Earth Satellites

Earth satellites orbit our planet for various purposes, such as communication, navigation, weather forecasting, and research. There are two types: natural satellites, like the Moon, and artificial satellites, which are human-made. Satellites play a crucial role in modern society, enabling weather predictions, climate monitoring, global communications, GPS, military surveillance, and space exploration. Many technologies and services we depend on would not be possible without them.

1.0Satellites

• A satellite is a small body that orbits a larger planet due to gravitational attraction. The Moon is Earth's natural satellite.
• A satellite is a body which continuously revolves on its own around a much larger body in a stable orbit.

2.0Key Requirements for Satellite Motion

• The center of the satellite's orbit must align with the center of Earth.
• The plane of the orbit of the satellite should pass through the centre of Earth.

This means that a satellite can only orbit Earth in circular paths whose centers align with the center of the Earth. These circles, drawn on the globe with the Earth at their center, are called "great circles." Thus, a satellite orbits Earth along paths that are concentric with these great circles.

3.0Launching of a Satellite

To Put the satellites into an orbit around the Earth,two velocities are required

1. A Minimum vertical velocity(escape velocity) to take the satellite to a suitable height.
2. After the satellite has ascended the required height ,it is given a suitable horizontal velocity to make it move in a circular orbit around the Earth.                             

4.0Multistage Rocket For Launching A Satellite

1. Escape Velocity and Air Resistance: The escape velocity at Earth's surface is 11.2 km/s, but air resistance is significant at this speed, requiring much higher velocities to reach the desired satellite altitude.
2. Multistage Rockets: To achieve these high speeds, multistage rockets are used. A typical rocket for this purpose is a 3-stage design.
3. Third Stage Placement: The satellite is placed on the third stage of the rocket, which is responsible for placing it in orbit.
4. Launch and Upthrust: When the rocket lifts off, the exhaust gases produce strong upward thrust, allowing the rocket to quickly accelerate through the denser lower atmosphere.
5. Stage Separations and Satellite Orbiting: After the first and second stages complete their tasks and are detached, the rocket's third stage adjusts the satellite's trajectory and provides the necessary velocity to place it in a stable orbit around Earth.

5.0Orbital Velocity

• The velocity needed to launch a satellite into its orbit around Earth.

M=Mass of Earth, m=Mass of Satellite, R=Radius of Earth, 

$v_o$=Orbital Velocity of Satellite, h=height of satellite above Earth's Surface

R+h=Orbital Radius of the satellite

According to the Law of Gravitation,

$F=\frac{GMm}{(R+h{)}^2}$

Centripetal force required by the satellite to keep it in its orbit is ,

$F=\frac{m{v}_o^2}{R+h}$

In equilibrium, the centripetal force is just provided by the gravitational pull of the Earth,

$\frac{m{v}_o^2}{R+h}=\frac{GMm}{(R+h{)}^2}$

$v_o=\sqrt{\frac{GM}{R+h}}$

$g=\frac{GM}{R^2}\Rightarrow GM=g{R}^2$

$v_o^2=\sqrt{\frac{g{R}^2}{R+h}}=R\sqrt{\frac{g}{R+h}}$

When the satellite close to the Earth surface,h=0 so orbital velocity becomes,

$v_o=\sqrt{gR}$

$g=9.8m/s^2,R=6.4\times 10^{6}m$

$v_o=\sqrt{gR}=\sqrt{9.8\times 6.4\times 10^6}=7.92\times 10^{3}m/s=7.92km/s$

The orbital velocity of a satellite:

• Is independent of the satellite's mass.
• Decreases as the radius of the orbit and the satellite's height increase.
• Relies on the mass and radius of the planet the satellite orbits.

Example-1.

Satellites A and B are orbiting the Earth in circular orbits with radii in the ratio of 1:4. What is the ratio of their orbital velocities?

Solution:

$v\propto \frac{1}{\sqrt{r}}$

$\frac{v_1}{v_2}=\sqrt{\frac{r_2}{r_1}}=\sqrt{\frac{4}{1}}\Rightarrow \frac{2}{1}$

Relation Between Orbital Velocity and Escape Velocity

$v_e=\sqrt{2gR}\Rightarrow \text{Escape Velocity}$

$v_o=\sqrt{gR}\Rightarrow \text{Orbital Velocity}$

$\frac{v_e}{v_o}=\frac{\sqrt{2gR}}{\sqrt{gR}}=\sqrt{2}$

$v_e=\sqrt{2}{v}_o$

The escape velocity of a body from Earth's surface is $\sqrt{2}$ times its velocity in a circular orbit just above the surface.

6.0General Expression Related To Satellite

1.Time period of a Satellite -It is the time taken by a satellite to complete one revolution around the Earth.

