Question

7. What is the gravitational field strength at a distance of 60.0 km above the surface of the Earth?

Answer 1

Gravitational field strength (in other words, gravitational acceleration) is given as follows:

$$g = \dfrac{GM}{R^2}$$

where $G = 6.674\times 10^{-11}\dfrac{m^3}{kg\cdot s^2}$ is the gravitational constant, $M = 5.972\times 10^{24}kg$ is the mass of the Earth, and $R = 6.371\times 10^{6}m + 0.06\times 10^{6}m = 6.431\times 10^{6}m$ is the distance from the center of the Earth to the required point above the surface (radius plus 60 km).

Thus, obtain:

$$g = \dfrac{6.674\times 10^{-11}\cdot 5.972\times 10^{24}}{(6.431\times 10^{6})^2} \approx 9.64m/s^2$$

Answer. $g = 9.64m/s^2$.