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

The correct answer is T.⁻⁶⁶⁷.

To find how the kinetic energy of a satellite relates to its time period of revolution, we can analyze the orbital mechanics of a satellite of mass m revolving in a circular orbit of radius r around a planet of mass M.

The gravitational force between the planet and the satellite provides the centripetal force required for circular motion:

$\frac{GMm}{r^2}=\frac{m{v}^2}{r}$

where v is the orbital velocity of the satellite, and G is the universal gravitational constant. Simplifying this equation gives the orbital velocity squared:

$v^2=\frac{GM}{r}$

The kinetic energy (K) of the satellite is given by the formula:

$K=\frac{1}{2}m{v}^2$

Substituting the expression for v² into the kinetic energy formula, we get:

$K=\frac{GMm}{2r}$

Since G, M, and m are constants, the kinetic energy is inversely proportional to the orbital radius:

$K\propto \frac{1}{r}$

Next, we relate the orbital radius r to the time period of revolution T. The time period T is the time taken to complete one full orbit of circumference 2πr:

$T=\frac{2\pi r}{v}$

Squaring both sides of the equation:

$T^2=\frac{4{\pi }^2{r}^2}{v^2}$

Substitute v² = GM/r into the squared time period equation:

$T^2=\frac{4{\pi }^2{r}^2}{GM/r}=\frac{4{\pi }^2{r}^3}{GM}$

This shows that T² is proportional to r³ (Kepler's Third Law):

$T^2\propto r^3$

Solving for r, we find:

$r\propto T^{2/3}$

Now, we substitute this relationship back into the proportionality for kinetic energy:

$K\propto \frac{1}{T^{2/3}}=T^{-2/3}$

Since 2/3 is approximately 0.667, we can write:

$K\propto T^{-0.667}$

This corresponds to T.⁻⁶⁶⁷.