Gravitation Notes

GRAVITATION Page # 1 GRAVITATION 1. Newtons's Law of Universal Gravitation All physical bodies are subjected 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 mI and m2 at separation r from each other   there exist attractive forces FAB and FBA directed from one 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 m1m2 ...(1) r2 Where G is called universal gravitational constant. Its value is equal to 6.67 × 10–11 Nm2/kg. 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.   FAB m2 m1 FBA FAB  FBA  G A        B r If the bodies are not very small sized, we can not directly apply the expression in equation-(1) to find their natural gravitational attraction. In this case we use the following procedure to find the same. The bodies are initially split into small parts or a large number of point masses. Now using equation-(1) the force of attraction exerted on a particle of one body by a particle of another body can be obtained. Now we add all forces vectorially which are exerted by all independent particles of second body on the particle of first body. Finally the resultants of these forces is summed over all particles of the first body to obtain the net force experinced by the bodies. In general we use integration or basic summation of these forces. Gravitational force is a conservative force. Gravitational force is a central force. Gravitational force is equal in magnitude & opposite in direction Gravitational forces are action - reaction pair. Gravitational force acts along the line joining the two masses. Gravitational force doesn't depend upon the medium Gravitational force is an attractive force.  –Gm m r F 1 2 | r |3  [Head of r is placed at that position where we have to evaluate force] 2. 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 field is recognized by the force that the mass M exerts another mass, such as m, brought into the region. 2.1 Strength of Gravitational Field We define gravitational field strength at any point in space to be the gravitational force per unit mass on a test mass (mass brought into the field for experimental observation). Thus for a point in space if a test mass m0, experiences a force  F , then at that point in space, gravitational field strength which is denoted   F  g  by g , is given as m0  Gravitational field strength g is a vector quantity and has same direction as that of the force on the test mass in field. GRAVITATION Page # 2 Generally magnitude of test mass is very small such that its gravitational field does not modify the field that is being measured. It should be also noted that gravitational field strength is just the acceleration that a unit mass would experience at that point in space. 2.2 Gravitational Field Strength of Point Mass As per our previous 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 Fg m0 Gmm0 x2 Now if at point P, gravitational field strength due to m is gp then it is given as x Fg  Fg Gm gp   2 m0 x m The expression written in above equation gives the gravitational field strength at a point due to a point mass. It should be noted that the expression in equation written above is only applicable for gravitational field strength due to point masses. It should not be used for extended bodies. However, the expression for the gravitational field strength produced by extended masses has already been derived in electrostatics section. 3. INTERACTION ENERGY This energy exists in a system of particles due to the interaction forces between the particles of system. Analytically this term is defined as the work done against the interaction of system forces in assembling the given configuration of particles. To understand this we take a simple example of interaction energy of two points masses. Figure (a) shows a system of two point masses m1 and m2 placed at a distance r apart in space. here if we wish to find the interaction potential energy of the two masses, this must be the work done in bringing the two masses from infinity (zero interaction state) to this configuration. For this we first fix m1 at its position and bring m2 slowly from infinity to its location. If in the process m2 is at a distance x from m1 then force on it is Gm1m2 ˆ F– i x2 m1 m2 m1 m2 r r Fdx (a) (b) This force is applied by the gravitational field of m1 to m2. If it is further displaced by a distance dx towards m1 then work done by the field is Gm1m2   dx dW  F .dx = x2 Now in bringing m2 from infinity to a position at a distance r from m1, the total work done by the field is r Gm1m2 dx = – G m m 1 2 x2  W   dW –  W r  1 – x    Gm1m2 r Gm1m2 amount of work. The work is positive because r the displacment of body is in the direction of force. Initially when the separation between m1 and m2 was very large (at infinity) there was no interaction between them. We conversely say that as a reference when there is no interaction the interaction energy of the system is zero and during the process system forces (gravitational forces) are doing work so system energy will decrease and becomes negative (as initial energy was zero). As a consequence we can state that in general if system forces are attractive, in Thus during the process field of system has done GRAVITATION Page # 3 assembling a system of particles work will be done by the system and it will spend energy in assembling itself. Thus finally the interaction energy of system will be negative. On the other hand if in a given system of particles, the system forces are repulsive, then in assembling a system some external forces have to be work against the system forces and in this case some work must be done by external forces on the system hence finally the interaction energy of the system of particles must be positive. In above example as work is done by the gravitaional forces of the system of two masses, the interaction energy of system must be negative and it can be given as Gm1m2 ...