derive an expression for escape velocity and orbital velocity escape velocity of earth formula derivation definition class 11 moon ?
Newton’s Law of Gravitation
Newton’s law of universal gravitation states as follows: ‘Any two particles of matter anywhere in the universe attract each other with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them, the direction of the force being along the line joining the particles, i.e.
where F is the magnitude of the force of attraction between two particles of masses m1 and m2 separated by a distance r. In the form of an equation the law is written as
where G is a constant called the universal gravitation constant. The value of this constant is to be determined experimentally and is found to be
Gravitational Force due to Multiple Masses
If a system consists of more than two masses, the gravitational force experienced by a given mass due to all other masses is obtained from the principle of superposition which states that ‘the gravitational force experienced by one mass is equal to the vector sum of the gravitational forces exerted on it by all other masses taken one at a time.’
The gravitational force on mass m due to masses m1, m2, m3, … mn is given by (Fig. )
note : 1. Gravitational force is alway attractive.
2. Gravitational force between two masses does not depend on the medium between them.
3. Gravitational force acts along the straight line joining the centres of the two bodies.
Acceleration due to Gravity
Considering the earth as an isolated mass, a force is experienced by a body near it. This force is directed towards the centre of the earth and has a magnitude mg, where g is the acceleration due to gravity
where M is the mass of the earth and R its radius (nearly constant for a body in the vicinity of the earth)
All bodies near the surface of the earth fall with the same acceleration which is directed towards the centre of the earth.
Variation of g
1. Variation with altitude
The acceleration due to gravity of a body at a height h above the surface of the earth is given by
where g is the acceleration due to gravity on the surface of the earth. If h is very small compared to R, we can use binomial expansion and retain terms of order h/R. We then get
Thus, the acceleration due to gravity decreases as the altitude (h) is increased.
2. Variation with depth
The acceleration due to gravity at a depth d below the surface of the earth is given by
This equation shows that the acceleration due to gravity decreases with depth. At the centre of the earth where d = R, gd = 0. Thus the acceleration due to gravity is maximum at the surface of the earth, decreases with increase in depth and becomes zero at the centre of the earth
3. Variation with Latitude
Due to the rotation of earth about its axis, the value of g varies with latitude, i.e. from one place to another on the earth’s surface. At poles, the effect of rotation on g is negligible. At the equator, the effects of rotation on g is the maximum. In general, the value of acceleration due to gravity at a place decreases with the decrease in the latitude of the place. The accelration due to gravity at a place on earth where the latitude is φ is given by
Thus the value of g varies slightly from place to place on earth. Variation of g with altitude and depth is shown in Fig.
Gravitational Field Intensity
Just as the region around a magnet has magnetic field and the region around a charge has electric field, the region around a mass has gravitational field. The gravitational field intensity (or simply gravitational field) at a point is defined as the gravitational force experienced by a unit mass placed at that point.
Consider the gravitational field of a body of mass M. To find the strength of the field at a point P at a distance r from M, we place a small mass m at P. The gravitational force exerted m by M is
By definition, the gravitational field (intensity) of M at P is given by
I is a vector quantity. In vector form
where r is a unit vector directed from M to P, i.e radially away from M. The negative sign indicates that I directed radially inwards towards M. The SI unit of I is N kg–1. In three dimensions, if mass M is located at the origin, the magnetic field at a P (x, y, z) is given by
where r = xi + y j + zk represents the position of point P with respect to mass M at the origin. For a many body system, the principle of superposition holds for gravitational field (intensities) just as it holds for gravitational forces, i.e.
where I1, I2, … In are the gravitational field intensities at a point due to bodies of masses M1, M2, … Mn. For continuous mass distributions (i.e rigid bodies), we divide the body into an infinitely large number of infinitesimally small elements. Then the gravitational field intensity is given by
Gravitational Field due to some continuous Mass Distributions
1. Gravitational field due to a circular ring of mass M and radius R at a point at a distance r from the centre and on the axis of the ring is given by
2. Gravitational field due to a thin spherical shell of mass M and radius R at a point P at a distance r > R from the centre of shell,
Inside the shell, I = 0
3. Gravitational field of a solid sphere of mass M and radius R is
Figure shows the variation of the gravitational field of a solid sphere.
6. Gravitational Potential Energy
Gravitational potential energy of a system of two masses M and m held a distance r apart is defined as the amount of work done to bring the masses from infinity to their respective locations along any path and without any acceleration.
