Time Period And Energy Of A Satellite

Satellites, whether natural like the Moon or man-made, orbit Earth due to a delicate balance of gravitational pull and velocity. The time period of a satellite, which is the time it takes to complete one full orbit, is a fundamental concept in understanding its motion. This period is influenced by the satellite's altitude and the mass of the Earth. The energy of a satellite, comprising both kinetic and potential energy, determines its speed and orbit stability. In real life, these principles are crucial for the operation of GPS satellites that guide our navigation systems, communications satellites that enable global connectivity, and weather satellites that monitor Earth's climate. Understanding the time period and energy of satellites helps engineers design efficient and reliable systems that are integral to modern technology, ensuring everything from accurate weather forecasts to uninterrupted phone calls.

This Story also Contains

1. The Time Period of the Satellite
2. Height of Satellite
3. The Energy of Satellite
4. Binding Energy (B.E.)
5. Solved Examples Based on Time Period And Energy of a Satellite

The Time Period of the Satellite

The time period of a satellite is the duration it takes for the satellite to complete one full orbit around a celestial body, such as Earth. This time period is determined by the satellite's altitude and the mass of the celestial body it orbits. According to Kepler's laws of planetary motion, the time period is directly related to the size of the orbit—the higher the satellite, the longer the time period.

The time period (T) of the satellite is given by
$\begin{array}{ll}T=\frac{2\pi r}{v}=2\pi r\sqrt{\frac{r}{GM}}& [\text{As}v=\sqrt{\frac{GM}{r}}]\\ T=2\pi \sqrt{\frac{r^3}{GM}}=2\pi \sqrt{\frac{r^3}{g{R}^2}}& [\text{As}GM=g{R}^2]\\ T=2\pi \sqrt{\frac{(R+h{)}^3}{g{R}^2}}=2\pi \sqrt{\frac{R}{g}}{(1+\frac{h}{R})}^{3/2}& [\text{As}r=R+h]\end{array}$
Where
$r=$ radius of orbit
$T\to$ Time period
$M\to$ Mass of planet
If the satellite is very close to the Earth's surface,
i.e., $h\ll \ll R$,

$\begin{array}{rl}& T=2\pi \sqrt{\frac{R}{g}}\cong 84.6\text{minutes}\\ & \text{then}\\ & \text{or}T\simeq 1.4hr\end{array}$

The time period of a satellite in terms of density
$T=\sqrt{\frac{3\pi }{G\rho }}$

$\rho \to$ Density of planet
$T\to$ Time period
$G\to$ Gravitational constant

$\rho =5478.4\mathrm{Kg}/{\mathrm{m}}_{\text{for earth}}^3$

For a satellite, the time interval between the two consecutive appearances overhead

If a satellite in the equilateral planes moves from west to east Angular velocity of the satellite with respect to an observer on earth will be $({\omega }_S-{\omega }_E)$
${\omega }_S\to$ Satellite angular velocity
${\omega }_E\to$ earth angular velocity
So $T=\frac{2\pi }{{\omega }_S-{\omega }_E}=\frac{T_S{T}_E}{T_E-T_S}$
if ${\omega }_S={\omega }_E,T=\mathrm{\infty }$

means the satellite will appear stationary relative to Earth.

Height of Satellite

As we know, time period of satellite $T=2\pi \sqrt{\frac{r^3}{GM}}=2\pi \sqrt{\frac{(R+h{)}^3}{g{R}^2}}$ By squaring and rearranging both sides $\frac{q{R}^2{T}^2}{4{\pi }^2}=(R+h{)}^3$ $\Rightarrow h={\left(\frac{T^{2}g{R}^2}{4{\pi }^2}\right)}^{1/3}-R$

Putting the value of the period in the above formula we can calculate the height of the satellite from the surface of the earth.

