## Artificial Satellites

The moon is the only satellite of the earth. It has already been discussed that the motions of different planets and satellites can be determined by Newton’s laws of motion and of gravitation. From these considerations, it was thought that, if a projectile from the surface of the earth was raised to an appropriate height and given an appropriate velocity, it would also revolve around the earth like the moon.

- After a considerable amount of research in this field, the first artificial satellite (Sputnik-I) of the Earth was launched on 4 October 1957. Presently technological expertise has advanced to such an extent that it has been possible to set up artificial satellites not only around the Earth but also around other planets and satellites.
- Any artificial satellite is projected vertically upwards or towards the east from the surface of the earth using rockets. The direction of motion of the satellite is changed using rockets arranged at the rear of the satellite in such a way that the satellite attains the desired horizontal velocity on reaching the predetermined height. In that case, it would set it in the desired orbit and start revolving around the Earth.

Obviously, once in an orbit, any satellite follows Kepler’s laws.

**Read and Learn More: Class 11 Physics Notes**

Hence, the orbit of an artificial satellite can also be elliptical or circular like the orbits of planets and satellites. However, these orbits are assumed to be circular as the eccentricities of such orbits are usually very low; the errors that may occur during calculations are negligible.

**Orbital speed And Period Of Revolution Of an Artificial Satellite:** The speed with which the satellite revolves around the earth is called its orbital speed. Let the mass of the earth = M, radius of the earth (OP) = R, mass of the satellite = m, orbital speed of the satellite = v and the height of the orbit from the surface of the earth (PA) = h. Hence, the distance of the orbit from the centre of the earth, i.e., the radius of the orbit, r = R+h

Considering the orbit to be circular, the centripetal force = \(\frac{m v^2}{r}=\frac{m v^2}{(R+h)}\)

The gravitational force between the earth and the satellite \(\frac{G M m}{r^2}\) supplies this centripetal force.

∴ \(\frac{m v^2}{r}=\frac{G M m}{r^2} \text { or, } v^2=\frac{G M}{r}=\frac{G M}{R+h}\)

or, \(v=\sqrt{\frac{G M}{R+h}}\)…(1)

If the value of the acceleration due to gravity on the earth’s surface is g, then

g = \(\frac{G M}{R^2} \text { or, } G M=g R^2\)

Inserting the value in equation (1)

v = \(\sqrt{\frac{g R^2}{r}}=R \sqrt{g}\)…(2)

If the time period of revolution of the satellite is T, the distance covered in time T by the satellite = the circumference of the orbit = \(2 \pi r\).

∴ T = \(\frac{2 \pi r}{v}=2 \pi r \cdot \frac{1}{R} \sqrt{\frac{r}{g}}=\frac{2 \pi}{R} \sqrt{\frac{r^3}{g}}\)…(3)

Equations (2) and (3) desired above can be written in terms of the distance of the satellite from the earth’s surface (h) as \(v=\sqrt{\frac{g R^2}{R+h}}=R \sqrt{\frac{g}{R+h}}\)…(4)

and T = \(\frac{2 \pi}{R} \sqrt{\frac{(R+h)^3}{g}}=\frac{2 \pi}{R} \sqrt{\frac{R^3\left(1+\frac{h}{R}\right)^3}{g}}=\frac{2 \pi}{R} \cdot R \sqrt{\frac{R\left(1+\frac{h}{R}\right)^3}{g}}\)

= \(2 \pi \sqrt{\frac{R}{g}\left(1+\frac{h}{R}\right)^{3 / 2}}\)….(5)

Equations (1) to (5) show clearly that the orbital speed or the time period of a satellite does not depend on the mass of the satellite at all. It should be mentioned that if the orbit of a satellite is an ellipse of eccentricity e, the highest and the lowest orbital speed will be

⇒ \(v=\sqrt{\frac{g R^2}{R+h}}=R \sqrt{\frac{g}{R+h}}\)

where a is the semi-major axis of the ellipse.

**Artificial Satellite Very Close To The Earth:** From

equations (4) and (5), it is evident that the lower the height (h) of the orbit from the earth’s surface, the higher the value of the orbital speed (v) and the lower the time period (T). Hence, to set a satellite in an orbit very close to the earth’s surface, it is necessary to impart a very high velocity to the satellite.

In the case of such orbits, the value of h is negligible in comparison to the radius R of the earth. That means R+h ≈ R and h = 0 can be substituted in equations (4) and (5) without introducing much error.