(1) r As gravitational forces are always attractive, the gravitational potential energy is always taken negative. U12  – 3.1 Interaction Energy of a System of Particles If in a system there are more than two particles then we can find the interaction energy of particle in pairs using equation (1) and finally sum up all the results to get the total energy of the system. For example in a system of N particles with masses m1, m2........mn separated from each other by a distance r12........ where r12 is the separation between m1 and m2 and so on. In the above case the total interaction energy of system is given as U– 1 N N Gmm i j   2 i1 j1 rij 1 is taken because the interaction energy for each possible pair of system is 2 taken twice during summation as for mass m1 and m3 In this expression the factor U– Gm1m3 Gm3m1 – = r13 r31 Now to understand the applications of interaction energy we take few examples. 4 GRAVITATIONAL POTENTIAL The gravitational potential at a point in gravitational field is the gravitional potenial energy per unit mass placed at that point in gravitational field. Thus at a certain point in gravitational field, a mass m0 has a potential energy U then the gravitational potential at that point is given as U V= m 0 or if at a point in gravitational field gravitational potential V is known then the interaction potential energy of a point mass m0 at that point in the field is given as U = m0v Interaction energy of a point mass m0 in a field is defined as work done in bringing that mass from infinity to that point. In the same fashion we can define gravitational potential at a point in field, alternatively as "Work done in bringing a unit mass from infinity to that point against gravitational forces."  When a unit mass is brought to a point in a gravitational field, force on the unit mass is g at a point in the field. Thus the work done in bringing this unit mass from infinity to a point P in gravitational field or gravitational potential at point P is given as P   VP  – g .dx  Here negative sign shown that VP is the negative of work done by gravitation field or it is the external required work for the purpose against gravitational forces. 4.1 Gravitational Potential due to a Point Mass We know that in the surrounding of a point mass it produces its gravitational field. If we wish to find the gravitational potential at a point P situated at a distance r from it as shown in figure, we place a test mass m0 at P and we find the interaction energy of m0 with the field of m, which is given as U– Gmm0 r 1 GRAVITATION Page # 4 Now the gravitational potential at P due to m can be written as V P U Gm – m0 r r The expression of gravitational potential in equation is a standard result due to a point mass which can be used as an elemental form to find other complex results, we'll see later. The same thing can also be obtained by using equation    g.dx P VP  m r VP  or  Gm Gm dx or VP  – 2 x r   4.2 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 values of g at different points due to earth. • Earth behaves as a non conducting solid sphere 4.3 Value of g on Earth's Surface : If gs 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 gs can A Me be given as gs gs  GMe R 2e Re ...(1) Earth 4.4 Value of g at a Height h Above the Earth's Surface: If we wish to find the value of g at a point P as shown in figure shown at a height h above the Eath's surface. Then the value can be obtained as P gs  GMe (R e  h)2 gs  or GMe 2  h  R 1   R e   2 e  g h gs  h  1   R e   Me 2 Re If point P is very close to Earth's surface then for h << Re we can rewrite the expression in given equation as –2  h  gh  gs  1   Re   ~ –  2h  gs 1 –  R e  [Using binomial approximation]  ...(2) 4.5 Value of g at a Depth h Below the Earth's Surface If we find the value of g inside the volume of earth at a depth h below the earth's surface at point P as shown in figure, then we can use the result of g inside a solid sphere as gin  GMe x R 3e Here x, the distance of point from centre of earth is given as x = Re – h h Re–h Me Re  GMe (R e – h) h  gh   gs 1 –  Thus we have ...(3) 3 Re  Re  From equation (1), (2) and (3) we can say that the value of g at eath's surface is maximum and as we move above the earth's surface or we go below the surface of earth, the value of g decrease. 4.6 Effect of Earth's Rotation on Value of g Let us consider a body of mass m placed on Earth's surface at a latitude  as shown in figure. This mass experiences a force mgs towards the centre of earth and a centrifugal force mwe2 Re sin  relative to Earth's surface as shown in figure. GRAVITATION Page # 5 If we consider geff as the effective value of g on earth surface at a latitude  then we can write  Fnet geff = = geff  (2eR e sin )2  g2  22eR e sin .g cos(90  ) m pole R e sin m m 2Re sin mg 4e is very very small So we can write e Re 2 2 e 2 geff  g  2 R e sin g equator 1/2  22eR e sin2   geff  g 1    g – e2 R sin2  ...