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Physics Topics are also essential for space exploration, allowing scientists to study phenomena such as gravitational waves and cosmic rays.
What is the Height and Speed of Geostationary Satellite?
The earth rotates about its own axis (line joining the north and south poles] once in every 24 h. This is the diurnal motion of earth.
We may consider an artificial satellite, set in an orbit in such a way that,
- the orbit of the satellite is circular,
- the plane of the orbit coincides with the equatorial plane,
- the satellite revolves in the direction of the diurnal motion of the earth (i.e., from west to east) and completes a revolution in 24 h.
Then, the satellite will seem to be stationary at a place in the sky over the equator, when observed from the earth’s surface.
Such a satellite is called a geostationary satellite and its orbit is called a geostationary or parking orbit. Clearly, the centres of such orbits coincide with the centre of the earth.
Definition: With reference to the diurnal motion of the earth, if the relative angular velocity of an artificial satellite is zero and the satellite is always on the equatorial plane so that it appears stationary at one place in the sky, as seen from the earth’s surface, the satellite is called a geostationary artificial satellite.
Height and orbital speed of a geostationary satellite: Let the mass of the earth = M, the radius of the earth = R, the mass of the geostationary satellite = m, the distance of the satellite from the centre of the earth = r, the orbital speed of the satellite = v, time period of revolution = T [Fig.]
As the force of gravitation provides the necessary centripetal force for the revolution in a circular orbit,
\(\frac{m v^2}{r}\) = \(\frac{G M m}{r^2}\) or, v2 = \(\frac{G M}{r}\)
The acceleration due to gravity on the earth’s surface is g.
For a geostationary satellite, T = 24 h = 24 × 60 × 60 s, g = 9.8 m ᐧ s-2, R = 6400 km = 6.4 × 106m.
Substituting these values, we obtain,
r = 4.234 × 107m
= 4.234 × 104 km = 42340 km (approx.).
Hence, the height of the geostationary satellite at the equator from the earth’s surface,
h = r – R = 42340 – 6400 = 35940 = 36000 km
It is to be noted that the height of a geostationary orbit does not depend on the mass of the satellite.
The orbital speed of a geostationary satellite,
v = \(\frac{2 \pi r}{T}\) = \(\frac{2 \pi \times 4.234 \times 10^2}{24 \times 60 \times 60}\) = 3079 m ᐧ s-1
= 3.079 km ᐧ s-1
Clearly, this orbital speed is much less than the escape velocity from the earth and the orbital speed for an artificial satellite close to the earth’s surface.
Discussions:
Uses: Nowadays, geostationary satellites are in extensive use specially for weather observation, TV and radio broadcasting, telecommunication, etc. Games and entertainment programmes performed anywhere in the world can be transmitted and telecast live at other places using parking relay satellites on a geostationary orbit.
Difficulties in use:
i) As geostationary satellites are placed at a height of about 36000 km above the earth’s surface, a signal has to travel a minimum distance of 72000 km before it is received at the other end. This causes a delay of about \(\frac{1}{4}\) s from the time of occurrence, which is negligible in the case of radio, TV broadcasting, but causes inconvenience in telephonic conversations between two coun-tries. Due to the high orbit, the spatial resolution of data is not as great as for the polar satellites.
ii) The plane of orbits of geostationary satellites coincides with the earth’s equatorial plane. Hence, they cannot function properly for the zones closer to the poles.
Polar satellite : Satellites placed 700-800 km above the surface of the earth and aligned with the polar plane are called polar satellites. Using these satellites, a signal can reach from one point on the earth to another point in approximately \(\frac{1}{100}\) s Because of such a small time difference, there is practically no time lag felt during telephonic communication. These satellites also do away with the problems of reception and transmission of signals in the polar regions. Obviously, these are not geostationary satellites and hence, are not apparently stationary above the earth’s surface. Polar satellites have a time period of 2 h, i.e., to revolve around the earth once, a polar satellite takes about 2 h.
Geosynchronous satellite: If the orbit of a satellite is inclined with the equatorial plane and the satellite revolves around the earth once in 24 h in the direction of the diurnal motion of the earth, the angular velocity of the satellite equals that of the earth. If observed from the surface of the earth, it appears that the artificial satellite is oscillating from north to south along a longitude and completes one oscillation in 24 h. Such satellites are called geosynchronous satellites.
Numerical Examples
Example 1.
An artificial satellite revolves around the earth in a circular orbit and is at a height of 300km above the surface of the earth. Find the orbital speed and the time period of the satellite. The radius of the earth = 6400 km, g = 980 cm ᐧ s-2. [HS 07]
Solution:
Radius of the earth, R = 6400 km = 64 × 107 cm; height, h = 1.6 × 105 m
Distance of the artificial satellite from the centre of the earth (r) = 6400 + 300 = 6700 km = 67 × 107 cm.
