Answer
385.5k+ views
Hint:Here given that the satellite is revolving on a circular radius around the earth having a radius of $4R$, hence to find the time period of the satellite in which it completes its revolution the concept of Kepler’s law is used. Here the Kepler’s third law should be used to evaluate the time period of the satellite.
Complete step by step answer:
As discussed this question can be solved and the time period can be evaluated by using Kepler’s third law. Kepler’s third law states that the square of the time period given as ${T^2}$ of the revolution of the planet around the sun is directly proportional to the cube of the semi-major axis of the orbit that is the radius of the orbit of the planet which can be assumed as circular. Hence the relation between the time period of the planet $T$ and the radius $R$ of the orbit of the revolution of the planet can be given as
$\begin{array}{*{20}{c}}
{{T^2}}& \propto &{{R^3}}
\end{array}$ ……….. $(1)$
Now as it is given that the radius becomes $4R$ which is four times from the radius of another satellite hence substituting this in equation $(1)$ results in
$\begin{array}{*{20}{c}}
{{T^2}}& \propto &{{{\left( {4R} \right)}^3}}
\end{array}$
$\begin{array}{*{20}{c}}
{ \Rightarrow {T^2}}& \propto &{64{R^3}}
\end{array}$
On comparing the values on both sides as the square of the time period is proportional to the cube of the radius. Therefore the time period would be given as
$\begin{array}{*{20}{c}}
{64{T^2}}& \propto &{64{R^3}}
\end{array}$
$\begin{array}{*{20}{c}}
{ \Rightarrow {{\left( {8T} \right)}^2}}& \propto &{{{\left( {4R} \right)}^3}}
\end{array}$
Hence the time period of the satellite when the radius of the orbit becomes $4R$ is given $8T$.
Therefore option C is the correct answer.
Note:There are three Kepler's law that governs the planetary motions. The first Kepler's law states that the planets move around the sun in an elliptical orbit. The second law states that planets move around the sun as its foci and the third law as we had already discussed which is used to evaluate the time period of the satellite for the given radius.
Complete step by step answer:
As discussed this question can be solved and the time period can be evaluated by using Kepler’s third law. Kepler’s third law states that the square of the time period given as ${T^2}$ of the revolution of the planet around the sun is directly proportional to the cube of the semi-major axis of the orbit that is the radius of the orbit of the planet which can be assumed as circular. Hence the relation between the time period of the planet $T$ and the radius $R$ of the orbit of the revolution of the planet can be given as
$\begin{array}{*{20}{c}}
{{T^2}}& \propto &{{R^3}}
\end{array}$ ……….. $(1)$
Now as it is given that the radius becomes $4R$ which is four times from the radius of another satellite hence substituting this in equation $(1)$ results in
$\begin{array}{*{20}{c}}
{{T^2}}& \propto &{{{\left( {4R} \right)}^3}}
\end{array}$
$\begin{array}{*{20}{c}}
{ \Rightarrow {T^2}}& \propto &{64{R^3}}
\end{array}$
On comparing the values on both sides as the square of the time period is proportional to the cube of the radius. Therefore the time period would be given as
$\begin{array}{*{20}{c}}
{64{T^2}}& \propto &{64{R^3}}
\end{array}$
$\begin{array}{*{20}{c}}
{ \Rightarrow {{\left( {8T} \right)}^2}}& \propto &{{{\left( {4R} \right)}^3}}
\end{array}$
Hence the time period of the satellite when the radius of the orbit becomes $4R$ is given $8T$.
Therefore option C is the correct answer.
Note:There are three Kepler's law that governs the planetary motions. The first Kepler's law states that the planets move around the sun in an elliptical orbit. The second law states that planets move around the sun as its foci and the third law as we had already discussed which is used to evaluate the time period of the satellite for the given radius.
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