Question

Let kinetic energy of a satellite be $X$. Then its time of revolution T is proportional to:
A. $X^{-3}$
B. $X^{-\;\dfrac{3}{2}}$
C. $X^{-1}$
D. $\sqrt{X}$

Answer 1

Hint: Try to think of the forces that a satellite experiences when its orbiting around a planet. In other words, it experiences both centripetal and gravitational forces that are equivalent to each other. And since it is a moving body, use the most general expression of kinetic energy to obtain its orbital velocity. Then using the distance travelled as the perimeter of the circular orbit obtains an expression from the time period.

Formula Used:
Centripetal force: $F_{centripetal} = \dfrac{m\;v^2}{r}$, where m is satellite mass, v is its velocity and r is the radius of its orbit.
Gravitational force: $F_{gravitational} = \dfrac{G\;M\;m}{r^2}$ , where G is the gravitational constant, M is the mass of the body around which satellite is orbiting, m is the satellite mass and r is the distance to the satellite from orbiting centre.
In general for a circular orbit, time period $T = \dfrac{Perimeter}{Velocity} = \dfrac{2 \pi r}{v}$

Complete step by step answer:
Here, we are required to establish a relation between the period of revolution $T$ of the satellite and its kinetic energy $X$.
In general, the kinetic energy of a moving body is given as $\dfrac{1}{2}mv^2$. Let’s say :
$X = \dfrac{1}{2}mv^2 \Rightarrow v^2 = \dfrac{2\;X}{m} $ and $v =\sqrt{\dfrac{2\;X}{m}}$
Now, for a satellite orbiting a planet, it experiences a centripetal force that is equivalent to the gravitational force it experiences i.e.,
$F_{centripetal} = F_{gravitational} \Rightarrow \dfrac{m\;v^2}{r} = \dfrac{G\;M\;m}{r^2}$
$\Rightarrow v^2 = \dfrac{G\;M}{r}$
Equating the two expressions we got for v^2 we obtain an expression for r i.e.,
$\dfrac{2\;X}{m} = \dfrac{G\;M}{r} \Rightarrow r = \dfrac{G\;M\;m}{2\;X}$
Now, we know that for a circular orbit, the time period can be given as:
$T = \dfrac{Perimeter}{Velocity} = \dfrac{2 \pi r}{v}$
Substituting the equations we got for r and v from above, we get:
$T = 2 \pi \times \dfrac{\dfrac{G\;M\;m}{2\;X}}{\sqrt{\dfrac{2\;X}{m}}} $
$\Rightarrow T = (\pi \dfrac{G\;M}{\sqrt{2}}m^{\dfrac{3}{2}}) \times \dfrac{1}{\sqrt{X^3}}$
$\Rightarrow T \propto X^{-\;\dfrac{3}{2}}$
Therefore, the correct option would be: B. $X^{-\;\dfrac{3}{2}}$

Note:
Always remember that for a body to maintain a stable circular orbit the centripetal forces and the gravitational forces it experiences should be equivalent, since both the forces are essential to keep the satellite in orbit. And we consider the perimeter of a circle since the satellite traces a circular path.