EDUACHIEVERS ACADEMY
9
Gravitation
GRAVITATION Introduction : Any two objects in the universe attract each other with force which is called the gravitational force. It is the gravitational force that binds us to the earth and makes the earth and the other planets revolve around the sun. It controls the structure and the evolution of the entire universe. Newton's Law of Gravitation : The gravitational force of attraction between any two particles is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. 1 F m1 m2 , F 2 i.e., r Thus the force F between two particles of masses m1 and m2, separated by a distance r, is given by F G
Newton’s law of gravitation is called gravitational mass. It is found that the two masses are equivalent and therefore, need not be distinguished. Earth’s Gravitational Field : Let us assume that the earth is a uniform sphere of mass M and radius R. It can be shown that a uniform sphere produces gravitational field outside it as if whole of its mass is concentrated at its centre. Therefore, the force experience by an object of mass m at a distance r (>R) from the centre of the earth is F =
F
A U D E g
D A C A
GMm
mg r2 This gives the acceleration due to earth's gravity as
S R
E V
IE H C
as
g
Inertial and Gravitational Masses : The mass that appears in Newton’s second law of motion, F = ma, measures the inertia of the body against the action of a force. Therefore, it is called inertial mass. On the other hand, the mass of an object appearing in
GM
Fig. 9.1 r Variation of 'g' : The acceleration due to gravity is expressed
g
F m
where F is the attractive force exerted by the earth on a body of mass m. This force is affected by a number of factors and hence g also depends on these factors. (A) Above the surface of earth : At a point which is located at a height h above the earth’s surface, acceleration due to gravity (g') is given as GMm force ( R h) 2 GM g mass m ( R h) 2
F m
This gives F = m0g Using Newton’s second law we see that the acceleration produced by the gravitational force on the test mass is g. Therefore, g is also called the acceleration due to gravity. The gravitational field strength due to a mass M at a distance r from it is given by GM g r
M E
Y
. r2 This force is also called the weight of the object. If 'g' is the acceleration produced by this force, then
m m
r Where G is called the universal gravitational constant. Its value in S.I. system is 6.67 10–11 Nm2/kg2. The constant G is scalar and possess the dimension [M–1L3T–2]. It is a universal constant because its value does not depend on the type of particles, the nature of the invening medium, or temperature etc. If m1 = m2 = 1 and r = 1, then G = F. Hence constant of gravitation is numerically equal to the force of attraction between two unit masses placed at unit distance apart. The gravitational force is a central force, as it acts along the line joining the two particles. Gravitational Field and Acceleration due to Gravity : A gravitational field is said to exist in a region if a gravitational force is experienced by a test mass placed at any point in that region. If a test mass mo placed at a point in a gravitational field experiences a force F, then the gravitational field strength g at that point is defined as
GMm
F R IJ g g G H R hK
2
where g = acceleration due to gravity on the surface of earth, i.e., h
g
g'
GM R
R–h
g h R
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Fig. 9.2 1
EDUACHIEVERS ACADEMY g ' g . When we move above the surface of earth, acceleration due to gravity goes on decreasing, e.g., at a height equal to radius of the earth g
g
FG1 R IJ H RK
2
g. 4
If the height h is negligible compared to radius R of the earth then expanding the above equation, by binomial theorem and terms containing higher powers of h/R are neglected h g ' g 1 R
2
2h g 1 R
(B) Below the surface of earth : Let us place a mass m at a depth h below the surface of earth. This mass m is attracted towards the centre of earth by gravitational force due to shaded sphere of radius (R – h). The net gravitational force on m due to the remaining portion of earth (outside the shaded portion) is zero. Let the mass of the shaded portion = M Assuming earth to be sphere of uniform density d, d
M 4 R3 3
=
( R h) d
=
M M ( R h) ( R h) R R
Force =