Hence, the orbital speed v = \(\sqrt{g R}\)….(6)

and the time period, T = \(2 \pi \sqrt{\frac{R}{g}}\)….(7)

The time period of revolution for such types of satellites depends only on the average density (ρ) of the earth.

As, \(\rho=\frac{3 g}{4 \pi G R} \quad therefore \frac{R}{g}=\frac{3}{4 \pi G \rho}\)

Putting the value of \(\frac{R}{g}\) in equation (7), we get \(T=2 \pi \sqrt{\frac{3}{4 \pi G \rho}} \text { or } T \propto \frac{1}{\sqrt{\rho}}\)

The radius of the earth R = 6400 km = 6.4 x 10^{6 }m, the acceleration due to gravity on the earth’s surface g = 9.8 m · s^{-2}. Substituting these values in equations (6) and (7),

v = \(\sqrt{9.8 \times 6.4 \times 10^6}=7.9 \times 10^3 \mathrm{~m} \cdot \mathrm{s}^{-1}=7.9 \mathrm{~km} \cdot \mathrm{s}^{-1}\)

and T = \(2 \times \pi \times \sqrt{\frac{6.4 \times 10^6}{9.8}}=5078 \mathrm{~s} \text { (approx.) }\)

= \(1 \mathrm{~h} 24 \mathrm{~min} 38 \mathrm{~s}\)

The escape velocity from the earth, \(v_e=\sqrt{2 g R}=11.2 \mathrm{~km} \cdot \mathrm{s}^{-1}\). It can be said that the orbital speed of an artificial satellite close to the earth’s surface \(v=\sqrt{g R}=7.9 \mathrm{~km} \cdot \mathrm{s}^{-1}\) is comparatively less than the escape velocity v_{e} from the earth’s surface.

The ratio of these two velocities, \(\frac{\nu}{v_e}=\frac{\sqrt{g R}}{\sqrt{2 g R}}=\frac{1}{\sqrt{2}}=0.707\)

As the value of h increases, the value of orbital speed v decreases. Hence, the orbital speed decreases further compared to the escape velocity. Hence, to set up an artificial satellite to revolve around the Earth, it is not necessary to impart an orbital speed equal to or greater than the escape velocity.

**Possible Trajectories Of Satellite:** For different velocities, the trajectory of the satellite would be different. Let us discuss these cases.

If v is the velocity given to a satellite, v_{0} represents the orbital speed of the satellite and v’_{e} be the escape velocity at a distance h from the earth’s surface, then

⇒ \(v_o=\sqrt{\frac{G M}{R+h}}, v_e^{\prime}=\sqrt{\frac{2 G M}{R+h}}\) where M and R are the mass and radius of the earth respectively.

**Possible Trajectories Of Satellite Notes: **When, v < v_{0}, the satellite follows an elliptical path with the centre of the earth as the further foci. In this case, if the satellite is projected from near the surface of the earth, it will fall on the earth without completing the orbit.

- If v = v
_{0}, the satellite follows a circular orbit with the centre of the earth as the centre of the orbit. - If v
_{0}< v < v’_{e}, the satellite follows an elliptical path with the centre of the earth as the nearer foci. - If v = v’
_{e}, then the satellite escapes the gravitational field of Earth along a parabolic trajectory. - If v > v’
_{e}, the satellite escapes the gravitational field of Earth along a hyperbolic trajectory.

**Uses Of An Artificial Satellite:** A few uses of artificial satellites are mentioned below:

- Determination of the air pressure, height and composition of the atmosphere at a higher altitude.
- Observation and forecast of weather.
- Defence surveillance.
- Study of the shape and size of the earth.
- Telecommunication. [Signals can be exchanged using artificial satellites to broadcast television shows, games, etc., and also to provide communication by telephone.]
- Collection of data about the ionosphere, cosmic rays, Van Allen radiation belts, effects of solar flares etc.

**Geostationary Or Parking Orbit Geostationary Satellite**

The earth rotates about its own axis (line joining the north and south poles) once in every 24 hours. 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

**Geostationary Satellite 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

As the force of gravitation provides the necessary centripetal force for the revolution in a circular orbit, \(\frac{m v^2}{r}=\frac{G M m}{r^2} \text { or, } v^2=\frac{G M}{r}\)

The acceleration due to gravity on the earth’s surface is g.

