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

but remember that h/R is small so h^{2}/R^{2}
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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