Contents
Many modern technologies, such as computers and smartphones, are built on the principles of Physics Topics such as quantum mechanics and information theory.
Why is Light Gases Rare in Earth Atmosphere? What are the Uses of Artificial Satellites?
When an object is thrown vertically upwards, it moves up to a certain height and then comes down to the earth’s surface due to the gravitational pull of the earth. At the maximum height, the velocity of the object becomes zero. More the projection velocity, greater the height to which the object rises before its velocity becomes zero. If the velocity of projection continuously increases, a state is reached when the velocity of the object (for a finite height) does not become zero. This implies that it does not return to the earth’s surface.
At this state, the object recedes from the earth’s surface and the force of gravity also decreases to almost a negligible value. Hence, it can be said that the body reaches an infinite distance from the earth. The minimum velocity of projection required for this to happen is called the escape velocity. It is to be noted that this concept of escape velocity is not only applicable to objects projected from the earth’s surface but also valid for objects projected from the surface of other celestial bodies.
Definition: The minimum velocity required to project a body from the surface of the earth or a planet or a satellite such that it can escape the gravitational attraction of the earth or the planet or the satellite is called escape velocity.
Receding from the gravitational attraction of a planet or satellite implies that the projected body does not return to the planet or satellite any more.
Let the mass of the earth = M and its radius = R. The force of gravity on a body of mass m at a distance x from the
centre of the earth = \(\frac{G M m}{x^2}\).
For a small displacement dx of the body against this force of gravity, work done = \(\frac{G M m}{x^2}\)dx.
Hence, the work done to displace the body from the earth’s surface to an infinite distance
= \(\int_R^{\infty} \frac{G M m}{x^2} d x\) = GMm\(\left[-\frac{1}{x}\right]_R^{\infty}\) = \(\frac{G M m}{R}\)
If the escape velocity from the earth is ve, the initial kinetic energy of the body is \(\frac{1}{2} m v_e^2\). When this energy does the work as body recedes from the earth’s gravitational attraction.
∴ \(\frac{1}{2} m v_e^2\) = \(\frac{G M m}{R}\) or, \(v_e^2\) = \(\frac{2 G M}{R}\)
or, ve = \(\sqrt{\frac{2 G M}{R}}\) ….. (1)
The acceleration due to gravity on the earth’s surface
g = \(\frac{G M}{R^2}\) or, GM = gR2
∴ ve = \(\sqrt{2 g R}\) ….. (2)
Equation (2) gives the value of the escape velocity for a body from the earth’s surface. Note that, the escape velocity does not depend on the mass of the body. This means that for a small and a massive body, the value of the escape velocity is the same.
It is known that the acceleration due to gravity on the earth’s surface g = 9.8 m ᐧ s-2 and the radius of the earth R = 6400 km = 6.4 × 106m.
∴ ve = \(\sqrt{2 \times 9.8 \times 6.4 \times 10^6}\)
= 11200 m ᐧ s-1 = 11.2 km ᐧ s-1
Hence, if a body of any mass is projected upwards from the surface of the earth with a velocity of 11.2 km ᐧ s-1, it moves out of the gravitational field of the earth, i.e., the body does not return to the earth’s surface.
Equation (2) can also be used for calculating the escape velocity from the surface of any planet or satellite by putting the corresponding values of g and R. For example, the value of g on the surface of the moon is \(\frac{1}{6}\) th of the value on the surface of the earth and the moon’s radius is \(\frac{1}{4}\) th that of the earth. Putting these values in equation (2), the value of the escape velocity from the moon’s surface can be evaluated and the value is about 2.4 km ᐧ s-1. This value is about \(\frac{1}{5}\) th the value of the escape velocity from the earth’s surface.
Scarcity of light gases in earth’s atmosphere:
When the earth was formed, the earth’s atmosphere was filled with lighter gases such as hydrogen and helium along with heavier gases like oxygen, nitrogen, etc. At present, there is practically no existence of lighter gases. Scarcity of lighter gases can be explained in the light of escape velocity.
From the kinetic theory of gases, it is known that the rms speed of hydrogen at STP is nearly 1.6 km ᐧ s-1. During the formation stage, the earth was warmer and the rms speed of hydrogen had a value close to 5 km ᐧ s-1. Obviously, many hydrogen molecules had velocities more than or equal to the escape velocity (11 km ᐧ s-1). Hence, gradually with the passage of time, most of the hydrogen molecules escaped the gravitational field of the earth and went into space. Same was the fate of helium. On the other hand, heavy molecules have very low rms velocity of thermal motion. As the value is much lower than the value of the escape velocity, heavy gas molecules remained confined to the earth’s atmosphere. Hence, at present, the earth’s atmosphere is primarily made up of heavier gases like oxygen, nitrogen and carbon dioxide.
In the case of the moon or of Mercury, owing to their low masses and shorter radii, the value of the escape velocity from the surface is low. As the rms speeds of both the lighter and heavier molecules are comparable to the escape velocity, these gas molecules could escape. Hence, there is no atmosphere on the moon or on Mercury. On the other hand, the escape velocity for Jupiter or Saturn is of much higher value because of the higher mass and size. Thus, neither lighter nor heavier gas molecules could escape the gravitational attraction of these planets. So, there is an abundance of hydrogen and helium gas on these planets.