So, g = \(\frac{G M}{R^2}\) or, \(G M=g R^2\)

∴ \(v^2=\frac{g R^2}{r} \text { or, } v=\sqrt{\frac{g R^2}{r}}\)….(1)

Also, during the time period of the revolution,

T = \(\frac{2 \pi r}{v}=2 \pi r \sqrt{\frac{r}{g R^2}}=2 \pi \sqrt{\frac{r^3}{g R^2}}\)

or, \(T^2=\frac{4 \pi^2 \cdot r^3}{g R^2} or, r^3=\frac{g R^2 T^2}{4 \pi^2}\)

or, \(r=\left(\frac{g R^2 T^2}{4 \pi^2}\right)^{1 / 3}\)….(2)

For a geostationary satellite, T = 24 h = 24 x 60 x 60 s , g = 9.8 m · s^{-2}, R = 6400 km = 6.4 x 10^{6}m.

Substituting these values, we obtain, r = 4.234 x 10^{7}m = 4.234 x 10^{4} 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, \(\nu=\frac{2 \pi r}{T}=\frac{2 \pi \times 4.234 \times 10^7}{24 \times 60 \times 60}=3079 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

= 3. 079 km · 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.

**Height And Orbital Speed Of A Geostationary Satellite Discussions:**

**Height And Orbital Speed Of A Geostationary Satellite Uses:** Nowadays, geostationary satellites are in extensive use, especially 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.

**Height And Orbital Speed Of A Geostationary Satellite Difficulties In Use:**

- 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 countries. Due to the high orbit, the spatial resolution of data is not as great as for the polar satellites.

- 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 lagfelt 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.

## Unit 6 Gravitation Chapter 1 Newtonian Gravitation And Planetary Motion

## Artificial 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 = 6400km, g = 980cm · s ^{-2}.**

**Solution:**

Radius of the earth, R = 6400 km = 64 x 10^{7} cm

Distance of the artificial satellite from the centre of the earth (r) = 6400 + 300 = 6700 km = 67 x 10^{7} cm.

The orbital speed of the artificial satellite, \(\nu=R \sqrt{\frac{g}{r}}=64 \times 10^7 \sqrt{\frac{980}{67 \times 10^7}}\)

= \(7.74 \times 10^5 \mathrm{~cm} \cdot \mathrm{s}^{-1}=7.74 \mathrm{~km} \cdot \mathrm{s}^{-1}\)

Time period of revolution, T = \(\frac{2 \pi r}{\nu}=\frac{2 \times \pi \times 67 \times 10^7}{7.74 \times 10^5}=5439 \mathrm{~s}\)

= \(1 \mathrm{~h} 30 \mathrm{~min} 39 \mathrm{~s}\)

**Example 2. A man is positioned in a circular orbit at a height of 1.6 x 10 ^{5} m above the surface of the earth. The radius of the earth is 6.37 x 10^{6} m and the mass is 5.98 x 10^{24} kg. What would be the orbital speed of the man? (G = 6.67 x 10^{-11} N · m^{2} · kg^{-2})**

**Solution:**

Radius of the earth, R = 6.37 x 10^{6 }m = 63.7 x 10^{5 }m ; height, h = 1.6 x 10^{5} m

Distance of the man from the centre of the earth (r) = R+h = 63.7 x 10^{5} + 1.6 x 10^{5} = 65.3 x 10^{5} 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 x 10^{4} m · s^{-1} = 7.8 km · s^{-1}.

**Example 3. Prove that, if the orbital speed of the moon increases by 42%, it will stop orbiting around the earth.**

**Solution:**

Let the mass of the earth = M and the radius of the moon’s orbit = r.

Orbital speed of the moon (v) = \(\sqrt{\frac{G M}{r}}\)

The acceleration due to gravity, \(g^{\prime}=\frac{G M}{r^2} \text { or, } G M=g^{\prime} r^2\)

∴ \(\nu=\sqrt{\frac{g^{\prime} r^2}{r}}=\sqrt{g^{\prime} r}\)

Also, the escape velocity of the moon with respect to the earth’s surface, \(v_e=\sqrt{2 g^{\prime} r}\)

∴ \(\frac{v_e}{v}=\sqrt{\frac{2 g^{\prime} r}{g^{\prime} r}}=\sqrt{2}=1.414\)

This implies, v_{e} = 1.414 v = 141.4% of the 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 · s ^{-2}.**

**Solution:**

Radius of the earth, R = 6400 km = 64 x 10^{7} cm

Distance of the artificial satellite from the centre of the earth, r = 6400 + 3400 = 9800 km = 98 x 10^{7} cm

Hence, the orbital speed of the satellite, \(v=R \sqrt{\frac{g}{r}}=64 \times 10^7 \sqrt{\frac{980}{98 \times 10^7}}\)

= 6.4 x 10^{5} cm · s^{-1} = 6.4 km · 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 · s ^{-2})**

**Solution:**

The radius of the earth (RF) = 6300 km = 63 x 10^{5} m.