Orbital speed of the artificial satellite,
v = R\(\sqrt{\frac{g}{r}}\) = 64 × 107 \(\sqrt{\frac{980}{67 \times 10^7}}\)
= 7.74 × 105 cm ᐧ s-1 = 7.74 km ᐧ s-1
Time period of revolution,
T = \(\frac{2 \pi r}{\nu}\) = \(\frac{2 \times \pi \times 67 \times 10^7}{7.74 \times 10^5}\) = 5439 s
= 1 h 30 min 39 s.
Example 2.
A man is positioned in a circular orbit at a height of 1.6 × 105 m above the surface of the earth. The radius of the earth is 6.37 × 106 m and the mass is 5.98 × 1024 kg. What would be the orbital speed of the man? (G = 6.67 × 10-11N ᐧ m2 ᐧ kg-2)
Solution:
Radius of the earth, R = 6.37 × 106m = 63.7 × 105m;
height, h = 1.6 × 105 m
Distance of the man from the centre of the earth (r) = R + h
= 63.7 × 105 + 1.6 × 105 = 65.3 × 105 m
∴ Orbital speed,
v = \(\sqrt{\frac{G M}{R+h}}\) = \(\sqrt{\frac{G M}{r}}\) = \(\sqrt{\frac{6.67 \times 10^{-11} \times 5.98 \times 10^{24}}{65.3 \times 10^5}}\)
= 0.78 × 104 mᐧs-1 = 7.8 kmᐧs-1.
Example 3.
Prove that, if the orbital speed of the moon increases by 42%, it would stop orbiting around the earth.
Solution:
Let mass of the earth = M and radius of the moon’s orbit = r.
Orbital speed of the moon (v) = \(\sqrt{\frac{G M}{r}}\).
The acceleration due to gravity,
g’ = \(\frac{G M}{r^2}\) or, GM = g’r2
∴ v = \(\sqrt{\frac{g^{\prime} r^2}{r}}\) = \(\sqrt{g^{\prime} r}\)
Also, the escape velocity of the moon with respect to the earth’s surface,
ve = \(\sqrt{2 g^{\prime} r}\)
∴ \(\frac{v_e}{v}\) = \(\sqrt{\frac{2 g^{\prime} r}{g^{\prime} r}}\) = \(\sqrt{2}\) = 1.414
This implies, ve = 1.414v = 141.4% of orbital speed (v).
Hence, if the orbital speed of the moon increases by 42%, which means that it changes to 142% of its present value, it crosses the value of the escape velocity from the earth. As a result, the moon will move out of the earth’s gravitational field and will not move around the earth anymore.
Example 4.
An artificial satellite is orbiting around the earth at a height of 3400 km above the earth’s surface in a circular orbit. Find the orbital speed of the satellite. The radius of the earth = 6400 km and g = 980 cm ᐧ s-2. [HS ’06]
Solution:
Radius of the earth, R = 6400 km = 64 × 107 cm
Distance of the artificial satellite from the centre of the earth,
r = 6400 + 3400 = 9800 km = 98 × 107 cm
Hence, the orbital speed of the satellite,
v = R\(\sqrt{\frac{g}{r}}\) = 64 × 107\(\sqrt{\frac{980}{98 \times 10^7}}\)
= 6.4 × 105 cm ᐧ s-1 = 6.4 km ᐧ s-1
Example 5.
A satellite at a height of 700 km is revolving around the earth in a circular orbit. Find the velocity of the satellite with respect to the earth’s surface. (Radius of the earth R = 6300 km, g = 9.8 m ᐧ s-2)
Solution:
Radius of the earth (R) = 6300 km = 63 × 105 m.
Distance of the satellite from the centre of the earth, r = 6300 + 700 = 7000 km = 7 × 106
Hence, velocity of the artificial satellite,
v = R\(\sqrt{\frac{g}{r}}\) = 63 × 105\(\sqrt{\frac{9.8}{7 \times 10^6}}\) = 74.543 × 102 m ᐧ s-1
= 7.454 km ᐧ s-1.
Example 6.
A small satellite revolves around a planet of average density 10 g ᐧ cm-3. The radius of the orbit of the satellite is slightly more than the radius of the planet. Find the time period of rotation of the satellite. G = 6.68 × 10-8 CGS unit. [HS ’02]
Solution:
Let mass of the planet = M, radius = R and density = ρ.
∴ M = \(\frac{4}{3} \pi R^3 \rho\)
As per the question, radius of the orbit of the satellite \(\simeq\) R. If its mass = m and orbital speed = v,
centripetal force = \(\frac{m v^2}{R}\)
Mutual force of gravitation between the planet and satellite provides this centripetal force.