Question

Calculate the effective value of g, the acceleration of gravity, at 7400m above the Earth's surface.

Calculate the effective value of g, the acceleration of gravity, at 7400 km above the Earth's surface.

Answer 1

.'. g = GM/R^2 .....(iii)
[Note:
The expression in equation (iii) give the value of 'g' on the surface of the earth and hence 'R' denotes mean radius of the earth]

Also by rearranging the terms we get,
g R^2 = GM .....(iv)


Now going back to the given problem,
consider a body which is at a distance 'h' from the SURFACE of the earth.
.'. The body is at a distance of (h+R) from the CENTRE of the earth.
Let,
g" be the acceleration due to gravity of a body at a distance of (h+R) from the centre of the earth.

.'. In analogus with equation (iii) we can write

g" = GM/(h + R)^2 .....(v)

So using equation (v) and putting all the known values, we can find g"

Rearranging the terms in above equation, we get
g" (h + R)^2 = GM .....(vi)
================================
However, to reduce the calculation, there is another method as given before.
We know that,
g = 9.81 m/sec^2

So,
using equations (iv) and (vi), we get

g" (h + R)^2 = g R^2

.'. g" = g R^2/(h + R)^2

.'. g" = 9.81 * [R/(h + R)]^2

=9.81*(7400/(7400+6378.1))^2=2.82 m/s^2