$T=\frac{\text{Circumference of the orbit}}{\text{Orbital Velocity}}=\frac{2\pi (R+h)}{v_o}$

$v_o=\sqrt{\frac{GM}{R+h}}$

$T=\frac{2\pi (R+h)}{\sqrt{\frac{GM}{(R+h{)}^3}}}=2\pi \sqrt{\frac{(R+h{)}^3}{GM}}$

$GM=g{R}^2$

$T=\frac{2\pi (R+h)}{\sqrt{\frac{GM}{(R+h{)}^3}}}=2\pi \sqrt{\frac{(R+h{)}^3}{g{R}^2}}\dots (1)$

If Earth is a sphere of mean density then mass,

$M=\text{Volume}\times \text{Density}=\frac{4}{3}\pi r^3\rho$

$T=2\pi \sqrt{\frac{(R+h{)}^3}{GM}}=T=2\pi \sqrt{\frac{(R+h{)}^3}{G\cdot \frac{4}{3}\pi r^3\rho }}=\sqrt{\frac{3\pi (R+h{)}^3}{G\rho R^3}}$

When the satellite revolves close to  the Earth,h=0 and time period will be 

$T=2\pi \sqrt{\frac{R^3}{GM}}=2\pi \sqrt{\frac{R}{g}}=\sqrt{\frac{3\pi }{G\rho }}$

$g=9.8m/s^2,R=6.4\times 10^{6}m$

$T=2\pi \sqrt{\frac{6.4\times 10^6}{9.8}}=5078s=84.6min$

2. Height of a Satellite 

$T=2\pi \sqrt{\frac{(R+h{)}^3}{GM}}$

$T^2=\frac{4{\pi }^2(R+h{)}^3}{g{R}^2}$

$(R+h{)}^3=\frac{T^2{R}^{2}g}{4{\pi }^2}\Rightarrow R+h={\left[\frac{T^2{R}^{2}g}{4{\pi }^2}\right]}^{\frac{1}{3}}$

$h={\left[\frac{T^2{R}^{2}g}{4{\pi }^2}\right]}^{\frac{1}{3}}-R$

3.Angular Momentum

$L=mvr=mv\sqrt{\frac{GM}{r}}=\sqrt{GM{m}^{2}r}$

Note: In case Gravitation Force is different $F\propto \frac{1}{r^n}$ then, Orbital Velocity $v\propto r^{\frac{1-n}{2}}$ and Time period $T\propto r^{\frac{1+n}{2}}$

Example-2.

The time period of a satellite in a circular orbit of radius $R_0$ is $T_O$. Find the time period of another satellite in a circular orbit of radius $4R_0$. 

Solution:

$T_0=2\pi \sqrt{\frac{R_0^3}{G{M}_s}}$

$R^{\prime }=4R_0\Rightarrow T^{\prime }=2\pi \sqrt{\frac{(4R_0{)}^3}{G{M}_s}}=2\pi \sqrt{\frac{64R_0^3}{G{M}_s}}$

$T^{\prime }=8\left(2\pi \sqrt{\frac{R_0^3}{G{M}_e}}\right)\Rightarrow T^{\prime }=8T_0$

7.0Types of Satellite

8.0Energy Analysis Of A Satellite

1.Kinetic energy of satellite

Due to Motion of Satellite,

$KE=\frac{1}{2}m{v}_o^2$

$v_o=\sqrt{\frac{GM}{r}}$

$K.E=\frac{GMm}{2r}$

2.Potential energy of satellite

Due to position of satellite

$P.E=-\frac{GMm}{r}$

Total Energy = Kinetic Energy + Potential Energy

$T.E=-\frac{GMm}{2r}$

$K.E=\frac{GMm}{2r},T.E=-\frac{GMm}{2r},P.E=-\frac{GMm}{r}$

$K.E=-T.E=-\frac{P.E}{2}$

As the total energy of a satellite is negative, it is called a bounded system.

Binding energy:

The minimum energy needed for a satellite to escape Earth's orbit and reach infinity.

B.E+T.E=0B.E=-T.E

B.E.=|T.E.| = $B.E=\frac{GMm}{2r}$

$T.E=-\frac{GMm}{2r},P.E=-\frac{GMm}{r},K.E=\frac{GMm}{2r}$

$v_o=\sqrt{\frac{GM}{r}},T\propto R^{3/2}$

As the radius of a satellite's orbit increases:

• Kinetic energy (KE) decreases
• Total energy (TE) increases
• Potential energy (PE) increases
• Orbital velocity decreases
• Orbital period increases

Example-3

What is the energy required to launch a satellite of mass m, initially at rest on the Earth's surface, into a circular orbit at a height equal to the Earth's radius?

Solution:

h=r

Necessary centripetal force is provided by gravitational force.

$\frac{m{v}^2}{2R}=\frac{GMm}{(2R{)}^2}\Rightarrow v=\sqrt{\frac{GM}{2R}}$

Now by conservation of mechanical energy.

$K_i+U_i=K_f+U_f$

$K_i+\frac{-GMm}{R}=\frac{1}{2}m{\left(\sqrt{\frac{GM}{2R}}\right)}^2+\left(-\frac{GMm}{2R}\right)$

$K_i=\frac{1}{2}\frac{GMm}{2R}+\frac{GMm}{R}-\frac{GMm}{2R}$

$K_i=\frac{GMm}{4R}+\frac{GMm}{2R}=\frac{3}{4}\frac{GMm}{R}$

$K_i=\frac{3}{4}mgR\left(\because g=\frac{GM}{R^2}\Rightarrow GM=g{R}^2\right)$

9.0Work Done To Shift Orbit Of Satellite

$W_{\text{ext}}=\Delta T.E.$

$W_{\text{ext}}=T.E_f-T.E_i$

$W_{\text{ext}}=\left(-\frac{GMm}{2r_2}\right)-\left(-\frac{GMm}{2r_1}\right)=-\frac{GMm}{2r_2}+\frac{GMm}{2r_1}$

$W_{\text{ext}}=\frac{GMm}{2}\left(\frac{1}{r_1}-\frac{1}{r_2}\right)$

Key factors for satellite motion

• The center of the satellite's orbit aligns with Earth's center.
• The plane of orbit of the satellite is passing through the center of Earth.