Question

A satellite is orbiting around the earth. In a particular orbit its time period is $ T $ and orbital speed is $ V $ . In another orbit the orbital speed is $ 2V $ , then time period will be
(A) $ 8T $
(B) $ 2T $
(C) $ T/2 $
(D) $ T/8 $

Answer 1

Hint: The orbital speed and the time period of a satellite in a particular orbit are given as, $ T \propto {V^{ - 3}} $ . So for the 2 cases we take the ratio of the time period and hence get the time period in the second case.

Formula Used In this solution we will be using the following formula,
 $ T \propto {V^{ - 3}} $
where $ T $ is the time period and $ V $ is the orbital velocity.

Complete Step by Step Solution
 In the problem we are given the time period and the orbital speed of a satellite in an orbit around the earth. The relation is given by,
 $ T \propto {V^{ - 3}} $
So in the first case let us consider the time period as $ T $ and the orbital speed as $ V $
Hence the equation remains as,
 $ T \propto {V^{ - 3}} $
In the second case let us consider the time period as $ T' $ and the orbital speed as $ V' $
So the relation in the second case will be,
 $ T' \propto {V'^{ - 3}} $
Now according to the question we can write,
 $ V' = 2V $
So substituting this value we get,
 $ T' \propto {\left( {2V} \right)^{ - 3}} $
Now we can take the ratio of the time period in the second case to the first case as,
 $ \dfrac{{T'}}{T} = \dfrac{{{{\left( {2V} \right)}^{ - 3}}}}{{{V^{ - 3}}}} $
Therefore, the $ V $ gets cancelled in this equation. So we get,
 $ \dfrac{{T'}}{T} = \dfrac{{{{\left( 2 \right)}^{ - 3}}}}{1} $
Therefore, the time period in the second case will be,
 $ T' = \dfrac{1}{{{2^3}}}T $
Hence the time period is, $ T' = \dfrac{T}{8} $

So the correct option is D.

Note:
The time period of a satellite in an orbit is given by,
 $ T = 2\pi \sqrt {\dfrac{{{r^3}}}{{GM}}} $
and the orbital speed is given by,
 $ V = \sqrt {\dfrac{{GM}}{r}} $
Hence we can write this as,
 $ V = {\left( {\dfrac{{GM}}{r}} \right)^{1/2}} $
Therefore, in the formula for the time period, we can multiply $ GM $ in the numerator and the denominator as,
 $ T = 2\pi \sqrt {\dfrac{{{r^3}}}{{GM}} \times \dfrac{{{{\left( {GM} \right)}^2}}}{{{{\left( {GM} \right)}^2}}}} $
Therefore, we can simplify this as,
 $ T = 2\pi GM\sqrt {\dfrac{{{r^3}}}{{{{\left( {GM} \right)}^3}}}} $
We can write the terms in the root as,
 $ T = 2\pi GM{\left[ {{{\left( {\dfrac{{GM}}{r}} \right)}^{1/2}}} \right]^{ - 3}} $
Therefore we can write this in terms of velocity as,
 $ T = 2\pi GM{V^{ - 3}} $
Since $ 2\pi GM $ are constant, hence
 $ T \propto {V^{ - 3}} $ .