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The time period of a satellite in a circular orbit of radius R is T, the period of another satellite in a circular orbit of radius $4R$ is-
(A) $4T$
(B) $T/4$
(C) $8T$
(D) $T/8$

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Last updated date: 13th Jun 2024
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Answer
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Hint Kepler’s laws of planetary motion define the orbit and motion of a planet around its star. This law holds true for any object orbiting the other due to gravitational force between them, thus it is also valid for a satellite that orbits the earth. To determine the time period of a satellite orbiting earth, Kepler’s third law is used.

Complete Step by step solution
The time period of a satellite is given by-
$\Rightarrow$ $t = \dfrac{{2\pi }}{r}\sqrt {\dfrac{{{{\left( {r + h} \right)}^3}}}{g}} $
Here, $t$ is the time taken by the satellite to complete one revolution around the earth.
$r$ is the radius of the earth (or any other planet around which the satellite revolves.
$h$ is the height at which the satellite is situated, taken perpendicular from the surface of the planet.
$g$ is the acceleration due to gravity.
In the question, it is given that,
The orbit of the first satellite is$R$ while the orbit of the second satellite is $4R$
Orbit is defined as the distance of the satellite from the center of the planet,
Therefore, if $r$is the radius of earth then,
For the first satellite,
$\Rightarrow$ $r + {h_1} = R$
For the second satellite,
$\Rightarrow$ $r + {h_2} = 4R$
We know that in the formula for the time period, all the terms other than the orbit of the satellite and the time period are constant and can be replaced with a constant value.
Thus,
$\Rightarrow$ $t = C{(r + h)^{3/2}}$
$t \propto {(r + h)^{3/2}}$
This relationship between the orbit and the time period was also given by Kepler in his laws of planetary motion.
Thus the orbits of both satellites can be represented by-
$\dfrac{{{T_1}}}{{{T_2}}} = {\left( {\dfrac{R}{{4R}}} \right)^{3/2}}$
$\dfrac{{{T_1}}}{{{T_2}}} = {\left( {\dfrac{1}{4}} \right)^{3/2}}$
On solving the RHS of the equation we get,
$\dfrac{{{T_1}}}{{{T_2}}} = \left( {\dfrac{1}{8}} \right)$
${T_2} = 8{T_1}$
Therefore the time period of the second satellite is $8$times more than that of the first satellite.

The correct option is (C).

Note Kepler’s three laws of planetary motion define that-
1. The orbit of a planet is elliptical, not circular.
2. A line that joins the planet and the sun sweeps equal areas in equal intervals of time.
3. The square of the time period of revolution is directly proportional to the cube of its mean orbit.