The Energy of Satellite

When a satellite revolves around a planet in its orbit, it possesses both kinetic energy (due to orbital motion) and potential energy (due to its position against the gravitational pull of Earth). These energies are given by

Potential energy: $U=mV=\frac{-GMm}{GMm}=\frac{-L^2}{m{r}^2}$
Kinetic energy : $K=\frac{1}{2}m{v}^2=\frac{GMm}{2r}=\frac{m{L}^2}{2m{r}^2}$
Total energy : $E=U+K=\frac{-GMm}{r}+\frac{(2Mm}{2r}=\frac{-GMm}{2r}=\frac{-L^2}{2m{r}^2}$
Where

$\begin{array}{rl}& M\to \text{mass of planet}\\ & m\to \text{mass of satellite}\end{array}$
And

$\begin{array}{rl}& K=-E\\ & U=2E\\ & U=-2K\end{array}$

Energy Graph of Satellite

Where

$E\to$ Energy of satellite
$K\to$ Kinetic energy
$U\to$ Potential energy

Energy Distribution in an Elliptical Orbit

In this Total Energy
$E=-\frac{GMm}{2a}=\text{const}$
Where $a=$ semi-major axis

• Binding Energy (B.E.)

The minimum energy required to remove the satellite from its orbit to infinity is called Binding Energy.

And It is given by
$B\cdot E=\frac{GMm}{2r}$

where
B.E Binding energy
$M\to$ mass of planet
$m\to$ mass of satellite

Work done in changing the orbit

When the satellite is transferred to a higher orbit i.e. $(r_2>r_1)$ as shown in the figure.

$\begin{array}{rl}W& =E_2-E_1\\ W& =\frac{GMm}{2}[\frac{1}{r_1}-\frac{1}{r_2}]\end{array}$
Where
$W\to$ work done
$r_1\to$ radius of 1st orbit
$r_2\to$ radius of $2nd$ orbit

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Solved Examples Based on Time Period And Energy of a Satellite

Example 1: A very long (length L) cylindrical galaxy is made of uniformly distributed mass and has a radius R (R<<L). A star outside the galaxy is orbiting the galaxy in a plane perpendicular to the galaxy and passing through its centre. If the period of the star is T and its distance from the Galaxies axis is r, then :

1) $T^2\alpha r^3$
2) $T\alpha r^2$
3) $T\alpha r$
4) $T\alpha \sqrt{r}$

Solution:

The period of satellite
$\begin{array}{rl}& T=2\pi \sqrt{\frac{r^3}{GM}}\\ & r=\text{radius of orbit}\\ & T\to \text{Period}\\ & M\to \text{Mass of planet}\\ & T=\frac{\text{Circum ference of orbit}}{\text{orbital velocity}}\\ & F=\frac{2GM}{Lr}m,F=\left(\frac{k}{r}\right)m\end{array}$
Here K is some constant

$\begin{array}{rl}& \left(\frac{m{v}^2}{r}\right)=\frac{km}{r}\Rightarrow v=\mathrm{constant}\\ & T=\frac{2\pi r}{v}\\ & \therefore T\propto r\end{array}$

Hence, the answer is the option (3).

Example 2: The relative uncertainty in the period of a satellite orbiting around the earth is ${10}^{-2}$. If the relative uncertainty in the radius of the orbit is negligible, the relative uncertainty in the mass of the earth is

1) ${10}^{-2}$
2) $2\times {10}^{-2}$
3) $3\times {10}^{-2}$
4) $6\times {10}^{-2}$

Solution:

Period of satellite

$\begin{array}{rl}& T=2\pi \sqrt{\frac{r^3}{GM}}\\ & r=\text{radius of orbit}\\ & T\to \underset{―}{\text{Time period}}\\ & M\to \text{Mass of planet}\\ & \text{wherein}\\ & T=\frac{\text{Circum ference of orbit}}{\text{orbital velocity}}\\ & v=\sqrt{\left(\frac{QM}{R}\right)}\Rightarrow T=\frac{2\pi R}{v}=\frac{2\pi }{\sqrt{(4m)}}\cdot R^{3/2}\\ & M=\frac{4{\pi }^2}{4}\cdot \frac{R^3}{T^2}\\ & \frac{\mathrm{\Delta }M}{M}=3\cdot \frac{\mathrm{\Delta }R}{R}+2\frac{\mathrm{\Delta }T}{T}=3.0+2\times {10}^{-2}\\ & \frac{\mathrm{\Delta }M}{M}=2\times {10}^{-2}\end{array}$

Hence, the answer is the option (2).

