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,
[tex]g_h=g(1-\frac{2h}{R})[/tex]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,
[tex]\begin{gathered} \frac{g_h}{g}=(1-\frac{2\times320\times10^3}{6.37\times10^6}) \\ =1-0.1 \\ =0.9 \end{gathered}[/tex]Thus the ratio of the acceleration due to gravity at the given height to that at the surface of the earth is 0.9
