In the previous section, we completed a discussion on energy of orbiting earth satellites
• In this section we will see Geostationary satellites
1. In fig.8.51 below, the earth is shown as a blue sphere
• The axis of rotation of the earth is shown in yellow color
♦ It passes through the N and S poles
• The direction of rotation is indicated by the orange curved arrow
2. A satellite shown in red color orbits around the earth
• The orbit of the satellite is circular in shape. It is shown in green color
3. In fig.8.52 below, the earth is shown to be transparent
• So center of the earth is visible. The center is indicated by a small white sphere
4. A magenta line is drawn from the center of earth to center of the satellite
• This magenta line intersects the surface of the earth at P
5. Suppose that, the 'time period TE of the earth (which is 24 hours)' and 'time period TS of the satellite' are the same
• That is., both earth and the satellite complete one rotation at the same time
• Then, both of them will be rotating as a single unit
• The point P will not change
• For a person standing at P, the satellite will appear to be stationary
6. Not only for a person at P. In fig.8.53 below, two more magenta lines are drawn in random directions
• Those new lines intersect the surface of the earth at P1 and P2
• For those standing at P1 and P2 also, the satellite will appear to be stationary
• In fact, since the earth and the satellite rotate as a single unit, the satellite will appear to be stationary for every person on earth
• Such a satellite is called a geostationary satellite
7. One important point has to be noted:
■ For a satellite to be geostationary, it's orbit must lie in the equatorial plane
• This can be explained in 9 steps:
(i) Let us assume that orbits of all the earth satellites are circles
(ii) The centers of all those circles must coincide with the center of the earth
• This is because, the gravitational force (which provides the centripetal force), acts towards the center of the earth
(iii) So we can have the green orbits shown in fig.8.54 below. Both of them have their centers at the center of the earth
• But the magenta orbit is not possible because, it's center is not at the center of the earth
(iv) The view in fig.8.55 below helps to understand the difference between the three orbits
We see that:
• The green orbits has centers at the earth's center
♦ So the green orbits are possible
• The center of the magenta orbit is away from the earth's center
♦ So the magenta orbit is not possible for satellites
(v) Out of the two green orbits, one is horizontal and the other is inclined
• Consider the horizontal green orbit
♦ We know that, center of all green orbits coincide with the center of the earth
♦ So the horizontal green orbit falls exactly on the equatorial plane
(We know that, the equator is a circle. The plane in which this circle lies, is the equatorial plane)
(vi) For a satellite to be geostationary, it's orbit should lie in the equatorial plane
• Let us see what happens if another green orbit is selected:
(vii) In the fig.8.56 below, the green orbit is inclined. It does not lie in the equatorial plane
• P is the familiar 'point of intersection of the magenta line' that we saw in the fig.8.52
(viii) A person standing initially at P will be moving along the horizontal red circle
• But the point P will be moving along the inclined yellow circle
• So the satellite will not appear to be stationary
(ix) Thus it is clear that, for a satellite to be geostationary, it's orbit must lie in the equatorial plane
• Let the time period TS of the satellite in fig.8.56 be 24 hours
• Then satellite will reach the exact same spot after 24 hours. But it will not appear to be stationary for a viewer on earth
• Such satellites are called geosynchronous satellites
• All geostationary satellites are geosynchronous satellites. But all geosynchronous satellites are not geostationary satellites
8. Now we can write the definition for geostationary satellites:
Satellites in circular orbits around the earth in the equatorial plane with T = 24 hours are called Geostationary satellites
9. Now we will see some mathematical calculations related to such satellites:
• We have Eq.8.26 that we derived in a previous section: $\mathbf\small{T={\frac{2\pi(R_E+h)^{3/2}}{\sqrt{G\,M_E}}}}$
• Squaring both sides, we get: $\mathbf\small{T^2={\frac{4\pi^2(R_E+h)^{3}}{{G\,M_E}}}}$
$\mathbf\small{\Rightarrow (R_E+h)^{3}={\frac{T^2\,G\,M_E}{4\pi^2}}}$
$\mathbf\small{\Rightarrow (R_E+h)=\left({\frac{T^2\,G\,M_E}{4\pi^2}}\right)^{1/3}}$
10. In the above equation, only h and T are the variables
• If we put T = 24 hours (converted into seconds), we will get: h = 35800 km
11. This is a very large height. Powerful rockets will be required to take the satellites to such heights
But having satellites at such heights have many practical advantages. Let us see some of them. We will write them in steps:
(i) The ionosphere is a particular layer in the earths atmosphere
• It's level above the surface of the earth is indicated by the red line in fig.8.57 below:
(ii) Antenna A is transmitting radio signals. Those signals are indicated by the blue arrows
• The radio signals get reflected by the ionosphere. So they can be received on a larger area on the earth's surface
(iii) Antenna B is transmitting television signals. Those signals are indicated by the yellow arrow
• The television signals have a higher frequency than radio signals
• Such high frequency signals are not reflected by the ionosphere
• So such signals cannot be received over a large area
(iv) It is clear that, the curvature of the earth's surface is playing a major role here
• If the surface was flat, we would be able to transmit the TV signals to a wider area, even if there is no reflection from the ionosphere
• But as seen from the fig., the opposite side of the curvature does not receive any signals