Numerical Examples
Example 1.
What is the escape velocity of a meteorite situated 1800 km above the surface of the earth? Given, the radius of the earth = 6300 km and the acceleration due to gravity on the earth’s surface = 9.8 m ᐧ s-2.
Solution:
Radius of the earth (R) = 6300 km = 63 × 105 m
Distance of the meteorite from the centre of the earth (r) = 6300 + 1800 = 8100 km = 81 × 105 km
Acceleration due to gravity on the earth’s surface, g = acceleration due to gravity at the position of the meteorite, g’ = \(\frac{G M}{r^2}\).
Example 2.
A man can jump up to a height of 1.5 m on the earth’s surface. What should be the radius of a planet having the same average density as that of the earth so that the man can come out of the gravitational field of that planet in one jump? Radius of the earth is 6400 km. [HS 05]
Solution:
The man can jump up to a height h on the earth’s surface.
Hence, his initial kinetic energy = his potential energy at height h.
Thus, \(\frac{1}{2}\)mv2 = mgh [m = mass of the man,
v = initial velocity]
or, v = \(\sqrt{2 g h}\) ……. (1)
The man can jump on the surface of any planet with this velocity v. If v equals the escape velocity from a planet, the man can go out of the gravitational pull of that planet. Let the acceleration due to gravity for a planet be g’, its radius R’ and average density ρ, then
g’ = \(\frac{4}{3} \pi G R^{\prime} \rho\)
As \(\frac{4}{3}\)πG and ρ are constants, g’ ∝ R’.
Hence, \(\frac{g^{\prime}}{g}\) = \(\frac{R^{\prime}}{R}\) or, g’ = gᐧ\(\frac{R^{\prime}}{R}\)
[R = radius of the earth]
Escape velocity for the planet is v = \(\sqrt{2 g^{\prime} R^{\prime}}\) ………. (3)
From (1) and (3),
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. Presendy technological expertise have 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 itself in the desired orbit and start revolving around the earth.
Obviously, once in an orbit, any satellite follows Kepler’s laws.
Hence, the orbit of an artificial satellite can also be elliptical or circular like the orbits of planets and satellites. But 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 [Fig.].
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.
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 = 2πr.
∴ T = \(\frac{2 \pi r}{v}\) = 2πr ᐧ \(\frac{1}{R} \sqrt{\frac{r}{g}}\) = \(\frac{2 \pi}{R} \sqrt{\frac{r^3}{g}}\) ….. (3)
Equations (2) and (3) derived above can be written in terms of the distance of the satellite from the earth’s surface (h) as
v = \(\sqrt{\frac{g R^2}{R+h}}\) = \(\sqrt{\frac{g}{R+h}}\) …….. (4)
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
vmax = \(\sqrt{\frac{G M}{a(1-e)}}\) and vmin = \(\sqrt{\frac{G M}{a(1+e)}}\),
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 will be 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 \(\simeq\) 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 = 2π\(\sqrt{\frac{R}{g}}\) …… (7)
The time period of revolution for such type of satellites depends only on the average density (ρ) of the earth.
Putting the value of – in equation (7), we get,
T = 2π\(\sqrt{\frac{3}{4 \pi G \rho}}\) or T ∝ \(\frac{1}{\sqrt{\rho}}\)
The radius of the earth R = 6400 km = 6.4 × 106m, the acceleration due to gravity on the earth’s surface g = 9.8 m ᐧ s-2. Substituting these values in equations (6) and (7),
v = \(\sqrt{9.8 \times 6.4 \times 10^6}\) = 7.9 × 103 m ᐧ s-1 = 7.9 km ᐧ s-1
and T = 2 × π × \(\sqrt{\frac{6.4 \times 10^6}{9.8}}\) = 5078 s (approx.)
= 1 h 24 min 38 s
The discussions in Section 1.9 established that the escape velocity from the earth, ve = \(\sqrt{2 g R}\) = 11.2 km ᐧ 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 km ᐧ s-1) is comparatively less than the escape velocity v<sub.e from the earth’s surface. Ratio of these two velocities,
\(\frac{v}{v_e}\) = \(\frac{\sqrt{g R}}{\sqrt{2 g R}}\) = \(\frac{1}{\sqrt{2}}\) = 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, v0 represents the orbital speed of the satellite and v’e be the escape velocity at distance h from the earth’s surface, then
v0 = \(\sqrt{\frac{G M}{R+h}}\), v’e = \(\sqrt{\frac{2 G M}{R+h}}\)
where M and R are the mass and radius of the earth respectively.
Notes:
- When, v < v0, the satellite follows an elliptical path with centre of the earth as the further foci. In this case, if sat-ellite is projected from near surface of the earth, it will fall on the earth without completing the orbit.
- If v = v0, the satellite follows a circular orbit with centre of earth as the center of the orbit.
- If v0 < v < v’e, the satellite follows an elliptical path with 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 flare etc.