Work done to bring mass M from infinity to A is W1 = 0. Work done to bring mass m from r = ∞ to r = r is
Since mass M will attract mass m, angle θ between F and dr is zero. Hence
The zero of potential energy is assumed to be at r = ∞. The negative sign indicates the potential energy is negative for any finite separation between the masses and increases to zero at infinite separation.
Expression for Increase in Gravitational Potential Energy
If the body m is moved away from M, the potential energy of the system increases.
If the body of mass M is the earth, then the increase in gravitational P.E. when a body of mass m is taken from the surface of the earth to a height h above the surface is given by (see Fig. 6.11), R = radius of the earth.
Gravitational potential at a point P in the gravitational field of a body of M is defined as the amount of work done to bring a unit mass from infinity to that point along any path and without any acceleration, i.e., (Fig. below )
Potential V is a scalar quantity. Hence the gravitational potential at a point P due to a number of masses m1, m2, … mn at distances r1, r2, … rn respectively from P is given by
Relation between Gravitational Field Intensity (I) and Gravitational Potential (V) Gravitational field intensity and gravitational potential at a point are related as
Gravitational Potential due to a Spherical Shell
Figure shows the variation V with r for a spherical shell.
Gravitational Potential due to a Solid Sphere of Mass M and Radius R
The escape velocity is the minimum velocity with which a body must be projected in order that it may escape the earth’s gravitational pull. The magnitude of the escape velocity is given by
where M is the mass of the earth and R its radius. Substituting the known values of G, M and R, we get ve = 11.2 kms–1. The escape velocity is independent of the mass of the body. The expression for the escape velocity can be written in terms of g as
The escape velocity is independent of the mass of the body and the direction of projection.
A body moving in an orbit around a much larger and massive body is called a satellite. The moon is the natural satellite of the earth.
Orbital Velocity Let us assume that a satellite of mass m goes around the earth in a circular orbit of radius r with a uniform speed v. If the height of the satellite above the earth’s surface is h, then r = (R + h), where R is the mean radius of the earth. The centripetal force m v2 /r necessary to keep the satellite in its circular orbit is provided by the gravitational force GmM/r2 between the earth and the satellite. This means that
If the satellite is a few hundred kilometres above the earth’s surface (say 100 to 300 km), we can replace (R + h) by R. The error involved in this approximation is negligible since the radius of the earth (R) = 6.4 x 106 m. Thus, we may write
The periodic time T of a satellite is the time it takes to complete one revolution and it is given by (since r = R + h)
note 1. E = P.E./2
2. E = – (K.E.)
3. The total energy is negative which implies that the satellite is bound by the gravitational field of the earth. The binding energy = GmM/2r . This energy must be given to the orbiting satellite to escape to infinity
The magnitude of angular momentum of a satellite is given by
Geostationary Satellites A geostationary satellite is a particular type used in communication. A number of communication satellites are launched which remain in fixed positions at a specified height above the equator. They are called geostationary or synchronous satellites used in international communication. For a satellite to appear fixed at a position above a certain place on the earth, it must corotate with the earth so that its orbital period around the earth is exactly equal to the rotational period of the earth about its axis of rotation. This condition is satisfied if the satellite is put in orbit at a height of about 36,000 km above the earth’s surface.
10. Kepler’s Laws of Planetary
Motion Johannes Kepler formulated three laws which describe planetary motion. They are as follows: 1. Law of orbits Each planet revolves about the sun in an elliptical orbit with the sun at one of the focii of the ellipse. The orbit of a planet is shown in Fig. (a) in which the two focii F1 and F2, are far apart. For the planet earth, F1 and F2 are very close together. In fact, the orbit of the earth is practically circular. 2. Law of areas A line drawn from the sun to the planet (termed the radius) sweeps out equal areas in
equal intervals of time. In Fig. 6.18(b) P1, P2, P3 and P4 represent positions of a planet at different times in its orbit and S, the position of the sun. According to Kepler’s second law, if the time interval between P1 and P2 equals the time interval between P3 and P4, then area A1 must be equal to area A2. Also the planet has the greater speed in its path from P1 to P2 than in its path from P3 to P4.
3. Law of periods The squares of the periods of the planets are proportional to the cubes of their mean distances from the sun. If T1 represents the period of a planet about the sun, and r1 its mean distance, then T1 2 ∝ r1 3 If T2 represents the period of a second planet about the sun, and r2 its mean distance, then for this planet
T2 2 ∝ r2 3
These two relations can be combined since the factor of proportionality is the same for both. Thus
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