Answer

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**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}} $ .

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