( R h)
( R h) R
A U D E
Acceleration due to gravity = g'
GMm ( R)
( R h)
FG H
force on m GMm ( R h ) GM h g' 2 1 m R R3 m R h g' g R
GMm
, which is the true weight of the body, and the R upward normal reaction, W. The apparent weight of the body is equal to W. Since the body is moving in a circle, there is a net force directed towards the centre of the earth. Therefore, the true weight F must be more than the apparent weight W. If is the angular velocity of the earth, we have F – W = mR 2 or , W = F – mR 2 GMm mg mR or, R or, g = g0 – R 2
D A C A
where g0
M E
Y
GM
R2 Substituting the value of R and 2, we get g0 – g = 0.034 m/s2. Actually the centripetal acceleration is not directed towards the centre of the earth except at the equator. It is directed towards the earth’s axis of rotation . It can be shown that at latitude g = g0 – R 2 cos2 At the poles = 90o and, therefore, there is no effect of rotation on the value of g. If the earth were a non-rotating perfect sphere, the value of acceleration due to gravity would have been g0 everywhere. Gravitational Potential Energy : Potential energy of a body at a point in a gravitational field is the work done by an external agent in moving the body from infinity to that point. P Consider the body at some point P at any instant such that OP = x. Then, gravitational Q dx
E V
IE H C
GMm ( R h)
attraction,
S R
M ' = (Volume of shaded part) (density)
GM ' m
9
Gravitation
The polar radius is less than the equatorial radius by about 21 km. Therefore, g increases from the equator to the poles. (ii) Rotation of the Earth : The earth rotates about an axis which passes through the poles. Therefore, bodies on earth are not in equilibrium because they experience centripetal acceleration as they rotate with the earth. Consider a body of mass m at the equator. The force on it are the downward pull F of the earth’s gravitational
IJ K
GM g R
where g = acceleration due to gravity on the surface of earth. g' decreases with depth. It is maximum at the earth's surface and becomes zero at the centre of the earth ( h = R). Note : It should be noted that value of g decreases if we move above the surface or below the surface of the earth. Variation of g on the Surface of the Earth : The value of g is maximum at the poles and minimum at the equator. It decreases gradually as one goes from the poles to the equator. This is due to two reasons: (i) Shape of the Earth : The earth is not a perfect sphere but is ellipsoidal. It is flattened at the poles and is bulging at the equator.
force on the body at P is given by F
GMm
x Now, small work done in moving the body through a very small distance PQ = dx, is GMm
dx ...(i) x Thus, total work done by the field in moving the body from to r is
dW = Fdx =
r
r
x
GMm dx GMm W = x
A
x
r
O
earth
M
Fig. 9.3
dx
r
GMm = GMm GMm = r x r Thus, the gravitational potential energy U of a body of mass m at a distance r from the centre of earth is
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EDUACHIEVERS ACADEMY GMm r Negative sign implies that gravitational potential energy is due to attractive force of earth on the body, i.e., to bring the body from to a distance r, work is done by the gravitational field of earth itself. Gravitational potential energy goes on increasing with increase in distance r (as it becomes less negative) and becomes zero or maximum at r = . Gravitational Potential : The gravitational potential at a point in the gravitational field of a body is defined as the work done in bringing a unit mass from infinity to that point. It is a scalar quantity. Its dimensions are [L2T–2]. The units of gravitational potential in SI system are joules/kg. The gravitational potential due to earth at distance r from its centre is
U=W=
GM r Thus, gravitational potential at a point (i) is always negative. (ii) increases as we move toward infinity and becomes zero or maximum at infinity. (iii) at the surface of earth, it is expressed as
V=
GM . r Escape Velocity : It is the minimum velocity with which a body must be projected from the surface of earth so that it permanently overcomes and escapes the gravitational field of the earth. We can also say that a body projected with escape velocity will be able to go to a point which is at infinite distance from the earth.
