Question

Two satellites of earth ${s_1}$ and ${s_2}$ are moving in the same orbit. The mass of ${s_1}$ is four times the mass of ${s_2}$. Which of the following is correct?
A) The time-period of ${s_1}$ is four times that of ${s_2}$.
B) The potential energies of earth and satellite in the two cases are equal.
C) ${s_1}$ and ${s_2}$ are moving at the same speed.
D) The kinetic energies of the two satellites are equal.

Answer 1

In the first step, we have to first learn about the cause of the earth’s satellites’ motion. Then we can determine the kinetic energy, potential energy, and the time period of the earth’s satellites. Finally, we will compare these quantities for the two satellites, and we will get our answer.

Formula used:
$F = G.\dfrac{{M.m}}{{{r^2}}}$,
${F_C} = \dfrac{{m.{v^2}}}{r}$, &
$U = - \dfrac{{G.M.m}}{r}$
The gravitational force of the earth is-
$F = G.\dfrac{{M.m}}{{{r^2}}}$
$M$ is the mass of the earth
$m$ is the mass of the satellite
$r$ is the distance of the satellite from the earth’s center
and the centripetal force on the satellite-
${F_C} = \dfrac{{m.{v^2}}}{r}$
$v$ is the orbital speed of the satellite.

Complete step by step solution:
The motion of the earth’s satellite around the earth is a combined action of the earth’s gravitational pull towards the satellite and the centripetal force acting on the satellite. The two forces balance each other.
So we have,
$\Rightarrow \dfrac{{m.{v^2}}}{r} = G.\dfrac{{M.m}}{{{r^2}}}$
$\Rightarrow {v^2} = \dfrac{{G.M}}{r}$
$\Rightarrow v = \sqrt {\dfrac{{G.M}}{r}}$……….$(1)$
Hence, it is clear that the satellites’ speeds do not depend on their masses; instead, they depend on the radius of their radius.
The potential energy of a satellite is given as-
$\Rightarrow U = - \dfrac{{G.M.m}}{r}$
So, the potential energy of a satellite does depend on its mass. Therefore, it is also not identical for the given two satellites.
The kinetic energy of a satellite can be derived from $(1)$-
$\Rightarrow v = \sqrt {\dfrac{{G.M}}{r}}$
Squaring both sides-
$\Rightarrow {v^2} = \dfrac{{G.M}}{r}$
We multiply $\dfrac{m}{2}$ both sides-
$\Rightarrow \dfrac{1}{2}m{v^2} = \dfrac{{G.M.m}}{{2r}}$
$\therefore {K_E} = \dfrac{{G.M.m}}{{2r}}$
So we can see it is also dependable on the mass of the satellite. Hence it is also not identical for the given two satellites.
The time period of a satellite is-
$\Rightarrow T = \sqrt {\dfrac{{G.M}}{{{r^3}}}}$
Therefore the two satellites will have the same time period.

Note: Unlike the centrifugal force, the centripetal force is not a pseudo force. In a uniform circular motion, the object in that motion experiences a net inward force. The object that experiences a centripetal force can never alter the direction of movement in orbit.