Variation of gravity with height

Click For Summary
SUMMARY

The variation of gravity with height can be approximated using the potential function V(z) = mgz(1 - z/R), where R represents the Earth's radius and z is the height above the surface. The derivation begins with the potential energy function V(r) = -GmM/r and involves expanding this function around z = 0. The correct formulation of V(z) ensures that it equals zero at z = 0, leading to the final expression V = mgz(1 - 2z/R) after correcting for the factorial in the series expansion.

PREREQUISITES
  • Understanding of gravitational potential energy and force equations
  • Familiarity with Taylor series expansion
  • Knowledge of constants such as G (gravitational constant) and M (mass of the Earth)
  • Basic concepts of physics related to gravity and height
NEXT STEPS
  • Study the derivation of gravitational potential energy in more detail
  • Learn about Taylor series and its applications in physics
  • Explore the implications of gravitational variation in different celestial bodies
  • Investigate the effects of height on gravitational force in practical scenarios
USEFUL FOR

Students studying physics, particularly those focusing on gravitational theory, as well as educators looking for clear explanations of gravitational potential energy variations with height.

KBriggs
Messages
29
Reaction score
0

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?
 
Last edited:
Physics news on Phys.org
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...
 
Last edited:

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
Replies
5
Views
2K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
Replies
3
Views
3K
Replies
6
Views
2K
  • · Replies 16 ·
Replies
16
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
2
Views
744
  • · Replies 5 ·
Replies
5
Views
2K