Geostationary and polar satellites

Geostationary Satellite

A satellite which appears stationary relative to the Earth is called a geostationary or geosynchronous satellite.

A geostationary satellite will stay the same place above the earth always, such a satellite is never at rest. Geostationary satellites appear stationary due to their zero relative velocity with respect to their place on Earth.

It is also known as parking orbit.

Now, you can say that the time period of the geostationary satellite is also 24 hours or 86400 sec.

So $\mathrm{T}=24$ hours or 86400 sec.
and we know that the Height of the Satellite is given by

$
h=\left(\frac{T^2 g R^2}{4 \pi^2}\right)^{\frac{1}{3}}-R
$
So putting $\mathrm{T}=24$ hours or 86400 sec.
We get the height of geostationary satellite from the surface of the earth

$
\text { as } h=6 R=36000 \mathrm{~km}
$

Its sense of rotation should be the same as that of Earth about its own axis. l.e., in an anti-clockwise direction (from west to east).
A geostationary satellite is used for telecommunication, weather forecasting, etc.

Polar Satellite

A polar satellite is a satellite whose orbit is perpendicular or at right angles to the equator. or in simple words, it passes over the north and south poles as it orbits the earth. It can also be used as a communication satellite for countries/areas near the poles where Geostationary satellites have no / Poor coverage.

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Solved Examples Based on Geostationary and Polar Satellites

Example 1: Which one of the following statements regarding geostationary satellites of the earth is incorrect?

1) It should revolve in an orbit concentric and coplanar with the equatorial plane

2) The time period of revolution is 16 hours.

3) The height of the geostationary satellite from the centre of the earth is 42400km.

4) The orbital velocity of a geostationary satellite is 3.08km/s.

Solution:

geostationary satellite

Satellites that appear stationary relative to Earth are geostationary satellites, also known as communication satellites or geosynchronous satellites

The period of revolution is the same as the revolution of the earth, T = 24hrs.

The height of a geostationary satellite from the earth's centre is 4200km. and the Orbital velocity of a geostationary satellite is 3.08km/s.

Hence, the answer is the option (2)

Example 2: A geostationary satellite has always to be at an altitude of (in kilometres) :

1) 36000

2) 24000

3) 30000

4) 20000

Solution:

As we know, the time period

Geostationary and polar satellites are given by
$
T=2 \pi \sqrt{\frac{r^3}{G M}}=2 \pi \sqrt{\frac{(R+h)^3}{g R^2}}
$
By squaring and rearranging both sides

$
\begin{aligned}
& \frac{g 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
\end{aligned}
$
So, put the value, $T=24 h=24 \times 60 \times 60=86400 s$

$
\begin{aligned}
& R=\text { Radius of the earth }=6400 \mathrm{~km} \\
& g=9.8 \times 10^{-3} \mathrm{~km} / \mathrm{s}^2
\end{aligned}
$
Substituting these values in the equation of the formula of height, we get-

$
h=(42400-6400) \mathrm{km}=36000 \mathrm{~km}
$

Hence, the answer is the option (1).

Example 3: Given the radius of earth ‘R’ and the length of a day ‘T’, the height of a geostationary satellite is [G = gravitational constant, M = mass of the earth]

1) $\left(\frac{4 \pi^2 G M}{T^2}\right)^{\frac{1}{3}}$
2) $\left(\frac{4 \pi G M}{T^2}\right)^{\frac{1}{3}}-R$
3) $\left(\frac{G M T^2}{4 \pi^2}\right)^{\frac{1}{3}}-R$
4) $\left(\frac{G M T^2}{4 \pi^2}\right)+R$

Solution:

The height of geostationary satellites from the surface of the Earth is $h=6 R=36000 \mathrm{Km}$

$
\begin{aligned}
T & =2 \pi \sqrt{\frac{r^3}{G M}} \\
T & =2 \pi \sqrt{\frac{(r+h)^3}{G M}} \\
h & =\left(\frac{G M T^2}{4 \pi^2}\right)^{\frac{1}{3}}-R
\end{aligned}
$

Hence, the answer is the option (3).

Example 4: A satellite of mass M is in a circular orbit of radius R about the centre of the Earth. A meteorite of the same mass, falling towards the earth, collides with the satellite completely inelastically. The speeds of the satellite and the meteorite are the same, just before the collision. The subsequent motion of the combined body will be :

1) in an elliptical orbit

2) such that it escapes to infinity

3) in a circular orbit of a different radius

4) in the same circular orbit of radius R

Solution:

The shape of the orbit of the satellite

If $V<V_o$, then the satellite does not remain in its circular path rather it traces a spiral path and falls on the earth

$
V=V_e
$

satellite move along a parabolic path
wherein

$
V=V_o
$
Satellite revolves in a circular path

$
V>V_e
$

Satellite will move along a hyperbolic path

Apply momentum conservation

$\begin{aligned} & m V \hat{l}-m V \hat{j}=2 m \rightarrow \overrightarrow{V_2} \\ & V_2=\frac{1}{2} \times \sqrt{2} V_0=\frac{V_0}{2} \\ & \text { So }_2<V_0<\sqrt{\frac{G M}{R}}\end{aligned}$

So it will move in an elliptical orbit

Hence, the answer is the option (1).

Summary

Geostationary satellites orbit at 36,000 km above Earth, remaining fixed over one location, making them ideal for continuous communication, broadcasting, and weather monitoring. Polar satellites, however, orbit from pole to pole, covering the entire globe and providing crucial data for environmental monitoring and scientific research. Both satellite types play vital roles in connecting and informing us about global events and natural phenomena.