become
V = mgz[1-2z/R]
Why do I accept an extra factor of two in my answer....?
Oops: forgot to divide by ii factorial in the expansion...
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Source: https://www.physicsforums.com/threads/variation-of-gravity-with-height.344477/
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