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Kinetic Energy and Potential Energy of an Artificial Satellite
If an artificial satellite of mass m revolves around the earth in a circular orbit of radius r with an orbital speed v, then
\(\frac{m v^2}{r}\) = \(\frac{G M m}{r^2}\) or, v2 = \(\frac{G M}{r}\)
Hence, the kinetic energy of the satellite,
K = \(\frac{1}{2} m v^2\) = \(\frac{G M m}{2 r}\) ……. (1)
As stated in Section 1.6.3, the potential energy of an object of mass m at a distance r from another mass M,
U = –\(\frac{G M m}{r}\) ………. (2)
The value of this potential energy is negative as the force of gravitation is an attractive force. It is seen from equations (1) and (2) that the potential energy of a satellite revolving around the earth is double its kinetic energy, but negative.
Thus, the total energy of the satellite in orbit,
E = K + U = \(\frac{G M m}{2 r}\) – \(\frac{G M m}{r}\) = –\(\frac{G M m}{2 r}\) …….. (3)
Due to friction in the earth’s atmosphere or due to collision with meteors and meteorites in space, the energy of a satellite (-\(\frac{G M m}{2 r}\)) decreases, that is, the magnitude of \(\frac{G M m}{2 r}\) increases.
Hence, the value of r decreases, which means that the satellite starts shifting towards the earth’s surface. As a result, the kinetic energy (\(\frac{G M m}{2 r}\)) starts increasing- in brief, as the distance of a satellite from the surface of the earth decreases, its orbital speed and kinetic energy increase, but the total energy decreases. This is possible, because a decrease in the potential energy of a satellite is twice the increase in kinetic energy when it transfers from one orbit to another. Thus, for any artificial satellite to be working in its orbit for a long time (e.g., geostationary satellite), energy has to be supplied from outside to compensate for its loss of energy. The use of solar energy for this purpose is indispensable.
It should be noted that for an elliptical orbit, the expression for the total energy of the artificial satellite is \(\frac{-G M m}{2 a}\) where a is the semimajor axis of the ellipse.
Binding energy of a satellite: The energy required by a satellite to leave its orbit around the earth and escape to infinity is called its binding energy.
The total energy of a satellite is \(-\frac{G M m}{2 r}\). In order to escape to infinity, it must be supplied an extra energy equal to +\(\frac{G M m}{2 r}\), so that, its total energy E becomes zero. Hence binding
energy of a satellite is \(\frac{G M m}{2 r}\)
The plot of energies of a satellite versus orbit radius is shown in the Fig.
Numerical Examples
Example 1.
Find the velocity with which a body can be projected vertically upwards so that it can reach a height equal to the radius of the earth. The radius of the earth = 6400 km, g = 980 cm ᐧ s-2.
Solution:
Initial distance of the body from the centre of the earth = radius of the earth = R.
Final distance of the body from the centre of the earth = R + R = 2R
Let mass of the earth = M; mass of the body = m; velocity of projection from the earth’s surface = v, Hence, its kinetic energy on the earth’s surface = \(\frac{1}{2}\)mv2
At a height R above the earth’s surface, the body stops momentarily and then falls. Hence, at that height R, kinetic energy = 0.
Potential energy on the earth’s surface = – \(\frac{G M m}{R}\)
Total energy of the body on the earth’s surface
= kinetic energy + potential energy = \(\frac{1}{2}\)mv2 – \(\frac{G M m}{R}\)
Again, potential energy at height R = –\(\frac{G M m}{2 R}\)
and its total energy at that height = 0 – \(\frac{G M m}{2 R}\) = –\(\frac{G M m}{2 R}\)
From the law of conservation of energy
Alternative Method
Let the radius of the earth be R, the potential energy of the body on the earth’s surface be 0, and the acceleration due to gravity at a height h above the earth’s surface be g’. Hence, the potential energy at height h is mg’ h. For a fur-ther increase dh in height, let the increase in potential energy be dW.
