Question

Show that the square of the period of revolution of a satellite is directly proportional to the cube of the orbital radius

Answer 1

we know,
orbital velocity of satellite is given by
$v=\sqrt{\frac{GM}{r}}$
where G is gravitational constant, M is mass of earth and r is the radius of circular orbit.
but we know, when a particle moves in circular path of radius r, then orbital velocity of the particle is given by , $\frac{2\pi r}{T}$ , here T is time period.

$\sqrt{\frac{GM}{r}}=\frac{2\pi r}{T}$

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

squaring both sides,

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

here it is clear that $T^2\propto r^3$
e.g., the square of the period of revolution of a satellite is directly proportional to the cube of the orbital radius