Answer:
[tex]K = \frac{GMm}{2r}[/tex]
[tex]T^2 = 4\pi^2(\frac{r^3}{GM})[/tex]
Explanation:
As we know that for a satellite the force of gravitation is equal to the centripetal force
so we will have
[tex]F = \frac{GMm}{r^2}[/tex]
[tex]\frac{mv^2}{r} = \frac{GMm}{r^2}[/tex]
so we know that kinetic energy is given as
[tex]K = \frac{1}{2}mv^2[/tex]
so we have
[tex]K = \frac{GMm}{2r}[/tex]
now for time period we know
[tex]T = \frac{2\pi r}{v}[/tex]
from above expression of kinetic energy we have
[tex]v = \sqrt{\frac{GM}{r}}[/tex]
so we have
[tex]T = \frac{2\pi r}{\sqrt{\frac{GM}{r}}}[/tex]
so square of time period is given as
[tex]T^2 = 4\pi^2(\frac{r^3}{GM})[/tex]
