Chapter 14, Problem 45A

The force of gravity between Earth and an Earth satellite is given by $F=G\frac{mM}{d^2}$, where m is the mass of the satellite, M is the mass, of Earth, and d is the distance between the satellite and the center of Earth. If the satellite follows a circular orbit, the force keeping it in orbit must, be the centripetal force, given by $F=\frac{m{v}^2}{r}$. Equate the two expressions for force to show that the speed $v-=\sqrt{\frac{GM}{d}}$.

To determine

To show:The speed of the satellite is $v=\sqrt{\frac{GM}{d}}$ .

Explanation of Solution

Given:

If the satellite follows circular orbit, then the centripetal force is

  $F=\frac{m{v}^2}{r}$

Here, $v$ is the velocity of the satellite and $r$ is the radius of the orbit.

Formula used:

The Newton’s equation for gravitational force is,

  $F_{\text{g}}=G\frac{mM}{d}$

Here, $m$ is the mass of the satellite, $M$ is the mass of the Earth, $G$ is the gravitational constant, and $d$ is the distance between the satellite and the centre of the Earth.

Calculation:

Consider a satellite is orbiting at a radius of $r$ from the centre of the Earth. The expression for the gravitation force acting on the satellite is,

  $F_{\text{g}}=G\frac{mM}{d}$ ….. (1)

The expression for the centripetal force that keeps the satellite in the orbit is,

  $F=\frac{m{v}^2}{r}$ ….. (2)

The radius of the orbit $\left(r\right)$ and the distance $\left(d\right)$ between the satellite and the centre of the Earth are the same.Replace $r$ by $d$ in Equation (2).

  $F=\frac{m{v}^2}{d}$ ….. (3)

Equate equation (1) and (3).

  $\begin{array}{l}F=F_{\text{g}}\\ \frac{m{v}^2}{d}=G\frac{mM}{d}\\ v^2=G\frac{M}{d}\\ v=\sqrt{\frac{GM}{d}}\end{array}$

Conclusion:

Thus, the speed of the satellite is shown as $v=\sqrt{\frac{GM}{d}}$ .