Gravitation
Kepler’s law of planetary motion: 1) All planets revolve around sun in elliptical orbit and the sun situated at any one foci of that ellipse.
2) The speed of planet varies in such a way that the radius vector drawn from the sun to a planet sweeps out equal area in equal interval of time.
3) The square of the time period of revolution of a planet around the sun isdirectly proportional to the cube of the semi-major axis of the ellipse tracedout by the planet.
Prove Kepler’s 2nd law: Let us consider a planet of mass m revolve around sun in a circular path of radius r with angular speed ω. The planet moves for a small distance dr from point P to point Q for time dt. As dr << r, dr is considered as the tangent at point P and the angle created at the centre dθ (very small). So, dθ = 
.
The area swept by the planet for travelling from P to Q is dA =
rdr = r2dθ
The rate of area swept by the planet is 
= r2 
= r2ω [as ω = ]
Or, =
r2 = 
[ L = mr2ω = angular momentum of the planet]
The planet revolves around sun due to the gravitational force between them. So there is no external torque is acting. Therefore the angular momentum of the planet is conserved. So, is constant.
Prove Kepler’s 3rd law: Let us consider a planet of mass m revolves around sun in a circular path of radius r with speed v. M is the mass of sun.
According to Newton’s law of gravitation, the force of attraction between the sun and planet is
F = 
—-(1)
This force provides the required centripetal force to the planet for circular motion.
So, = 
or, v2 = 
——-(2)
If T is the time period of revolution of planet, then T = 
or, T2 = 
—–(3)
From equations (2) and (3), we get, T2 =
or,T2 = Kr3 [K = 
= constant] or, T2 
r3.
Concepts on Kepler’s law:
- According to Kepler’s law planet (P) revolve around sun in an elliptical orbit. The motion of a planet around sun is described in the given figure. Point O is the centre of the ellipse and XX/ and YY/ are called the principal axes. S and S/are the two focus of the ellipse and sun is situated at S/.Point A is the nearest position of the planet from sun is known as perihelion whereas point B is the farthest position of planet from sun is known as aphelion. AB is called the major axis and its length is 2a. The length of minor axis is 2b.The distance of each focus from the centre of ellipse is S/O = SO = ea where e is the dimensionless number between 0 to 1 is called eccentricity.A line through the focus of the ellipse and perpendicular to its major axis is called the latus rectum L =
. The nearest distance between sun and planet is S/A = a – ea = a(1 – e) and the farthest distance between them is S/B = S/O + OB = ea + a = a(1 + e).
- A spaceship is projected towards a planet of mass M and radius R from a distance of 4R from the centre of planet with speed v0 at an angle θ with the line joining spaceship and the centre of planet. Find the value of θ for which it just graze the surface of the planet.
Let us consider the mass of the spaceship is m and its speed near the surface of planet is v.As there is no external torque acting on the system (planet and spaceship) then the angular momentum of the system remain conserved.The angular momentum of spaceship is equal to linear momentum of spaceship × perpendicular distance of linear momentum vector from point O. Therefore, 
× 4R sinθ = mvR
Or, v = 4
sin θ
Using conservation of energy at points A and B we get, 
– 
= 
– 
Or, 
= 
= 
– 
Or, – = – = 
Or, – 1 = 
× = 
Or, 
= 1 +
Or, 4sin θ = 
∴θ = 
[ 
]
Newton’s law of gravitation: Every particle of matter in the universe attracts every other particle with a force which is directly proportion to the product of their masses and inversely proportional to the square of the distance between them.
Let us consider two bodies of masses m1 and m2 with their centers separated by a distance of r. F is the gravitational force (in magnitude) of attraction between two bodies. According to Newton’s law of gravitation,
F m1m2 and F 
so, For, F 
or, F= 
Where G is the universal constant of gravitation.
G = 6.6710-8 dynecm2g-2 in C.G.S. unit or, G = 6.6710-11 Nm2kg-2 in SI unit.
Dimension of G =
= 
= [M-1L3T-2].
