Question

(a) Calculate the effective value of g, the acceleration of gravity, at 7100 m above the Earth's surface. (b) Calculate the effective value of g, the acceleration of gravity, at 7100 km above the Earth's surface.

Answer 1

The equation to calculate the acceleration due to gravity at a given height is,

$g^{\prime}=\frac{GM}{(R+h)^2}$

Here, G is the gravitational constant, M is the mass of earth, R is the radius of earth and h is the given height.

Part (a)

The given height is

$h=7100\text{ m}$

Substitute the known values in equation,

$\begin{gathered} g^{\prime}=\frac{(6.67\times10^{-11}Nm^2kg^{-2})(\frac{1kgms^{-2}}{1\text{ N}})(5.97\times10^{24}\text{ kg)}}{(6.37\times10^6m+7100m)^2} \\ =\frac{39.82\times10^{13}m^3s^{-2}}{(6377100m)^2} \\ =9.79m/s^2 \end{gathered}$

Thus, the effective value of g at the given height is

$9.79m/s^2$

Part (b)

The given height is

$\begin{gathered} h=7100\text{ km} \\ =7.1\times10^6\text{ m} \end{gathered}$

Substitute the known values in the same equation,

$\begin{gathered} g^{\prime}=\frac{(6.67\times10^{-11}Nm^2kg^{-2})(5.97\times10^{24}\text{ kg)}}{(6.37\times10^6\text{ m+}7.1\times10^6m)^2} \\ =\frac{39.82\times10^{13}m^3s^{-2}}{(13.47\times10^6m)^2} \\ =2.19m/s^2 \end{gathered}$

Thus, the effective value of g at the given height is

$2.19m/s^2$