Near Earth Gravitation


When doing calculations on gravitation, one equation is g = go (1 - 2h/R) where g is the gravitational field above the earth surface, go is the gravitational field just on the earth surface, R is the radius of the earth. What is the meaning of (1 - 2h/R)? What does it represent physically?


The acceleration due to gravity at any altitude h above the surface of the Earth is actually given by g=g0*(R/(R+h))2, since gravitation follows an inverse square law. From this you can see that if h=0, g=g0. For large h, g approaches zero but never gets there. If we carry out the squaring operations in the formula we get:

If h is small compared to R, then h2 will be even smaller compared to R2 and even 2*R*h will be much larger than h2. In that case we may neglect the h2 square term and say that g=g0*(R2)/(R2+2*R*h).

Then g=g0*1/(1+2*h/R). Now if we multiply both numerator and denominator by (1-2*h/r) we get:
but remember that h/R is small so h2/R2 will be tiny and even 4 times it may be neglected compared to 1. That leaves us the formula you started with, so you can see it works only for h very small compared to R.

Frankly I think that using the real formula, g=g0*(R/(R+h))2, is easy enough that this approximation is hardly worth the trouble.

This information is brought to you by M. Casco Associates, a company dedicated to helping humankind reach the stars through understanding how the universe works. My name is James D. Jones. If I can be of more help, please let me know. JDJ