Gravitational force of attraction is equal in magnitude and opposite in direction:
Let m1 and m2 are the two stationary masses are placed at point A and B where 
=
and 
=
, where O is the origin and ,, are the position vectors of A and B respectively.
The force acting on m1 due to m2 is 
=
̂r̂12
Where r̂12 is the unit displacement vector from m1 to m2.
From ΔOAB, +
= or, +
= or, = – also r̂12 = 
=
So, = 
– – – -(1)
The gravitational force acting on m2 due to m1 is
=
̂r̂21
Where r̂21 is the unit displacement vector from m2 to m1.
From ΔOAB, + 
= or, +
= or, = – also r̂21 =
=
So, =
– – – -(2)
So, = – 
= – —-[using equation (1)]
So, gravitational force of attraction is equal in magnitude and opposite in direction.
Properties of gravitational force: 1) The gravitational force between two bodies constitutes an action and reaction force. I.e. forces are equal and opposite direction.
2) The gravitational force between two bodies does not depend on the intervening medium.
3) The gravitational force between two bodies does not depend on the presence or absence of other bodies.
4) The gravitational force is extremely small in the case of light bodies and it can be appreciable in case of massive bodies.
5) The gravitational force is conservative force. i.e. the work done by this force does not depend on the path between the initial and final position.
6) The gravitational force is a central force. It acts along the line joining the centers of the bodies.
Concept on gravitational force:
- A point mass m is placed at a distance d from one end of a uniform rod of mass M and length L.What is the force of attraction on point mass due to rod?
[As Newton’s law for gravitation is applicable for point masses then we have to consider the elementary part of the rod]
The mass per unit length of the rod is 
Let us consider an elementary part of the rod of thickness dx at a distance x from point mass m.
The mass of the elementary part of the rod is dM = dx.
The force of attraction between m and dM is dF =
=
Therefore, the total force acting on m due to the entire rod is F =∫dF =
Or, F = 
=[ 
– 
] = 
.
- Three particle of mass m are placed at three vertices of an equilateral triangle of each side a. If the system rotates about the centroid due to the mutual gravitational force, then calculate the velocity of each particle.
The gravitation force of attraction acting on each mass due to other is F = 
.
The net gravitation force is F/ =
= 
= 
.
As the system rotates about the centroid due to the mutual gravitational force, then net gravitational force provides the required centripetal force for rotation. So, =
= 
[AO = 
AD = 
= (
) = 
]
∴ v = 
.
- A body of mass m is placed at a distance of 4R from the centre of a solid sphere of mass M and radius R. A spherical part of radius
is cut out as shown in figure.Calculate the gravitational attraction experienced by the body due to the remaining part of the sphere.
The gravitational force of attraction on the body of mass m due to the sphere of mass M is F1 = 
.
If M/ is the mass of the cut out part of the sphere then M/ = 
× 
= 
.
The distance between the centre of cut out part with m is 4R – = 
.
The force of attraction between the centre of a sphere of mass with the body isF2 =
= 
.
The gravitational force on m due to the remaining part of the sphere is F = F1 – F2 =
– 
= 
.
Concepts on ring:
- Calculate the gravitational field intensity at a point on the axis of a uniform ring of mass M and radius R.
Let us consider a unit mass is placed at point P on the axis of the ring at a distance x from the centre of the ring O.
Let us consider an elementary part of the ring with mass dM.
At point P the gravitational field intensity due to dM is dE = 
The vertical component of field intensity i.e. dEsin is cancelled due to the field intensity of the opposite elementary part of the ring.
Only the horizontal component of field intensity dEcosθ is acting at point P towards the centre O.
Therefore the net gravitational field intensity at P due to dM is dE/ = 
= 
The gravitational field intensity at P due to the ring is E =∫DE/ = 
= 
.
- A point mass m is placed at point A on the axis of a uniform ring of mass M and radius R.If the distance between centre of the ring and A is x then the force acting on the m due to the ring towards its centre is F = Em =
.
- The force acting on the m due to the ring towards its centre isF = .
