Answer to Question #174312 in Astronomy | Astrophysics for Devadharshan

Question

Acceleration due to gravity at the earth s surface is 9.8ms-2At what height above the earth is the g equal to 9.8/4ms-2

Answer 1

The acceleration due to gravity at the Earth's surface can be written as follows:

$$g_1=\dfrac{GM}{R^2}.$$

The acceleration due to gravity at the height $h$ above the Earth can be written as follows:

$$g_2=\dfrac{GM}{(R+h)^2}.$$

Let's divide $g_1$by $g_2$:

$$\dfrac{g_1}{g_2}=\dfrac{(R+h)^2}{R^2},$$ $$\sqrt{\dfrac{g_1}{g_2}}=\dfrac{(R+h)}{R},$$ $$h=R(\sqrt{\dfrac{g_1}{g_2}}-1),$$ $$h=6.37\cdot10^6\ m\cdot(\sqrt{\dfrac{9.8\ \dfrac{m}{s^2}}{2.45\ \dfrac{m}{s^2}}}-1)=6.37\cdot10^6\ m.$$