If a satellite takes time T for revolution, the the kinetic energy is proportional to

The correct option is T⁻²/³.

To understand why the kinetic energy of a satellite is proportional to T⁻²/³, we can derive the relationship step-by-step using the principles of circular orbital motion.

First, consider a satellite of mass m revolving around a planet of mass M in a circular orbit of radius r. The gravitational force between the planet and the satellite provides the necessary centripetal force for its circular motion.
Therefore, we can write the equation of motion as:
$$\frac{m{v}^2}{r}=\frac{GMm}{r^2}$$ where:
- v is the orbital speed of the satellite,
- G is the universal gravitational constant, and
- r is the radius of the orbit.

Simplifying this equation by cancelling the mass of the satellite m and one factor of r from both sides, we get the expression for the square of the orbital velocity:
$$v^2=\frac{GM}{r}$$

The kinetic energy (K.E.) of the satellite is given by the formula:
$$\text{K.E.}=\frac{1}{2}m{v}^2$$ Substituting the value of $v^2$ into the kinetic energy equation, we obtain:
$$\text{K.E.}=\frac{GMm}{2r}$$ Since G, M, and m are constant values for a given satellite and planet system, we can see that:
$$\text{K.E.}\propto \frac{1}{r}$$

Next, we relate the orbital radius r to the time period of revolution T using Kepler's Third Law of Planetary Motion, which states that the square of the time period is proportional to the cube of the orbital radius:
$$T^2\propto r^3$$ Taking the cube root of both sides, we find the relationship for r:
$$r\propto T^{2/3}$$

Now, substituting this proportional relationship for r back into the expression for kinetic energy:
$$\text{K.E.}\propto \frac{1}{T^{2/3}}$$ Writing this fraction with a negative exponent, we get:
$$\text{K.E.}\propto T^{-2/3}$$ Thus, the kinetic energy of the satellite is proportional to T⁻²/³.