CLASS 11th
Gravitation
Gravitation
01. Newton’s Law of Universal Gravitation All physical bodies are subject to the action of the forces of mutual gravitational attraction. The basic law describing the gravitational forces was stated by Sir Issac Newton and it is called Newton’s Law of Universal gravitation. The law is stated as : “Between any two particles of masses m1 and m2 at separation r from and (as shown in figure) directed from one each other there exist attractive forces body to the other and equal in magnitude which is directly proportional to the product of the masses of the bodies and inversely proportional to the square of the distance between the two”. Thus we can write ...(i) Where G is called universal gravitational constant. The law of gravitation can be applied to the bodies whose dimensions are small as compared to the separation between the two or when bodies can be treated as point particles.
02. Gravitational Field We can state by Newton’s universal law of gravitation that every mass M produces, in the region around it, a physical situation in which, whenever any other mass is placed, force acts on it, is called gravitational field. This filed is recognized by the force that the mass M exerts another mass, such as m, brought into the region.
03. Strength of Gravitational Field We define gravitational filed strength at any point in space to be the gravitational force per unit mass on a test mass (mass brought into the filed for experimental observation). Thus for , then at that point in space a point in space if a test mass m , experiences a force 0
is given as gravitational field strength which is denoted by ,
...(i)
Gravitational field strength is a vector quantity and has same direction as that of the force on the test mass in field.
3
Gravitation
04. Gravitational Field Strength of a Point Mass As per our pervious discussion we can state that every point mass also produces a gravitational field in its surrounding. To find the gravitational field strength due to a point mass, we put a test mass m0 at a point P at distance x from a point mass m then force on m0 is given as ...(i)
Now if at point P, gravitational field strength due to m is then it is given as
...(ii)
The expression in equation- (ii) gives the gravitational field strength at a point due to a point mass.
05. Gravitational Field Strength due to a Ring Case-I : At the centre of ring To find gravitational field strength at the centre of a ring of mass M and radius R, we consider an elemental mass dm on it as shown in figure. Let be the gravitational field at the centre of ring C due to the element dm.
Here we can simply state that another element of same mass exactly opposite to dm on other half of ring will produce an equal gravitational field at C in opposite direction. Thus due to all the elements on ring, the net gravitational field at centre C will be vectorially nullified and hence net gravitational field strength at C will be 0.
4
Gravitation Case-II: At a point on the axis of ring Figure shows a ring of mass M and radius R placed in YZ plane with centre at origin. Here we wish to find the gravitational filed strength at a point P on its axis at a distance x from its centre. To find this we consider an element of length dl on ring as shown in figure. the mass dm of this element can be given as
...(i) Let the gravitational field strength at point P due to the element dm is then it is given as or This elemental gravitational field strength dg has two rectangular components, one along the axis of ring cosθ and other perpendicular to the axis of ring, sinθ. Here when we integrate the result for the complete ring, sinθ component will be cancelled out by symmetry and cosθ will be summed up to give the net gravitational filed strength at P.
Thus here net gravitational field strength at P is given as cos ×
or
5
Gravitation
06. Gravitational Field Strength due to a Sphere Care-I : Hollow Sphere Figure below shown as hollow sphere of mass M and radius R. If we think about the gravitational field in its surrounding, the direction must be along the arrows shown in its surrounding. Every mass placed in its surrounding must experience the gravitational force toward the centre of the shell. Thus we can state that it is because of its symmetrical geometrical shape and its uniform mass distribution.
Thus we can state that for the case of a hollow spherical shell we consider its whole mass is concentrated at its centre and for outside points it behaves like a point mass. Thus the gravitational field strength at different points due to a hollow spherical shell can be given as shown in figure.
For outer points
For point on surface
[Behaving as a point mass]
For inner points
[As no mass is enclosed within it]
[Behaving as a point mass]
If we plot a variation graph for values of with distance from centre, it is shown in figure below.
6
Gravitation
Case-II : Solid Sphere In case of a solid sphere also the direction of at the nearby points is radially inward as shown in figure. So here also we can consider it to be a point mass for outer points.
