Question

A satellite is launched into a circular orbit of radius R around the earth. A second satellite is launched into an orbit of radius $1.01R$. The period of the second satellite is longer than that of the first by approximately:
A) $0.5\% $
B) $1.0\% $
C) $1.5\% $
D) $3.0\% $

Answer 1

Hint: Recall that if the time period of satellites or planetary motion is to be studied then Kepler’s Laws of planetary motion are used. As per Kepler’s law of time period, the square of the time period of the planet or satellite is directly proportional to the cube of the semi major axis or radius of the orbit.

Complete step by step solution:
Step I: As per Kepler’s law, the time period of a satellite is given by
\[ \Rightarrow T = \dfrac{{2\pi R}}{{\sqrt {\dfrac{{GM}}{R}} }}\]
Or $T \propto {R^{\dfrac{3}{2}}}$---(i)
Where R is the radius of Earth
Step II: When the first satellite is launched, the radius of satellite around Earth is R. So the time period is given by
$\Rightarrow {T_1} \propto {R^{\dfrac{3}{2}}}$
$\Rightarrow {T_1} = {R^{\dfrac{3}{2}}}$---(ii)
Step III: When the second satellite is launched its radius will be 1.01R. This means that for the second satellite the radius is increased by 0.1R of the original radius. So the time period will be given by
$\Rightarrow {T_2} \propto {(R + 0.01R)^{\dfrac{3}{2}}}$
Or ${T_2} = {R^{\dfrac{3}{2}}}{\left[ {1 + \dfrac{1}{{100}}} \right]^{\dfrac{3}{2}}}$
Using Binomial theorem, the above equation can be written as
$\Rightarrow {T_2} = {R^{\dfrac{3}{2}}}[1 + \dfrac{3}{2} \times \dfrac{1}{{100}}]$---(iii)
Step IV: Substituting the value of R from equation (ii) to equation (iii),
$ \Rightarrow {T_2} = {T_1}[1 + \dfrac{3}{2} \times \dfrac{1}{{100}}]$
$ \Rightarrow {T_2} = {T_1} + \dfrac{3}{{200}}{T_1}$
$ \Rightarrow {T_2} - {T_1} = \dfrac{3}{{200}}{T_1}$
$ \Rightarrow \dfrac{{{T_2} - {T_1}}}{{{T_1}}} = \dfrac{3}{2} \times \dfrac{1}{{100}}$
$ \Rightarrow \dfrac{{{T_2} - {T_1}}}{{{T_1}}} = 1.5\% $
Step V: It is clear that the period of the second satellite is longer than that of the first by approximately $1.5\% $.

Option C is the right answer.

Note: It is important to remember that the Kepler’s law stated above is also known as the law of periods. It is the third law of Kepler. This law explains that if the orbit of the satellite around the planet is shorter, then it will take less time to complete one revolution.