Distance of the satellite from the centre of the earth, r = 6300 + 700 = 7000 km = 7 x 10^{6} m

Hence, the velocity of the artificial satellite, \(\nu=R \sqrt{\frac{g}{r}}=63 \times 10^5 \sqrt{\frac{9.8}{7 \times 10^6}}=74.543 \times 10^2 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

= \(7.454 \mathrm{~km} \cdot \mathrm{s}^{-1}\)

**Example 6. A small satellite revolves around a plane of density 10 g · 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 x 10^{-8} CGS unit.**

**Solution:**

Let the mass of the planet = M, radius = R and density = ρ.

∴ \(M=\frac{4}{3} \pi R^3 \rho\)

As per the question, the radius of the orbit of the satellite ≅ R. If its mass = m and orbital speed = v, centripetal force = \(\frac{m v^2}{R}\)

The mutual force of gravitation between the planet and satellite provides this centripetal force.

∴ \(\frac{m v^2}{R}=\frac{G M m}{R^2}\)

or, \(v^2=\frac{G M}{R}=\frac{G}{R} \frac{4}{3} \pi R^3 \rho=\frac{4}{3} \pi G \rho R^2\)

or, \(\nu=R \sqrt{\frac{4}{3} \pi G \rho}\)

Time period = \(\frac{\text { circumference of orbit }}{\text { orbital speed }}=\frac{2 \pi R}{\nu}\)

= \(\frac{2 \pi}{\sqrt{\frac{4}{3} \pi G \rho}}=\sqrt{\frac{3 \pi}{G \rho}}=\sqrt{\frac{3 \times \pi}{6.68 \times 10^{-8} \times 10}}\)

= \(3756 \mathrm{~s}=1 \mathrm{~h} 2 \mathrm{~min} 36 \mathrm{~s} .\)

**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 \(\frac{m v^2}{r}=\frac{G M m}{r^2} \text { or, } v^2=\frac{G M}{r}\)

Hence, the kinetic energy of the satellite,

K = \(\frac{1}{2} m v^2=\frac{G M m}{2 r}\)….(1)

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, \(\)…..(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 (for example, a 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 –\(\frac{G M m}{2 4}\), 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.

## Unit 6 Gravitation Chapter 1 Newtonian Gravitation And Planetary Motion

## Kinetic Energy Of Artificial Satellites 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 · 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 = 2 R

Let the mass of the earth = M; the mass of the body = m; velocity of projection from the earth’s surface = v, Hence, its kinetic energy on the earth’s surface = \(\frac{1}{2} m v^2\)

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 = \(\frac{1}{2} m v^2\)–\(\frac{G M m}{R}\)

Again, potential energy at height R = –\(\frac{G M m}{2R}\)

and its total energy at that height = 0 – \(\frac{G M m}{2R}\) = –\(\frac{G M m}{2R}\)

From the law of conservation of energy \(\frac{1}{2} m v^2-\frac{G M m}{R}=-\frac{G M m}{2 R}\)

or, \(\frac{1}{2} m v^2=\frac{G M m}{R}-\frac{G M m}{2 R}=\frac{G M m}{2 R}\)

= \(\frac{G M}{R^2} \cdot \frac{m R}{2}=g \cdot \frac{m R}{2}=\frac{1}{2} m g R\)

∴ \(v^2=g R\)

or, \(v=\sqrt{g R}=\sqrt{980 \times 64 \times 10^7}=79.2 \times 10^4 \mathrm{~cm} \cdot \mathrm{s}^{-1}\)

= \(7.92 \times 10^5 \mathrm{~cm} \cdot \mathrm{s}^{-1}=7.92 \mathrm{~km} \cdot \mathrm{s}^{-1}\)

**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 further increase dh in height, let the increase in potential energy be dW.