(i) g   From equation (1) we can find the value of effective gravity at poles and equatorial points on Earth as At poles  = 0  gpoles = gs = 9.83 m/s2   gequator = gs – 2Re = 9.78 m/s2 2 Thus we can see that the body if placed at poles of Earth, it will only have a spin, not circular motion so there is no reduction in value of g at poles due to rotation of earth. Thus at poles value of g on Earth surface is maximum and at equator it is minimum. But an average we take 9.8 m/s2, the value of g everywhere on earth's surface. At equator  = 4.7 Effect of Shape of Earth on Value of g Till now we considered that earth is spherical in its shape but this is not actually true. Due to some geological and astromonical reasons, the shape of earth is not exact spherical. It is ellipsoidal. As we've discussed that the value of g at a point on earth surface depends on radius of Earth.It is observed that the approximate difference in earth's radius at different points on equator and poles is re – rp ~ – gp – ge 21 to 34 km. Due to this the difference in value of g at poles and equatorial points is approximately ~ – 0.02 to 0.04 m/s2, which is very small. So for numerical calculations, generally, we ignore this factor while taking the value of g and we assume Earth is spherical in shape. 5. SATELLITE AND PLANETARY MOTION 5.1 Motion of a Satellite in a Circular Orbit To understand how a satellite continously 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 shown. This figure was given by newton in his explanantion of laws of gravitation. it shows different trajectories for diferent launching speeds. As the launching 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. 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. v Earth 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. 1 GRAVITATION Page # 6 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 GMem mv2 GMe  v or r r2 r Expression in above equation gives the speed of a statellite in a stable circular orbit of radius r. Fe m FG v r Me Re Earth 5.2 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 above equation. During its motion the kinetic energy of the satellite can be given as 1 1 GMem mv2  ...(1) 2 2 r As gravitational force on satellite due to earth is the only force it experiences during motion, it has gravitational interaction energy in the field of earth, which is given as K GMem ...(2) r 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 U –  1 GMem GMem – or 2 r r E– 1 GMem 2 r ...(3) U | E | 2 The above relation in magnitude of total, kinetic and potential energies of a satelline is very useful in numerical problem so it is advised to keep this relation in mind while handing satellite problems related to energy. Now to understand satellite and planetary motion in detail, we take few example. From equation (1), (2) and (3) we can see that | k | 6. KEPLER'S LAWS OF PLANETARY MOTION The motions of planet in universe have always been a 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 formulated his laws, which are kinematical description of planetary motion. Now we discuss these laws step by step. 6.1 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 ellipitcal orbits with sun at one of the focus not at centre of orbit." It is observed that the orbits of planets around sun are very less ecentric or approximately circular Planet Sun Focus 6.2 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 know from angular momentum conservation, in elliptical orbit plane will move faster when it is nearer to 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 seperation 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. GRAVITATION Page # 7 v  C D  A r1 Sun d r E r2 v2 Aphelion Sun B S Perihelion v 1 (b) (a) 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). The above statement of Kepler's second law can be verified by the law of conservation of angular momentum. To verify this consider the moving planet around the sun at a general point C in the orbit at speed v. Let at this instant the distance of planet from sun is r. If  be the angle between position vector  r of planet and its velocity vector then the angular momentum of planet at this instant is L = m v r sin  ...(1) In an elemental time the planet will cover a small distance CD = dl and will travel to another adjacent point D as shown in figure (a), thus the distance CD = vdt. In this duration dt, the position vector  r sweeps out an area equal to that of triangle SCD, which is calculated as 1 1 Area of triangle SCD is dA = × r × vdt sin ( – ) = r v sin . dt 2 2 Thus the rate of sweeping area by the position vector  r is dA 1  rv sin  dt 2 Now from equation (1) dA L   cons tan t ...(2) dt 2m The expression in equation (2) verifies the statement of Kepler' II law of planetary motion. 6.3 Kepler's Third law [The Law of Periods] Kepler's Third Law 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" 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 T 2  a3 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 GMs r Where Ms is the mass of sun. Here you can recall that this speed is independent from the mass of planet. Here the time period of revolution can be given as v T  2r v T  or 2r GMs r Squaring equation written above, we get 42 3 T2  r ...(1) GMs Equation (1) verifies the statement of Kepler's third law for circular orbits. Similarly we can also verify it for elliptical orbits. For this we start from the relation we've derived earlier for rate of sweeping area by the position vector of planet with respect to sun which is given as dA L  dt 2m 1