Question

For a satellite revolving around earth in circular motion PE=X(KE) then X=​

Answer 1

Let mass of satellite be m , mass of earth be M , radius of earth be r .

So potential energy of the satellite flying just above the Earth surface is the product of gravitational potential at that height and the mass of satellite.

$PE = - G \dfrac{mM}{r}$

Let orbital velocity of satellite be v

We know that $v = \sqrt{\dfrac{GM}{r}}$

So kinetic energy will be :

$KE = \dfrac{1}{2} m {v}^{2}$

$= > KE = \dfrac{1}{2} m { \bigg(\sqrt{ \dfrac{GM}{r} } \bigg)}^{2}$

$=>KE= \dfrac{1}{2} \dfrac{GmM}{r}$

Comparing the equation , we can easily understand that X = -½.

So final answer :

$\boxed{ \red{ \huge{ \bold{X = - \dfrac{1}{2} }}}}$