∴ dW = \(m g^{\prime} d h=m g \frac{R^2}{(R+h)^2} d h\left[because g^{\prime}=\frac{R^2}{(R+h)^2} \cdot g\right]\)

Hence, the total increase in the potential energy for an increase in height R from the surface of the earth,

⇒ \(\int_0^W d W=\int_0^R m g \frac{R^2}{(R+h)^2} d h\)

or, \( W=m g R^2\left[-\frac{1}{R+h}\right]_0^R=m g R^2\left(-\frac{1}{R+R}+\frac{1}{R}\right)\)

= \(\frac{m g R^2}{2 R}=\frac{1}{2} m g R\)

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, \(\nu^2=g R\)

or, \(v=\sqrt{g R}=7.92 \times 10^5 \mathrm{~cm} \cdot \mathrm{s}^{-1}=7.92 \mathrm{~km} \cdot \mathrm{s}^{-1}\)

**Example 2. Two bodies of masses M and m 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 an infinite distance from each other, both bodies did not have any potential energy. Also, their velocities were zero.

Hence, the 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 = \(\frac{m}{M}\)v

∴ 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.

= \(\frac{1}{2} m \nu^2+\frac{1}{2} M \nu^2-\frac{1}{2} m v^2+\frac{1}{2} M \cdot \frac{m^2}{M^2} \cdot \nu^2\)

= \(\frac{1}{2} m \nu^2\left(1+\frac{m}{M}\right)=\frac{1}{2} m \cdot \frac{v^2\left(1+\frac{m}{M}\right)^2}{1+\frac{m}{M}}=\frac{1}{2} m \cdot \frac{c^2}{\frac{M+m}{M}}\)

= \(\frac{1}{2} \cdot \frac{M m c^2}{M+m}\)

Also, potential energy at a distance r (due to gravitation) = \(-\frac{G M m}{r}\)

Hence, from the law of conservation of energy, \( 0=\frac{1}{2} \frac{m M c^2}{M+m}-\frac{G M m}{r}\)

or, \(\frac{1}{2} \frac{m M c^2}{M+m}=\frac{G M m}{r} \text { or, } c^2=\frac{2 G}{r}(M+m)\)

c = \(\sqrt{\frac{2 G}{r}(M+m)}\)

**Example 3. An artificial satellite revolves around the Earth in a circular orbit Its velocity is half the value of the escape velocity from the Earth,**

- What is the height of the satellite from the earth’s surface?
- If its revolution around the earth is stopped and the satellite is allowed to fall freely towards the earth, what will be the velocity with which it will strike the earth’s surface? (Radius of the earth = 6.4 x 10
^{6}m; g = 9.8 m · s^{-2})

**Solution:**

1. Orbital speed of the satellite = \(u=\frac{1}{2} \times \text { escape velocity }\) = \(\frac{1}{2} \times \sqrt{2 g R}=\sqrt{\frac{g R}{2}}\)

Again, if it is at a height h above the earth’s surface,

u = \(R \sqrt{\frac{g}{R+h}}=\sqrt{\frac{g R^2}{R+h}}\)

∴ \(\sqrt{\frac{g R}{2}}=\sqrt{\frac{g R^2}{R+h}} \text { or, } \frac{g R}{2}=\frac{g R^2}{R+h} \text { or, } R+h=2 R\)

∴ h = \(R=6.4 \times 10^6 \mathrm{~m}=6400 \mathrm{~km} .\)

2. 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 potential energy = \(\frac{G M m}{2R}\) (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 = \(\frac{1}{2} mv^2\). 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 = \(\frac{1}{2} mv^2\)\(-\frac{G M m}{R}\)

From the law of conservation of energy, \(-\frac{G M m}{2 R}=\frac{1}{2} m v^2-\frac{G M m}{R}\)

or, \(\frac{1}{2} m v^2=\frac{G M m}{R}-\frac{G M m}{2 R}=\frac{G M m}{2 R}\)

or, \(v^2=\frac{G M}{R}=\frac{g R^2}{R}=g R\)

or, \(v=\sqrt{g R}=\sqrt{9.8 \times 6.4 \times 10^6}\)

= \(7.9 \times 10^3 \mathrm{~m} \cdot \mathrm{s}^{-1}=7.9 \mathrm{~km} \cdot \mathrm{s}^{-1} .\)

**Example 4. Calculate the kinetic energy, potential energy and total energy of a geostationary satellite of mass 100 tonnes. The radius of the earth is 6400 km and the radius of the orbit of the satellite is 42400 km.**

**Solution:**

100 tone = 100 x 10^{3} kg = 10^{5} kg. The radius of the earth = 6400 km = 6.4 x 10^{6} m and the distance of the geostationary satellite from the centre of the earth, r = 42400 km = 4.24 x 10^{7} m.