When x >> R, then F =
= 
[using binomial theorem]
Neglecting the term 
we get, F = 
.
- The force acting from A to O where the displacement of the point mass is measured from O to A. Then the force is represented as F = –
When R >> x, then F = – 
= – 
[using binomial theorem]
Neglecting the term 
we get, F = –
. So, F – x.
Therefore, the particle is in simple harmonic motion andthe angular frequency of the point mass is ω = 
[acceleration a = 
= –
= –
].
The time period of oscillation of the point mass is T =
= 
.
- As the force on the point mass depends on displacement x, then there is a particular position where the force of attraction is maximum.To get the position (in terms of x) where the force is maximum, we can use maxima minima concept.
Therefore,
= 0
Or, 
[] = 0
Or, [
] = 0 [As G, M and m are not equal to zero.]
Or, 
+ x
[] = 0
Or, + x(-
)
= 0 [ 
= 
× 
= – 
× 2x]
Or, – 
= 0
Or, =
Or, 
+ 
=3
Or, = 
∴ x = 
.
The maximum value of force is Fmax =
= 
=
=
= 
.
- Calculate gravitational potential at a point on the axis of a uniform ring: Let us consider the mass of the ring is M and radius is R.We want to calculate the gravitational potential at point A on the axis of the ring at a distance x from the centre O.We consider an elementary part of mass dM of the ring which is at a distance r from point A.
The gravitational potential at point A due to elementary mass dM is
dV = – 
= –
.
The gravitational potential at point A due to the ring is V =∫dv = –
= – 
.
- A particle of mass m is placed at point A on the axis of the ring.Calculate the work done required to move the particle from A to the centre of the ring. Also calculate the speed of the particle at the centre of the ring if initially it is at rest.
The gravitational potential at point A due to the ring is VA = – 
.
The gravitational potential at point O due to the ring is VO = –
.
The potential difference between the points A and O is V = VO – VA = – – (- )
The work done required to move particle m from point A to O is W = mV = GMm[
– 
].
Using energy conservation at points A and O we get, – 
= – 
+ 
∴v = 
.
Weight: Weight of the body is defined as the force with which a body is attracted towards the centre of earth.
Force of gravity: If the force of attraction between a body and earth, then the force of attraction is called force of gravity.
Acceleration due to gravity: The acceleration produced in a body due to the force of gravity is called acceleration due to gravity. It is denoted by ‘g’.
Let us consider earth is a perfect sphere of radius R of mass M. A body of mass m is placed on the earth surface. According to Newton’s law of gravitation, the force of attraction between the body and earth is F = 
—-(1)
But the force with which a body is attracted towards the earth gives the weight of the body.
So, F =mg —-(2)(g = acceleration due to gravity)
From equation (1) and (2) mg = or, g = 
.
Factors on which g depends:
1) Variation of g with height: Let us consider earth is a perfect sphere of radius R of mass M. A bodyof mass m is placed on the earth surface at point A.
According to Newton’s law of gravitation, the force of attraction between the body and earth is
F == mg ——–(1). Where g is the acceleration due to gravity on earth’s surface.
Now the body is moved at a height h from earth’s surface and g/ is the acceleration due to gravity at that place.
According to Newton’s law of gravitation, the force of attraction between the body and earth is now
F/ =
= mg/——-(2).
From equation (1) and (2), we get = ()2 = ()2 = (1+) -2
g/ = g(1 – ) [as hR binomial theorem is used]
2) Variation of g with depth from: Let us consider earth is a perfect sphere of radius R of density ρ.
A body of mass m is placed on the earth surface at point A.
According to Newton’s law of gravitation, the force of attraction between the body and earth is
F = 
= mg ——–(1). Where g is the acceleration due to gravity on earth’s surface.
Now the body is moved at a depth h from earth’s surface and g/ is the acceleration due to gravity at that place.
According to Newton’s law of gravitation, the force of attraction between the body and earth is now
F/ = = mg/ ——-(2).
From equation (1) and (2), we get = () = (1 – )
So, g/= g(1 – ). At the centre of earth g/ is 0 (as h = R).
