Question

The geostationary orbit of the earth is at a distance of about 36000 km from the earth's surface. Find the weight of a 120-kg equipment placed in a geostationary satellite. The radius of the earth is 6400 km.

Answer 1

The geostationary orbit of earth is at a distance of about 3600 km

We know, $g^{\prime }=\frac{Gm}{(R+h{)}^2}$

At $h=36000km,$

$g^{\prime }=\frac{Gm}{(36000+6400{)}^2}$

At the surface,

$g^{\prime }=\frac{Gm}{(6400{)}^2}$

$\therefore \frac{g^{\prime }}{g}=\frac{6400\times 6400}{42400\times 42400}$

$=\frac{256}{106\times 106}=0.0227$

$\Rightarrow g^{\prime }=0.0227\times 9.8=0.223$

[taking g = 9.8 $m/s^2$ at the surface of the earth.]

A 120 kg equipment placed in a geostationary satellite will have weight

$m{g}^{\prime }=0.233\times 120$

$=26.79\approx 27N$