Hence, kinetic energy

= \(\frac{G M m}{2 r}=\frac{1}{2} m \frac{G M}{R^2} \cdot \frac{R^2}{r}\)

= \(\frac{m g R^2}{2 r}=\frac{10^5 \times 9.8 \times\left(6.4 \times 10^6\right)^2}{2 \times 4.24 \times 10^7}\)

= \(4.73 \times 10^{11} \mathrm{~J}\)

Potential energy = \(-\frac{G M m}{r}=-2 \times \frac{G M m}{2 r}\)

= \(-2 \times \frac{10^5 \times 9.8 \times\left(6.4 \times 10^6\right)^2}{2 \times 4.24 \times 10^7}\)

= \(-9.47 \times 10^{11} \mathrm{~J}\)

Total energy =\(-\frac{G M m}{2 r}=-4.73 \times 10^{11} \mathrm{~J} \text {. }\)

**Weightlessness In Artificial Satellites:** It is known that the weight of a body of mass m is mg, i.e., the earth pulls the body towards its centre with this force. When this body is placed on a non-accelerating plane, it exerts a force of mg on that plane.

- The plane also exerts an upward reaction force on the body R(= mg). This upward reaction force is the apparent weight of the body. The body feels only this upward reaction force as its weight. Hence, any person or body cannot feel its own weight while standing on a plane if the plane does not offer any upward reaction; this means that the person or body feels weightless.
- Let the plane along with the body have an acceleration vertically upwards. In this case, the upward reaction of the plane does not become equal to the weight. Because of this a passenger in a lift with an acceleration feels his weight to be increased or decreased.
- There are a few cases in which the apparent weight of a body is not related to the existence of the plane (on which the body is placed). But it is convenient to measure the apparent weight of a body taking into consideration the existence of such a plane.

**Weight Experienced By An Astronaut Before The Spaceship Is Parked In Its Orbit:** The velocity of the spaceship is increased rapidly by the rocket immediately after its launch from the surface of the earth. Sometimes the value of its upward acceleration may be 15 times the value of the acceleration due to gravity. If the mass of the astronaut is m, the upward reaction of the spaceship on the man is R, and the acceleration of the spaceship is 15g, then

R – mg – ma = m · 15g or, R = 16mg

This is the apparent weight of the astronaut. Thus he feels himself to be 16 times heavier than his actual weight.

**Weight Experienced By An Astronaut When The Spaceship Is Moving In Its Orbit:**

Let r = radius of the circular orbit of the spaceship with respect to the centre of the earth,

g’ = acceleration due to gravity at the position of the spaceship,

v = instantaneous velocity of the spaceship,

M = mass of the spaceship, m = mass of the astronaut,

R = upward reaction of the floor of the spaceship on the astronaut

For the rotation of the spaceship or of the astronaut around the earth, the force acting towards the centre of the earth supplies the necessary centripetal force. So, for the spaceship and for the astronaut, respectively,

⇒ \(M g^{\prime}=\frac{M v^2}{r}, \text { or } g^{\prime}=\frac{v^2}{r}\)

and \(m g^{\prime}-R=\frac{m v^2}{r} \text { or, } m \frac{v^2}{r}-R=\frac{m v^2}{r}\)

Thus, R = 0

The astronaut cannot feel his own weight as the reaction force is zero. He floats in the spaceship. This is called the weightlessness of man or any object in an artificial satellite.

Similarly, if it is assumed that the spaceship is resting on an imaginary plane and the motion of the spaceship along with its plane is considered, the spaceship also appears to be weightless.

