Question

An earth's satellite near the surface of the earth takes about 90 min per revolution. A satellite orbiting the moon also takes about 90 min per revolution. Then which of the following is true? (\(\rho\)_m is the density of the moon and ρe is the density of the earth).

• A. \(\rho_m<\rho_e\)
• B. \(\rho_m>\rho_e\)
• C. \(\rho_m=\rho_e\)
• D. No conclusion can be made about the densities

Answer 1

The Correct Option is C

The time period of a satellite in orbit depends on the mass of the celestial body it is orbiting and the radius of the orbit.

The formula for the time period is: 

$T = 2\pi \sqrt{\frac{a^3}{GM}}$

Squaring both sides:

$T^2 = \frac{4\pi^2 a^3}{GM}$

Rearranging gives:

$\frac{a^3}{M} = \text{constant}$

Since both satellites (one orbiting Earth and the other orbiting the Moon) have the same time period, we get:

$\frac{a_e^3}{M_e} = \frac{a_m^3}{M_m}$

We know that mass $M$ of a spherical body is related to density $\rho$ by:

$M = \rho \cdot \frac{4}{3} \pi R^3$

Assuming the satellites are close to the surfaces, we take $a \approx R$, so:

$\frac{M}{a^3} \approx \rho \cdot \frac{4}{3} \pi$

This implies:

$\frac{M_e}{a_e^3} = \frac{M_m}{a_m^3} \Rightarrow \rho_e = \rho_m$

Therefore, the correct option is (C): $\rho_m = \rho_e$