3) Variation of g due to rotation of earth: Let us consider earth is a perfect sphere of radius R rotates with angular speed ω. A body of mass m is placed on the earth surface at point P of latitude λ.
The weight of the body mg acts towards the centre O of the earth.
The body rotates with angular speed ω with a radius r = Rcos λ
The centrifugal force acts on the body is mω2r.
The component of centrifugal force acts away from the centre O is mω2rcos λ
So, the net force acts on the body towards the centre O of the earth is
mg/= mg – mω2rcos λ (where g/is the acceleration due to gravity at that place)
So, g/= g – ω2r cos λ= g – ω2R(cos λ)2 [as r = Rcos λ]
At the equator g/= g – ω2R [as λ = 00] and at pole g/= g [as λ = 900]
Gravitational field and field intensity: The gravitational field is the space around a mass or an assembly of masses over which it can exert gravitational force on other masses.
The gravitational field intensity at a point in the gravitational field is the force experience by a unit mass placed at that point. It is directed towards the particle producing the field. If E is the gravitational field intensity then E = 
=
.
Gravitational potential: Gravitational potential at a point in the gravitational field is the work done to bring a unit mass from infinity to that point.
Potential at a point due to a point mass: Let us consider a mass M is placed at point O. To calculate gravitational potential at a point P at a distance r from O, we consider a unit mass is moved from infinity to point P.
The gravitational force acting on unit mass when it is at P/ is F = 
——(1)
The work done by F to move unit mass from P/ to P// (P/P// = dr/) is dw = Fdr/ ——–(2)
From equation (1) and(2) we get, dw =
The total work done by the gravitational force F to move unit mass from infinity to point P is
W =
dw = 
= GM
= GM[
– 
] = – 
.
Gravitational potential energy: Gravitational potential energy at a point in the gravitational field is the work done to bring a mass from infinity to that point.
Gravitational potential energy or the work done to bring a mass m at a point in the gravitational field at a distance rfrom mass M is V = –
.
Relation between gravitational potential and gravitational field: Let us consider two points A and B are very close to each other, separated by a distance dr in gravitational field E of a particle of mass M placed at point P.
If the gravitational potential at point A and B are respectively –(V+ΔV) and –V, then the
gravitational potential difference between points A and B is –(V+ΔV) –( –V) = -ΔV.
The work done in displacing a test mass m0 from B to A is ΔW = FΔr
Therefore -ΔV =
= 
= EΔr. So, E = – 
= – 
.
Concept on gravitational potential energy:
- Three particles each of mass m are placed at three vertices of an equilateral triangle of each side l from infinity. What is the work done required in this process?
We know that the work done to move a particle from infinity to any point in the gravitational field = the potential energy = potential at that point × mass.
Initially there is no mass at point A. Therefore, potential at point A is VA = 0
So, the work done to bring a mass m from infinity to point A is WA = 0.
The potential at point B due to mass m at point A is VB = – 
The work done to bring mass m from infinity to point B is WB = mVB = – 
.
The potential at point C due to the masses of point A and B is VC = – + (-) = – 
The work done to bring mass m from infinity to point C is WC = mVC = –
.
Therefore, the total work done to bring three masses at point A, B and C is W = WA + WB + WC = – 
.
- Four particles of mass m are placed at four vertices of a square of side a.Find the gravitational potential energy of the system.
Total number of particle is n = 4. Total number of pair of particle are 
= 6.
4 pairs having distance between two particles are a and 2 pairs have the distance between particle are 
.
Therefore, the gravitational potential energy of the system is U = –
– 
.
- 8 pairs each of mass m are placed at the vertices of a cube of side a. Find gravitational potential energy of the system.
Total number of particle is n = 8. Total number of pair of particle are = 28.
12 pairs having distance between two particles are a. 12 pairs have the distance between particle is . 4 pairs have the distance between particle is 
.
Therefore, the gravitational potential energy of the system is U = 
– 
– 
.
Escape velocity: it is defined as the least velocity with which a body must be thrown vertically upwards in order that it may just escape the gravitational pull of earth. The value of escape velocity is 11.2 kms-1.