- As the plane is imaginary, it can be concluded that all objects (planet, satellite, etc.) orbiting under the action of gravitation do not experience any force directed towards the centre of the orbit, and therefore, all such objects are weightless.
- An astronaut feels much heavier than his actual weight at the time of launching the spaceship, but he feels weightless as soon as the spaceship is placed in its orbit.
- The phenomenon of weightlessness is widely used in many cases of scientific and technological requirements. For example, crystal formation on the earth’s surface due to gravitation is imperfect, but perfect crystals can be developed in orbiting spaceships easily and such crystals have immense importance in making transistors and other such electronic devices.
- Again, oil and water cannot really be mixed on the earth’s surface, but in a spaceship, in the weightless condition, it is possible to make a homogeneous mixture of oil and water. In addition, a pure alloy can be produced by cooling a homogeneous mixture of molten metals in a spaceship.
- As in the case of materials kept in an artificial satellite, objects on the moon’s surface also have no weight due to the gravitational pull of the Earth. But nobody is weightless on the moon’s surface as the attraction of the moon acts on the body.
- This weight is about| th of its weight on the earth. It is to be noted that bodies in an artificial satellite are also subject to the force of attraction of the satellite or spaceship. But this force is too small to be felt.
- The gravitational force of attraction of the earth on an artificial satellite or on an astronaut can never be zero. If it had been zero, the necessary centripetal force could not have been supplied and the satellite would not have been able to revolve around the earth in a circular orbit.

Thus the statement that a body is weightless implies that only the reaction force acting on it is zero, but the gravitational force is not zero.

## Unit 6 Gravitation Chapter 1 Newtonian Gravitation And Planetary Motion

## Momentum Of Satellite Numerical Examples

**Example 1. Two particles of equal mass revolve in a circular path of radius R due to their mutual force of attraction. Find the velocity of each particle.**

**Solution:**

At any time, during the revolution, the two particles stay at the two ends of any diameter of the circular path. The mutual force of gravitation acts along the diameter.

Considering the motion of any one of the particles, \(\frac{G m \times m}{(2 R)^2}=\frac{m v^2}{R} \text { or, } v^2=\frac{G m}{4 R} \text { or, } \nu=\sqrt{\frac{G m}{4 R}} \text {. }\)

**Example 2. The acceleration due to gravity at two places are g and g’ respectively. A body is dropped from the same height at both places. In the second place, the required time to touch the ground is t seconds less than that in the first place, while the velocity attained in reaching the ground is higher by a value of v than that in the first place. Show that gg’ = \(\frac{v^2}{t^2}\).**

**Solution:**

Let the time taken by the body to fall at the first place be T and the velocity with which the body touches the ground be V.

Hence, the time taken to reach the ground and the corresponding velocity are (T- t) and V+ v in the second place. Let the body fall from a height h in each case. Considering the motion in the first place,

h = \(\frac{1}{2} g T^2 \text { or, } T=\sqrt{\frac{2 h}{g}}\)…(1)

and \(V^2=2 g h or, V=\sqrt{2 g h}\)….(2)

Similarly, for motion at the second place, h = \(\frac{1}{2} g^{\prime}(T-t)^2 \text { or, } T-t=\sqrt{\frac{2 h}{g^{\prime}}}\)

and \((V+v)^2=2 g^{\prime} h or, V+v=\sqrt{2 g^{\prime} h}\)

Subtracting (3) from (1) \(t=\sqrt{\frac{2 h}{g}}-\sqrt{\frac{2 h}{g^{\prime}}}=\sqrt{2 h}\left(\frac{1}{\sqrt{g}}-\frac{1}{\sqrt{g}}\right)\)

Subtracting (2) from (4) \(v=\sqrt{2 g^{\prime} h}-\sqrt{2 g h}=\sqrt{2 h}\left(\sqrt{g^{\prime}}-\sqrt{g}\right)\)

∴ \(\frac{v^2}{t^2}=\frac{2 h\left(\sqrt{g^{\prime}}-\sqrt{g}\right)^2}{2 h\left(\frac{\sqrt{g^{\prime}}-\sqrt{g}}{\sqrt{g g^{\prime}}}\right)^2}=g g^{\prime} .\)

**Example 3. A satellite of mass m revolves around the earth in a circular orbit of radius r. Find the angular momentum of the satellite with respect to the centre of the orbit.**

**Solution:**

The angular momentum of the satellite about the centre of the orbit is L = mvr, where v is the orbital speed of the satellite. For motion in a fixed orbit, centripetal force = force of attraction due to gravity

or, \(\frac{m v^2}{r} = \frac{G M m}{r^2}\)

or, \(m v^2 r=G M m \text { or, }(m v r)^2=G M m^2 r\)

or,\(m v r=m \sqrt{G M r}\)

∴ L = \(m \sqrt{G M r}\)

