Question

The time period of revolution of a satellite close to earth is 90min. The time period of another satellite in an orbit at a distance of three times the radius of earth from its surface will be
A.$90\sqrt{8}\min $
B. $360\min $
C.$270\sqrt{3}\min $
D. $720\min $

Answer 1

Hint: Firstly, you could recall Kepler’s law of period in the case of planetary motion. Consider the given question in two different parts one where the orbit is approximately equal to earth’s radius and other where the orbit is at a distance of three times the earth’s radius from earth surface. Now by applying Kepler’s law to both cases and combining both and then substituting accordingly will give you the required time period.

Formula used:
Kepler’s law of period,
${{T}^{2}}\propto {{a}^{3}}$

Complete step-by-step answer:
In the question, we are given the time period of revolution of a satellite close to Earth as 90min. We are asked to find the time period of another satellite that is orbiting at a distance of three times the radius of earth from Earth’s surface.
In order to answer the question, let us recall Kepler’s third law of planetary motion which states that the square of the time period of the planet is directly proportional to the cube of the semi-major axis of its orbit. That is,
${{T}^{2}}\propto {{a}^{3}}$ ……………………………………. (1)
Where, T is the time period and$a$is the semi-major axis of the orbit.
This law could be applied to the situation given in the above question. On doing so, T will be the time period of the satellite and $a$ will be the radius of the circular orbit.
Now for the first case, the radius of the circular orbit will be approximately proportional to the radius of earth (R), so,
${{T}_{1}}^{2}\propto {{R}^{3}}$
$\Rightarrow {{T}_{1}}\propto {{R}^{\dfrac{3}{2}}}$ …………………………………… (2)
Now for the second case where the orbit is at a distance of three times the radius of earth,
$a=R+3R=4R$
Since, $a$ is measured from the centre of earth.
So, (1) will become,
${{T}_{2}}\propto {{\left( 4R \right)}^{\dfrac{3}{2}}}$ ……………………………………. (3)
Combining (2) and (3),
$\dfrac{{{T}_{1}}}{{{T}_{2}}}={{\left( \dfrac{R}{4R} \right)}^{\dfrac{3}{2}}}$
But we are given the value of ${{T}_{1}}$ as,
${{T}_{1}}=90\min $
$\Rightarrow \dfrac{90}{{{T}_{2}}}={{\left( \dfrac{1}{4} \right)}^{\dfrac{3}{2}}}$
$\Rightarrow {{T}_{2}}=8\times 90\min $
$\therefore {{T}_{2}}=720\min $
Therefore, the time period of the satellite in the orbit at a distance of three times the radius of earth from its surface will be 720min. Hence, option D is the correct answer.

So, the correct answer is “Option D”.

Note: You may have noted that while stating the law we have defined $a$as the semi-major axis of the planet’s orbit. This is because the law was based on our solar system where the orbits are elliptical. But here in the given question the orbits are circular and hence we have substituted for $a$ accordingly.