Near Earth Gravitation

Question:

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?

Answer:

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:
g=g0*(R²)/(R²+2*R*h+h²).

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:
g=g0*(1-2*h/R)/(1-4*h²/R²),
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.

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