**Example 4. A satellite is orbiting in a circular path around the earth close to its surface. What additional velocity Is to be imparted to the satellite so that it escapes the Earth’s gravitational pull? The radius of the earth = 6400 km and g = 9.8 m · s ^{-2}.**

**Solution:**

The orbital speed of a satellite in an orbit close to the surface of the earth \(v=\sqrt{g R}\) escape velocity \(v_e=\sqrt{g R}\)

∴Required additional velocity

= \(v_e-v=\sqrt{2 g R}-\sqrt{g R}\)

= \((\sqrt{2}-1) \sqrt{g R}=0.41 \times \sqrt{9.8 \times 6400 \times 1000}\)

= \(3.25 \times 10^3 \mathrm{~m} \cdot \mathrm{s}^{-1}=3.25 \mathrm{~km} \cdot \mathrm{s}^{-1} \).

**Example 5. Two satellites A and B of the same mass revolve around the Earth in circular orbits. They are at heights of R and 3R respectively from the surface of the earth (R = radius of the earth). Find the ratio of their kinetic energies and potential energies.**

**Solution:**

The height of satellite A from the centre of the earth = R + R = 2R and the height of satellite B from the centre of the earth = R + 3R = 4R. If the mass of each satellite is m and the mass of the earth is M,

potential energy of satellite A, U_{A} = –\(\frac{G M m}{2 R}\)

and potential energy of satellite B, U_{B} = –\(\frac{G M m}{4 R}\)

∴ \(\frac{U_A}{U_B}=\frac{G M m / 2 R}{G M m / 4 R}=\frac{2}{1}\)

If the orbital speeds of the two satellites are v_{1}^{ }and v_{2}, then the kinetic energy of satellite A, K_{A} = \(\frac{1}{2} m v_1^2\)

and kinetic energy of B, K_{B} = \(\frac{1}{2} m v_2^2\)

A, \( v_1=\sqrt{\frac{G M}{2 R}} \text { and } v_2=\sqrt{\frac{G M}{4 R}}\)

∴ \(\frac{K_A}{K_B}=\frac{v_1^2}{v_2^2}=\frac{G M / 2 R}{G M / 4 R}=\frac{2}{1} .\)

**Example 6. Two satellites S _{1} and S_{2} are orbiting around the Earth in circular orbits in the same direction. Time period for the two satellites is 1h and 8h respectively. The radius of the orbit of satellite S_{1} is 10^{4} km. If satellites S_{1} and S_{2} are on the same side of the earth, find the linear and angular speeds of S_{2} with respect to S_{1}.**

**Solution:**

If the radii of satellites S_{1}_{ }and S_{2} are r_{1 }and r_{2} and their respective time periods are T_{1} and T_{2}, then

⇒ \(T_1^2 \propto r_1^3 \text { and } T_2^2 \propto r_2^3\)

∴ \(\frac{T_1^2}{T_2^2}=\frac{r_1^3}{r_2^3} \text { or, }\left(\frac{1}{8}\right)^2=\left(\frac{10^4}{r_2}\right)^3\)

or, \(\frac{1}{64}=\left(\frac{10^4}{r_2}\right)^3 \text { or, }\left(\frac{1}{4}\right)^3=\left(\frac{10^4}{r_2}\right)^3\)

∴ \(r_2=4 \times 10^4 \mathrm{~km}\)

The speed of satellite \(S_1\), \(v_1=\frac{2 \pi r_1}{T_1}\) = \(\frac{2 \pi \times 10^4}{1}\)

= \(2 \pi \times 10^4 \mathrm{~km} \cdot \mathrm{h}^{-1}\)

and the speed of satellite \(S_2\), \(v_2=\frac{2 \pi r_2}{T_2}=\frac{2 \pi \times 4 \times 10^4}{8}\)

= \(\pi \times 10^4 \mathrm{~km} \cdot \mathrm{h}^{-1}\)

∴ Linear speed of \(S_2\) with respect to \(S_1\)

= \(\left|v_2-v_1\right|=\pi \times 10^4 \mathrm{~km} \cdot \mathrm{h}^{-1}\)

and angular speed of \(S_2\) with respect to \(S_1\),

ω = \(\frac{\left|v_2-v_1\right|}{r_2-r_1}=\frac{\pi \times 10^4}{4 \times 10^4-10^4}\)

= \(\frac{\pi}{3} \mathrm{rad} \cdot \mathrm{s}^{-1}.\)