Two satellites P and Q are moving in different circular orbits around the Earth (radius $$R$$). The heights of P and Q from the Earth surface are $$h_P$$ and $$h_Q$$, respectively, where $$h_P = R/3$$. The accelerations of P and Q due to Earth's gravity are $$g_P$$ and $$g_Q$$, respectively. If $$g_P/g_Q = 36/25$$, what is the value of $$h_Q$$?

The acceleration due to gravity at a distance $$r$$ from the centre of the Earth is

$$g(r)=\frac{GM}{r^{2}}$$

where $$G$$ is the gravitational constant and $$M$$ is the mass of the Earth.

For a satellite at height $$h$$ above the Earth’s surface (Earth’s radius $$R$$), the distance from the centre of the Earth is

$$r = R + h$$

Hence, for satellites P and Q we have

$$g_P = \frac{GM}{(R+h_P)^{2}}, \qquad g_Q = \frac{GM}{(R+h_Q)^{2}}$$

Taking their ratio, the common factor $$GM$$ cancels:

$$\frac{g_P}{g_Q} = \frac{(R+h_Q)^{2}}{(R+h_P)^{2}}$$ $$-(1)$$

Given data:

• $$h_P = \frac{R}{3} \;\;\Rightarrow\;\; R+h_P = R + \frac{R}{3} = \frac{4R}{3}$$
• $$\dfrac{g_P}{g_Q} = \dfrac{36}{25}$$

Substitute in equation $$(1)$$:

$$\frac{36}{25} = \frac{(R+h_Q)^{2}}{\left(\dfrac{4R}{3}\right)^{2}}$$

Solve for $$R+h_Q$$:

$$R+h_Q = \frac{6}{5}\left(\frac{4R}{3}\right) = \frac{24R}{15} = \frac{8R}{5}$$

Therefore,

$$h_Q = \frac{8R}{5} - R = \frac{8R}{5} - \frac{5R}{5} = \frac{3R}{5}$$

Hence, the height of satellite Q above the Earth’s surface is $$\displaystyle \frac{3R}{5}$$.

Option A which is: $$3R/5$$