Example 3: A spaceship orbits around a planet at a height of 20 Km from its surface. Assuming that only the gravitational field of the planet acts on the spaceship, what will be the number of complete revolutions made by the spaceship in 24 hours around the planet? (Given : Mass of planet $=8\times {10}^{22}\mathrm{Kg}$, Radius of planet $=2\times {10}^6\mathrm{m}$ Gravitational Constant $G=6.67\times {10}^{-11}{\mathrm{Nm}}^2{\mathrm{Kg}}^{-2}$ )

1) 11

2) 17

3) 13

4) 9

Solution:

( $T=$ time,$r=$ radius, $G=$ gravitational force,$m=$ mass of planet $)$
using r=R+h

$\begin{array}{rl}& T=2\pi \sqrt{\frac{r^3}{GM}}=2\pi \sqrt{\frac{{(2.02\times {10}^6)}^3}{6.67\ast {10}^{-11}\times 8\times {10}^{22}}}=7800\sec .\\ & \text{Now, No of revolution}=\frac{24\times 3600}{7800}=11.07=11\end{array}$

Hence, the answer is the Option (1).

Example 4: If the satellite whose mass m is revolving in a circular orbit of radius r around the earth (mass of earth =M). The time of the revolution of satellites is:

1) $T\alpha \frac{r^3}{\sqrt{GM}}$
2) $T\propto \frac{r^{\frac{3}{2}}}{\sqrt{GM}}$
3) $T\alpha \frac{\sqrt{GM}}{r3}$
4) $T\alpha \frac{\sqrt{GM}}{r^{\frac{3}{2}}}$

Solution:

$\begin{array}{rl}T& =\frac{2\pi r}{v_0}=\frac{2\pi r}{\sqrt{\frac{GM}{r}}}\\ T& =\frac{2\pi r}{\sqrt{\frac{GM}{r}}}\\ T& \propto \frac{r^{\frac{3}{2}}}{\sqrt{GM}}\end{array}$

Hence, the answer is the option (2).

Example 5: If the density of the planet is $\rho$ and a satellite is revolving in a circular orbit of radius r then the time of revolution varies with density $\rho$ as :

1) $T\propto \sqrt{\rho }$
2) $T\propto \frac{1}{\rho }$
3) $T\propto \rho$
4) $T\propto \frac{1}{\sqrt{\rho }}$

Solution

The time period of a satellite in terms of density
$T=\sqrt{\frac{3\pi }{G\rho }}$

$\rho \to$ Density of planet
$T\to$ Time period
$G\to$ Gravitational constant
wherein

$\rho =5478.4\mathrm{Kg}/{\mathrm{m}}^3\text{for earth}$
As we known

$\begin{array}{rl}& T=\frac{2\pi r^{\frac{3}{2}}}{\sqrt{GM}}\\ & M=\rho \ast \frac{4}{3}\pi r^3\\ & T=\frac{2\pi r^{\frac{3}{2}}}{\sqrt{G\rho \ast \frac{4}{3}\pi R^3}}\\ & \text{so,}\\ & T=\sqrt{\frac{3\pi }{G\rho }}\end{array}$

$T\propto \frac{1}{\sqrt{\rho }}$

Hence, the answer is the option (4).

Summary

The time period of a satellite depends on its orbit radius and the mass of the planet, following Kepler's laws. It is influenced by gravitational forces and can be calculated using specific formulas that account for satellite altitude and Earth's mass. The satellite's energy consists of kinetic and potential energy, which together determine orbit stability. Understanding these factors helps in applications like GPS, communication, and weather satellite