Question

find the retio of acceleration due to gravity at the position of a satelite located 320 km BOVE THE EARTHS SURFACE TO THE FREE FALL ACCELERATION AT THE SURFACE OF THE EARTH?

Answer 1

The height at which the satellite is situated, h=320 km=320×10³ m

The acceleration due to gravity at a height h is given by,

`g_h=g(1-(2h)/(R))`

Where R is the radius of the earth and g is the acceleration due to gravity at the surface of the earth.

The radius of the earth is 6.37×10⁶ m.

On substituting the known values in the above equation,

`\begin{gathered} (g_h)/(g)=(1-(2*320*10^3)/(6.37*10^6)) \\ =1-0.1 \\ =0.9 \end{gathered}`

Thus the ratio of the acceleration due to gravity at the given height to that at the surface of the earth is 0.9