In this case also the expression for gravitational field strength for outer and surface point remains same. Thus we have
For outer point
For point on surface
For points inside the sphere now is nonzero as there is mass content inside. To calculate at interior points at a distance x from its centre, we consider an inner sphere of radius x as shown in figure below. Say its mass is m, on the surface of which a point P exist where we wish to find the gravitational field, strength. The mass m is given as × ...(i)
7
Gravitation Now we can say that the given solid sphere is divided in two parts. One is an inner solid sphere of radius x and other is the outer shell of inner radius x and outer radius R. Here at point P gravitational field exist only due to the inner sphere as due to outer shell, we’ve discussed in previous section that, no gravitational field exist at interior points due to outer shell. Thus net gravitational field strength at P can be obtained by considering the inner sphere of radius x as a point mass at the centre. So gravitational field at P can be given as ...(ii) In this case the graph of variation of as a function of distance from centre of sphere is shown in figure.
07. Gravitational Lines of Forces Gravitational field can also be represented by lines of force. A line of force is drawn in such a way that at each point the direction of field is tangent to line that passes through the point. Thus tangent to any point on a line of force gives the direction of gravitational filed at that point. By convention lines of force are drawn in such a way that their density is proportional to the strength of field. Figure (i) below shows the field of a point mass in its surrounding. We can see that the lines of force are radially inward giving direction of field and as we go closer to the mass the density of lines is more which shows that field strength is increasing.
Fig. (i)
8
Gravitation
Fig. (ii) Shows the configuration of field lines for a system of two equal masses separated by a given distance.
08. Gravitational Field Strength of Earth We can consider earth to be a very large sphere of mass Me and radius Re. Gravitational field strength due to earth is also regarded as acceleration due to gravity or gravitational acceleration. Now we find the values of at different points due to earth. (a) Value of g on Earth’s Surface If be the gravitational field strength at a point A on the surface of earth, then it can be easily obtained by using the result of a solid sphere. Thus for earth, value of can be given as
...(i)
(b) Value of at Height h Above the Earth’s Surface If we wish to find the value of at a point P as shown in figure below at a height h above the Earth’s surface. Then the value can be obtained as
...(ii)
or
9
...(iii)
Gravitation
If point P is very close to Earth’s surface then for h R]
[For points on surface x = R] [For interior points x < R] Gravitational potential at centre [For x = 0]
...(xx) ...(xxi) ...(xxii) ...(xxiii)
12. Satellite and Planetary Motion (a) Motion of a Satellite in a Circular Orbit To understand how a satellite continually moves in its orbit, we consider the projection of a body horizontally from the top of a high mountain on earth as shown in figure. Here till our discussion ends we neglect air friction. The distance the projectile travels before hitting the ground depends on the launching speed. The greater the speed, the greater the distance. The distance the projectile travels before hitting the ground is also affected by the curvature of earth as shown in figure below.
This figure was given by newton in his explanation of laws of gravitation. It shown different trajectories fro different launching speeds. As the lunching speed is made greater, a speed is reached where by the projectile’s path follow the curvature of the earth. This is the launching speed which places the projectile in a circular orbit.
17
Gravitation Thus an object in circular orbit may be regarded as falling, but as it falls its path is concentric with the earth’s spherical surface and the object maintains a fixed distance from the earth’s centre. Since the motion may continue indefinitely, we may say that the orbit is stable. Let’s find the speed of a satellite of mass m in a circular orbit around the earth. Consider a satellite revolving around the earth in a circular orbit of radius r as shown in figure.
If its orbit is stable during its motion, the net gravitational force on it must be balanced by the centrifugal force on it relative to the rotating frame as or ...(i)
(b) Energies of a Satellite in a Circular Orbit When there is a satellite revolving in a stable circular orbit of radius r around the earth, its speed is given by equation- (i). During its motion the kinetic energy of the satellite can be given as ...(ii) As gravitational force on satellite due to earth is the only force it experiences during motion, it has gravitational interaction energy in the filed of earth, which is given as ...(iii) Thus the total energy of a satellite in an orbit of radius r can be given as Total energy E = Kinetic energy K + Potential Energy U
or
From equation- (i), (ii) and (iii) we can see that
18
...(iv)
...(v)
Gravitation
13. Kepler’s Laws of Planetary Motion The motions of planet in universe have always been puzzle. In 17th century Johannes Kepler, after a life time of study worded out some empirical laws based on the analysis of astronomical measurements of Tycho Brahe. Kepler formulate his laws, which are kinematical description of planetary motion. Now we discuss these laws step by step. (a) Kepler’s First Law [The Law of Orbits] Kepler’s first law is illustrated in the image shown in figure. It states that “All the planets move around the sun in elliptical orbits with sun at one of the focus not at centre of orbit.”