∴ dW= mg’dh = mg\(\frac{R^2}{(R+h)^2}\)dh [∵ g’ = \(\frac{R^2}{(R+h)^2}\) ᐧ h]
Hence, the total increase in the potential energy for an increase in height R from the surface of the earth,
Let the kinetic energy of the body on the earth’s surface = \(\frac{1}{2} m v^2\)
As per the question, the kinetic energy at a height R from the earth’s surface is 0. Hence, from the law of conservation of energy,
\(\frac{1}{2} m v^2\) = \(\frac{1}{2} m g R\) or, v2 = gR
or v = \(\sqrt{g R}\) = 7.92 × 105 ᐧ s-1
Example 2.
Two bodies of masses M and rn were initially at infinite distance from each other and they started approaching each other due to their mutual force of gravitation. Prove that their velocity of approach becomes \(\sqrt{\frac{2 G}{r}(M+m)}\) when they are at a distance of r from each other.
Solution:
At infinite distance from each other, both the bodies did not have any potential energy. Also, their velocities were zero.
Hence, initial momentum of both the bodies = 0.
Total initial energy = kinetic energy + potential energy = 0 + 0 = 0.
At a distance r from each other, let the velocity of the body of mass m be v and that of mass M be V.
Hence, their velocity of approach (c) = v + V
As the velocities v and V are in opposite directions, the total momentum at this stage = mv – MV
Hence, from the law of conservation of momentum
0 = mv – MV or, V = \(\frac{m}{M}\)M
∴ c = v + V = v + \(\frac{m}{V}\)v = v(1 + \(\frac{m}{M}\))
Hence, the kinetic energy of the bodies when they are at a distance r from each other
Also, potential energy at a distance r (due to gravitation)
= –\(\frac{G M m}{r}\)
Hence, from the law of conservation of energy,
Example 3.
An artificial satellite is revolving around the earth in a circular orbit. Its velocity is half the value of the escape velocity from the earth,
(i) What is the height of the satellite from the earth’s surface?
(ii) If its revolution around the earth is stopped and the satellite is allowed to fail freely towards the earth, what will be the velocity with which it will strike the earth’s surface? (Radius of the earth = 6.4 × 106 m; g = 9.8 m ᐧ s-2)
Solution:
i) Orbital speed of the satellite
u = \(\frac{1}{2}\) × escape velocity = \(\frac{1}{2}\) × \(\sqrt{2 g R}\) = \(\sqrt{\frac{g R}{2}}\)
Again, if it is at a height h above the earth’s surface,
ii) When the satellite stops revolving and falls freely towards the earth, its initial velocity of fall = 0. Hence, its initial kinetic energy = 0.
Since at this stage, the distance of the satellite from the centre of the earth is r = R+ h = 2R, its initial poten-tial energy = –\(\frac{G M m}{2 R}\) (M = mass of the earth, m = mass of the satellite).
Hence, the total initial energy of the satellite
= 0 – \(\frac{G M m}{2 R}\) = \(\frac{G M m}{2 R}\)
Let the velocity of the satellite be v, just before touching the earth’s surface.
So its kinetic energy = \(\frac{1}{2} m v^2\). At the same time, its distance from the centre of the earth is r = R and its potential energy = –\(\frac{G M m}{R}\)
Hence, total energy of the satellite = \(\frac{1}{2} m v^2\) – \(\frac{G M m}{R}\)
From the law of conservation of energy,
Example 4.
Calculate the kinetic energy, potential energy and total energy of a geostationary satellite of mass 100 tonne. The radius of the earth is 6400 km and the radius of the orbit of the satellite is 42400 km.
Solution:
100 tone = 100 × 103 kg = 105 kg. Radius of the earth = 6400 km = 6.4 × 106 m and the distance of the geostationary satellite from the centre of the earth, r = 42400 km = 4.24 × 107 m.
Hence, kinetic energy