Question

In a satellite , if the time of revolution is T, then KE is proportional to
\[\begin{align}
  & \text{A}\text{. }\dfrac{1}{T} \\
 & \text{B}\text{. }\dfrac{1}{{{T}^{2}}} \\
 & \text{C}\text{. }\dfrac{1}{{{T}^{3}}} \\
 & \text{D}\text{. }{{T}^{-\dfrac{2}{3}}} \\
\end{align}\]

Answer 1

Hint: Kepler's laws of motion for planets are three scientific laws describing the motion of planets around the Sun. Kepler’s third law has improved the model of Copernicus.

Formula used: Kepler’s third law: \[T\propto {{r}^{\dfrac{3}{2}}}\] where, T = the time period and r = distance and, satellitic kinetic energy \[K=\dfrac{GMm}{2r}\], where M = mass of earth, m= mass of satellite, G = gravitational constant.

Complete step by step solution:
We have to use Kepler’s third law which is given by,
\[\begin{align}
  & T\propto {{r}^{\dfrac{3}{2}}} \\
 & \Rightarrow r\propto {{T}^{\dfrac{2}{3}}}............(i) \\
\end{align}\]
Now, from kinetic energy formula we get,
\[K=\dfrac{GMm}{2r}\] which means \[K\propto \dfrac{1}{r}...................(ii)\]
Substituting, equation (i) in (ii) we get,
\[K\propto \dfrac{1}{{{T}^{\dfrac{2}{3}}}}\propto {{T}^{-\dfrac{2}{3}}}\]

So, the answer is option D.

Note: Kepler's laws of motion for planets is the concept dealing with the distance of planets from the sun and their orbital periods.