(b) Kepler’s Second Law [The Law of Areas] Kepler’s second Law is basically an alternative statement of law of conservation of momentum. It is illustrated in the image shown in figure (a). We known from angular momentum conservation, in elliptical orbit planet will move faster when it is nearer the sun. Thus when a planet executes elliptical orbit its angular speed changes continuously as it moves in the orbit. The point of nearest approach of the planet to the sun is termed perihelion. The point of greatest separation is termed aphelion. hence by angular momentum conservation we can state that the planet moves with maximum speed when it
is near perihelion and moves with slowest speed when it is near aphelion. Kepler’s second law states that “The line joining the sun and planet sweeps out equal areas in equal time or the rate of sweeping area by the position vector of the planet with respect to sun remains constant.” This is shown in figure (b). (c) Kepler’s Third Law [The Law of Periods] Kepler’s Third Laws is concerned with the time period of revolution of planets. It states that “The time period of revolution of a planet in its orbit around the sun is directly proportional to the cube of semi-major axis of the elliptical path around the sun.”
19
Gravitation If ‘T’ is the period of revolution and ‘a’ be the semi-major axis of the path of planet then according to Kepler’s III Law, we have ∝ ...(i) For circular orbits, it is a special case of ellipse when its major and minor axis are equal. If a planet is in a circular orbit of radius r around the sun then its revolution speed must be given as ...(ii)
Where Ms is the mass of sun. There you can recall that this speed is independent from the mass of planet. Here the time period of revolution can be given as or ...(iii)
Squaring equation no- (iii), we get
20
...(iv)
Gravitation
AIIMS Exercise (1) 1. Two solid spheres of the same material and same radius r touch each other, then the force between them varies as : (a) (b)
(c) (d)
2. The gravitational field intensity in region is given by N/m Find the work done by the gravitational force to shift a point mass 5 kg from (2, 2, 2). The distances measured are in metre. (a) 20 J (b) 60 J
(0, 0, 0) to
(c) 10 J (d) 40 J
3. Two heavy equal masses M each are separated by a distance r. A particle of mass m is projected from the midpoint of the line joining the particles. Find the velocity of m to escape from the gravitational pull of the heavy masses.
(b) (a)
(c)
(d)
4. Two, particles of masses m and M are at rest and are at infinite distance from each other. Due to mutual gravitational force, they move towards each other. Find their relative velocity when they are at a distance r from each other. (a) (b)
21
(d) (c)
Gravitation 5. A satellite of mass m is revolving in an orbit of radius 3R. Find the energy required to put in into another orbit or radius 5R. Take radius of earth = R and acceleration due to gravity = g. (a) (b)
(c) (d)
6. The radius of the earth is 6400 km. At what depth from the surface of the earth is the period of a simple pendulum two percent more than on the surface of earth? (a) 98 km (b) 980 m
(c) 128 m (d) 256 m
7. The variation of acceleration due to gravity as one moves away from earth’s centre is given by the graphs. Which is the correct graph?
(a)
(c)
(b)
(d)
8. The escape velocity from the surface of a planet is 104 ms−1. If a mass of 2 kg falls from infinity to the surface of the planet, its kinetic energy and magnitude of its potential energy on reaching the surface will be : (a) 108 J, 0 (b) 108 J,−108 J (c) 0.5 × 108 J, 0.5 × 108 J (d) 2 × 108 J,−2 × 108 J 9. A satellite moves in an elliptical orbit around the earth. The rate of change of arc length swept by the radius vector of a satellite is proportional to : (a) (b)
22
(c) (d)
Gravitation 10. If the earth were to suddenly contract to half the present radius (without any external torque acting on it) by how much would the day be decreased? (Assume earth to be a perfect solid sphere of moment of inertia = ) (a) 18 hours (b) 6 hours (c) 4 hours
(d) 2 hours (e) 1 hours
In each of the following questions, a statement of Assertion (A) is followed by a corresponding statement of Reason (R). Of the following statements, choose the correct one. (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is not correct explanation of A. (c) A is true but R is false. (d) A is false but R is true. (e) Both A and R are false. 11. (A) : Due to rotation of earth about the polar axis, the magnitude of acceleration due to gravity is maximum at the poles and minimum at the equator. (R) : If earth stops rotating about its axis, the value of g will become the same at the poles and at the equator. 12. (A) : Orbit of a planet revolving around the sun is in a plane. (R) : The angular momentum of a planet revolving round the sun is conserved.
23