V=
earth
A U D E radius = R mass = M
9
Gravitation ve =
GM or ve = R
gR
h
m
earth r
orbit
Fig. 9.5 Substituting the value of g = 9.81 m/s2 and R = 6400 km, we get ve = 11.3 km/s. Hence any object thrown with a velocity of 11.3 km/s or more will escape the gravitational field of the earth and will never come back to the earth. Satellite : The celestial bodies which revolve around the planets in close and stable orbits are called as satellites, i.e., A body revolving around a large body under the influence of the later. For example : moon is a satellite of earth. This is natural satellite. A man made satellite is called artificial satellite. These satellites are launched from the earth so as to move round it. The multistage rocket carries the satellite to the required vertical height and then gives it appropriate horizontal velocity required for stable orbit around the earth. Motion of a Satellite around Earth : Consider a satellite of mass m revolving in a circle around the earth (mass M) which is located at the centre of its orbit. If the satellite is at a height h above the earth's surface, the radius of its orbit is r = R + h, where R is the radius of earth. The gravitational force between m and M provides the centripetal force necessary for circular motion. Orbital Velocity : The velocity of a satellite in its orbit is called orbital velocity. Let v be the orbital velocity of satellite, then
D A C A
S R
M E
Y
E V
IE H C m
ve
Fig. 9.4 Let us imagine what happens to a body of mass m if it is thrown from the earth with a velocity ve (escape velocity). As the body moves away from the earth, it slows down (due to gravitational pull of the earth) and hence its kinetic energy is converted into gravitational potential energy of the mass-earth system. Let us imagine that it is just able to reach upto infinity (where G.P.E. is zero). K.E. lost by mass m = gain in G.P.E. of mass – earth system mve = (G.P.E)f – (G.P.E.)i GMm mve = R
GMm r
v
GM r
or
mv r
v
GM Rh
Hence the orbital velocity of a satellite is decided by the radius of its orbit or its height above the earth's surface. Time Period : The time taken to complete one revolution is called the time period. It is given by: T=
r r v
T
2r r GM
r GM
or T
r . GM
Total Energy of the Satellite : A satellite revolving around the earth has potential energy as well as kinetic energy. It has potential energy because of attractive force acting on it due to earth and kinetic energy because of its motion.
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EDUACHIEVERS ACADEMY
9
Gravitation
Height of the satellite = r – R = 35.93 106 m. Thus, the radius of the geostationary orbit is 4.2 104 km and the height above the surface of the earth is about 3.6 104 km. KEPLER’S LAWS 1. Law of Orbits : Each planet revolves around the sun in an elliptical orbit with the sun at one focus of the ellipse.
Total Energy = Kinetic energy + Potential energy =
GMm mv r
=
1 GM GMm m 2 r r
GMm GMm = r r
GMm 2r The total energy of the satellite is negative implies that in order to send the satellite to infinity, we have to provide an extra energy to satellite. If satellite is not given extra energy, it will go on revolving in closed orbit, i.e., the satellite is bound to earth.
Total energy =
2.
GMm Now, binding energy = – Total Energy = r For a satellite very close to the earth's surface,
v=
GM r
GM R
gR
Geostationary Satellite : An earth satellite so positioned that it appears stationary to an observer on earth is called a geostationary satellite. Its time period of revolution is 1 day. It serves as a fixed relay station for long distance TV transmission. If r is the radius of the geostationary orbit. M is the mass of earth and T is the time period, then
3.
D A C A
M E
Y
Fig. 9.7 Law of periods : It states that the square of the time taken by the planet about the sun is proportional to the cube of the planet’s mean distance from the sun. If T be time period of the planet and r be the mean distance of planet from the sun (average of maximum and minimum
S R
E V
IE H C
GM GM R T T T r or r GM R
Fig. 9.6 Law of Areas : This law states that radius vector from the sun to the planet sweeps out equal areas in equal time intervals. Both shaded areas are equal if time from A to B is equal to the time from P to Q.
distances from the sun =
rmin rmax ). Then 2
T 2 r 3 or T 2 K r 3 where K is a constant
R being the radius of the earth = 6400 km. This gives r=
gR
T
A U D E
(. . ) ( ) or r = (.)
or
K=
T2 r3
is same for all planets.
. km
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