# Question #47a24

Nov 22, 2016

Let

$m \to \text{mass of satellite}$

$M \to \text{mass of Earth}$

$R \to \text{radius of Earth}$

$G \to \text{Gravitational constant}$

$T \to \text{time period of satellite}$

$\omega \to \text{angular speed of satellite} = \frac{2 \pi}{T}$

$g \to \text{acceleration due to gravity on }$
$\text{ earth surface}$

Equating weight $\left(m g\right)$ of satellite revolving round the earth along the orbit of radius nrearly equal to the radius $\left(R\right)$ of the earth with the gravitational pull on it, we get

$m g = \frac{G m M}{R} ^ 2$

$\implies G M = g {R}^{2.} \ldots \ldots . \left(1\right)$

The centripetal force (${F}_{c}$) acting on the satellite revolving round the Earth along the orbit having radius nearly equal to radius of Earth is given by

${F}_{c} = m {\omega}^{2} R$

The gravitaional force $\left({F}_{g}\right)$ acting on the satellite will provide the required centripetal force.

${F}_{g} = G \frac{m M}{R} ^ 2$

Now ${F}_{c} = {F}_{g}$

$\implies m {\omega}^{2} R = \frac{G m M}{R} ^ 2$

$\implies {\left(\frac{2 \pi}{T}\right)}^{2} R = \frac{G M}{R} ^ 2$

$\implies {T}^{2} = \frac{4 {\pi}^{2} {R}^{3}}{G M}$

$\implies T = 2 \pi \sqrt{{R}^{3} / \left(G M\right)}$

[using relation (1)]

$\implies T = 2 \pi \sqrt{{R}^{3} / \left(g {R}^{2}\right)}$

$\implies T = 2 \pi \sqrt{\frac{R}{g}}$