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# The time period of an artificial satellite in a circular orbit of radius R is T. The radius of the orbit in which time period is 8T is(A) 2R(B) 3R(C) 4R(D) 5R

Last updated date: 14th Aug 2024
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Answer
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Hint: According to Kepler’s third law, the square of the time period of revolution of a planet around the sun is directly proportional to the cube of the semi-major axis of the orbit. Which can be expressed as ${T^2} \propto {R^3}$. We will use this law and compare the situation to find the radius when the time period is 8T.

Complete step by step answer:
It is given in the question that the time period of an artificial satellite in a circular orbit of radius R is T. Then we have to find the radius of the orbit in which the time period is 8T.
According to Kepler’s third law, the square of the time period of revolution of a planet around the sun is directly proportional to the cube of the semi-major axis of the orbit. Which can be expressed as ${T^2} \propto {R^3}$. (Here, the orbit is assumed to be circular)
Here, T is the time period of the revolution of the planet around the sun but we will consider it as the time period of the revolution of the artificial satellite around the planet, R is the radius of the orbit.
On comparing both the situation when the time period is T and 8T we get-
$\dfrac{{{T^2}}}{{{{(8T)}^2}}} = \dfrac{{{{(R)}^3}}}{{{{\left( x \right)}^3}}}$
On cross-multiplying both the sides we get-
${T^2}{\left( x \right)^3} = {(R)^3}{(8T)^2}$
${\left( x \right)^3} = \dfrac{{{{(R)}^3}{{(8T)}^2}}}{{{T^2}}}$
$\left( x \right) = \sqrt[3]{{\dfrac{{{{(R)}^3}{{(8T)}^2}}}{{{T^2}}}}}$
$\left( x \right) = \sqrt[3]{{64{{(R)}^3}}}$
x = 4R
Thus, the time period of the artificial satellite having length x = 4R will take 8T.

Therefore, option C is correct.

Additional information:
A satellite is an object in space that revolves around a bigger object. There are two types of satellites, the first is the natural satellite and the second is an artificial satellite. Moon is a natural satellite whereas the satellite used in our communication, GPS is an artificial satellite.

Note:
One can do a mistake in the last step of the calculation they may find the cubic of the $8{R^3}$which is 2R and tick option A as their response but it is Note:d that we have to find the cubic root of ${8^2}{R^3}$ which is equal to 4R.