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)=(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*10^(-11)Nm^2kg^(-2))(\frac{1kgms^(-2)}{1\text{ N}})(5.97*10^(24)\text{ kg)}}{(6.37*10^6m+7100m)^2} \\ =(39.82*10^(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*10^6\text{ m} \end{gathered}`

Substitute the known values in the same equation,

`\begin{gathered} g^(\prime)=\frac{(6.67*10^(-11)Nm^2kg^(-2))(5.97*10^(24)\text{ kg)}}{(6.37*10^6\text{ m+}7.1*10^6m)^2} \\ =(39.82*10^(13)m^3s^(-2))/((13.47*10^6m)^2) \\ =2.19m/s^2 \end{gathered}`

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

`2.19m/s^2`