A satellite is in a circular Earth orbit of radius $\mathbf{r}$. The area $\mathbf{A}$ enclosed by the orbit depends on ${\mathbf{r}}^{\mathbf{2}}$ because $\mathbf{A}\mathbf{=}{\mathbf{\pi r}}^{\mathbf{2}}$. Determine how the following properties of the satellite depend on r:

(a) period,

(b) kinetic energy,

(c) angular momentum, and

(d) speed.

The satellite is orbiting around the earth with a radius $\mathrm{r}$

The area enclosed by the orbit $\text{A}={\text{πr}}^{\text{2}}$

Using gravitational force and centripetal force, we can find out the relation of r with T and speed. Using critical velocity in the kinetic energy equation, we can findtherelation between r and K. Usingthemoment of inertia and angular velocity; we will gettherelation between r and L.

Formula:

$\begin{array}{l}{\text{T}}^{\text{2}}\text{=}\frac{{\text{4π}}^{\text{2}}{\text{r}}^{\text{3}}}{\text{Gm}}\\ \text{K=}\frac{\text{GMm}}{\text{2}\left(\text{r}\right)}\\ \text{F=}\frac{\text{GMm}}{{\text{r}}^{\text{2}}}\end{array}$

$\begin{array}{l}\text{CentripetalForce=}\frac{{\text{mv}}^{\text{2}}}{\text{r}}\\ \frac{{\text{mv}}^{\text{2}}}{\text{r}}{\text{=mrω}}^{\text{2}}\\ \mathrm{L}=\mathrm{I\omega }\\ \mathrm{I}={\mathrm{mr}}^2\end{array}$

For the orbiting satellite in a circular orbit, centripetal force is provided by the gravitational force of the earth.

${\text{mrω}}^{\text{2}}=\frac{\text{GMm}}{{\text{r}}^{\text{2}}}$

${\text{ω}}^{\text{2}}=\frac{\text{GM}}{{\text{r}}^{\text{3}}}$ (1)

$\text{ω}=\frac{\text{2π}}{\text{T}}$ ……………………………(2)

Using this in equation 1 and adjusting variables,

$\begin{array}{l}{\text{T}}^{\text{2}}=\frac{{\text{4π}}^{\text{2}}{\text{r}}^{\text{3}}}{\text{Gm}}\\ {\text{T}}^{\text{2}}\propto {\mathrm{r}}^3\end{array}$

The velocity required to rotate the satellite in the orbit with radius r is the critical velocity

${\text{V}}_{\text{c}}\text{=}\sqrt{\frac{\text{GM}}{\text{r}}}$

Using this in the kinetic energy equation,

$\text{K}=\frac{\text{1}}{\text{2}}{{\text{mV}}_{\text{c}}}^{\text{2}}$

$\text{K}=\frac{\text{GMm}}{\text{2}\left(\text{r}\right)}$

$\text{K}\propto \frac{\text{1}}{\text{r}}$

To findtherelation between L and r:

Angular momentum is given by

$\text{L=Iω}$

I is the moment of inertia and is the angular velocity.

$\text{I}={\text{mr}}^{\text{2}}$

$\text{L}={\text{mr}}^{\text{2}}\text{ω}$

$\text{L}=\text{m}\frac{{\text{r}}^{\text{2}}\text{v}}{\text{r}}$

From part b),

${\text{V}}_{\text{c}}\propto \sqrt{\frac{\text{1}}{\text{r}}}$

$\text{L}\propto \frac{{\text{r}}^{\text{2}}\left(\sqrt{\frac{\text{1}}{\text{r}}}\right)}{\text{r}}$

$\text{L}\propto \sqrt{\text{r}}$

For the orbiting satellite in a circular orbit, centripetal force is provided bythegravitational force of the earth.

$\frac{{\text{mV}}_{\text{c}}^{\text{2}}}{\text{r}}=\frac{\text{GMm}}{{\text{r}}^{\text{2}}}$

$\begin{array}{l}{\text{V}}_{\text{c}}^{\text{2}}=\frac{\text{GM}}{\text{r}}\\ {\text{V}}_{\mathrm{c}}=\sqrt{\frac{\text{GM}}{\text{r}}}\\ {\text{V}}_{\mathrm{c}}\propto \frac{1}{\sqrt{\mathrm{r}}}\end{array}$

We can find the relationship of r with different quantities using Newton’s law of gravitation and the formula for the respective quantities.