(v) If we put a geostationary satellite up in the sky, the signals can be picked up and send to the opposite side of the curvature. It is shown in fig.8.58(a) below:
• Note that, the satellite must be a geostationary one. Otherwise, the antenna B will have to continuously change the direction of transmission
• In this section we will see Geostationary satellites
1. In fig.8.51 below, the earth is shown as a blue sphere
 |
| Fig.8.51 |
♦ It passes through the N and S poles
• The direction of rotation is indicated by the orange curved arrow
2. A satellite shown in red color orbits around the earth
• The orbit of the satellite is circular in shape. It is shown in green color
3. In fig.8.52 below, the earth is shown to be transparent
• So center of the earth is visible. The center is indicated by a small white sphere
 |
| Fig.8.52 |
• This magenta line intersects the surface of the earth at P
5. Suppose that, the 'time period TE of the earth (which is 24 hours)' and 'time period TS of the satellite' are the same
• That is., both earth and the satellite complete one rotation at the same time
• Then, both of them will be rotating as a single unit
• The point P will not change
• For a person standing at P, the satellite will appear to be stationary
6. Not only for a person at P. In fig.8.53 below, two more magenta lines are drawn in random directions
 |
| Fig.8.53 |
• For those standing at P1 and P2 also, the satellite will appear to be stationary
• In fact, since the earth and the satellite rotate as a single unit, the satellite will appear to be stationary for every person on earth
• Such a satellite is called a geostationary satellite
7. One important point has to be noted:
■ For a satellite to be geostationary, it's orbit must lie in the equatorial plane
• This can be explained in 9 steps:
(i) Let us assume that orbits of all the earth satellites are circles
(ii) The centers of all those circles must coincide with the center of the earth
• This is because, the gravitational force (which provides the centripetal force), acts towards the center of the earth
(iii) So we can have the green orbits shown in fig.8.54 below. Both of them have their centers at the center of the earth
 |
| Fig.8.54 |
(iv) The view in fig.8.55 below helps to understand the difference between the three orbits
 |
| Fig.8.55 |
• The green orbits has centers at the earth's center
♦ So the green orbits are possible
• The center of the magenta orbit is away from the earth's center
♦ So the magenta orbit is not possible for satellites
(v) Out of the two green orbits, one is horizontal and the other is inclined
• Consider the horizontal green orbit
♦ We know that, center of all green orbits coincide with the center of the earth
♦ So the horizontal green orbit falls exactly on the equatorial plane
(We know that, the equator is a circle. The plane in which this circle lies, is the equatorial plane)
(vi) For a satellite to be geostationary, it's orbit should lie in the equatorial plane
• Let us see what happens if another green orbit is selected:
(vii) In the fig.8.56 below, the green orbit is inclined. It does not lie in the equatorial plane
 |
| Fig.8.56 |
(viii) A person standing initially at P will be moving along the horizontal red circle
• But the point P will be moving along the inclined yellow circle
• So the satellite will not appear to be stationary
(ix) Thus it is clear that, for a satellite to be geostationary, it's orbit must lie in the equatorial plane
• Let the time period TS of the satellite in fig.8.56 be 24 hours
• Then satellite will reach the exact same spot after 24 hours. But it will not appear to be stationary for a viewer on earth
• Such satellites are called geosynchronous satellites
• All geostationary satellites are geosynchronous satellites. But all geosynchronous satellites are not geostationary satellites
8. Now we can write the definition for geostationary satellites:
Satellites in circular orbits around the earth in the equatorial plane with T = 24 hours are called Geostationary satellites
9. Now we will see some mathematical calculations related to such satellites:
• We have Eq.8.26 that we derived in a previous section: $\mathbf\small{T={\frac{2\pi(R_E+h)^{3/2}}{\sqrt{G\,M_E}}}}$
• Squaring both sides, we get: $\mathbf\small{T^2={\frac{4\pi^2(R_E+h)^{3}}{{G\,M_E}}}}$
$\mathbf\small{\Rightarrow (R_E+h)^{3}={\frac{T^2\,G\,M_E}{4\pi^2}}}$
$\mathbf\small{\Rightarrow (R_E+h)=\left({\frac{T^2\,G\,M_E}{4\pi^2}}\right)^{1/3}}$
10. In the above equation, only h and T are the variables
• If we put T = 24 hours (converted into seconds), we will get: h = 35800 km
11. This is a very large height. Powerful rockets will be required to take the satellites to such heights
But having satellites at such heights have many practical advantages. Let us see some of them. We will write them in steps:
(i) The ionosphere is a particular layer in the earths atmosphere
• It's level above the surface of the earth is indicated by the red line in fig.8.57 below:
 |
| Fig.8.57 |
• The radio signals get reflected by the ionosphere. So they can be received on a larger area on the earth's surface
(iii) Antenna B is transmitting television signals. Those signals are indicated by the yellow arrow
• The television signals have a higher frequency than radio signals
• Such high frequency signals are not reflected by the ionosphere
• So such signals cannot be received over a large area
(iv) It is clear that, the curvature of the earth's surface is playing a major role here
• If the surface was flat, we would be able to transmit the TV signals to a wider area, even if there is no reflection from the ionosphere
• But as seen from the fig., the opposite side of the curvature does not receive any signals
(v) If we put a geostationary satellite up in the sky, the signals can be picked up and send to the opposite side of the curvature. It is shown in fig.8.58(a) below:
 |
| Fig.7.58 |
• In the next section we will see Polar satellites
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