Question

In a satellite , if the time of revolution is T, then KE is proportional to

& \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{aligned}$$

Answer 1

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,

& T\propto {{r}^{\dfrac{3}{2}}} \\\ & \Rightarrow r\propto {{T}^{\dfrac{2}{3}}}............(i) \\\ \end{aligned}$$ Now, from kinetic energy formula we get, $$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.