Let us consider a body of mass m is placed on earth surface. The mass and radius of earth is M and R respectively. The body is projected upwards with escape velocity ve to move beyond gravitational influence (at infinity).
The energy of the body on earth’s surface = gravitational potential energy + kinetic energy of the body
Esurface = – +
At infinity gravitational potential energy = –
=0 and the body become rest, so kinetic energy of the body = 0
The energy of the body beyond gravitational influence (at infinity) is Einfinity = 0
Using conservation of energy, – + = 0 or,ve =
If g is the acceleration due to gravity, then GM = gR2 so, ve = 
g = 9.8 ms -2 and R = 6.4106m so, ve= 
ms -1 = 11.2×103ms -1.
Orbital velocity and time period of a satellite: Let us consider a satellite of mass m revolves around earth in a circular orbital path of radius r with speed v. M is the mass and R is the radius of earth.
According to Newton’s law of gravitation, the force of attraction between the earth and satellite is F = 
This force provides the required centripetal force to the satellite for circular motion. So, = 
Or, v2 = 
or, v = 
If g is the acceleration due to gravity, then GM = gR2
If the satellite at a height h from the earth’s surface then, r = R +h
So, v = = .
If the satellite revolves very close to earth then h<< R, so, v =
If T is the time period of revolution of satellite, then T = 
.
Or, T = 2πr 
=2π
=2π [as, GM = gR2]
If the satellite revolves very close to earth then h << R, so, T = 2π
.
Energy of an artificial satellite: Let us consider a satellite of mass m revolves around earth in a circular orbital path of radius r with speed v. M is the mass of earth.
According to Newton’s law of gravitation, the force of attraction between the earth and satelliteprovides the required centripetal force to the satellite for circular motion. So, = ———(1)
The kinetic energy of the satellite is K.E. =
mv2 = 
[from equation (1)]
The gravitational potential energy of the satellite is P.E. = –
The total energy of the satellite is T.E. = K.E. + P.E. = – = –
So, T.E. = P.E. Or, T.E. = – K.E.
Conservation of energy: The distance between the centres of two planets is 15R. The masses of the planets are M and 16M and their radii are R and 2R respectively.A body of mass m is projected from the surface of smaller planet towards the surface of larger planet.What should be the minimum initial speed of the body so that it can reach the surface of the larger planet?
Let us consider O is the point where the net gravitational field intensity for two planets is zero. If the body is moved from the surface of lower planet (point A) to the point O then, due to inertia the body enters the gravitational field of the larger planet and it is attracted by the larger planet and moves towards the surface of it.
The distance of O from the centre of smaller planet is x.
Therefore,
= 
Or, 4x = 15R – x
∴x = 3R
Using conservation of energy at point A and O we get, (P.E. at A due to M) + (P.E. at A due to 16M) + K.E. of m at A = (P.E. at O due to M) + (P.E. at O due to 16M) + K.E. of m at O
Or, – + (- 
) + 
= – 
+ (-
) + 0
Or, – 
+ = – 
Or, = – +
Or, = 
∴ v = 
.
.
Geostationary satellite: The satellite which is in a circular orbit around the earth in the equatorial plane appears stationary to an observer on the earth is called geostationary satellite.
Geostationary satellite revolves around the earth from west to east with orbital velocity nearly equal to 3.1kms -1. Time period of revolution of geostationary satellite is 24 hour.
Let us consider a geostationary satellite of mass m revolves around earth in a circular orbital path of radius r with speed v on equatorial plane of earth. M is the mass and R is the radius of earth.
According to Newton’s law of gravitation, the force of attraction between the earth and satelliteprovides the required centripetal force to the satellite for circular motion. So,=
Or, v2 = ——(1)
So, v = = .
If T is the time period of revolution of planet, then T = 
or, v2 = 
Or, = [using equation (1)]
Or, r3 = 
If g is the acceleration due to gravity, then GM = gR2 then, r3 = 
G = 9.8ms -2, R = 6.38106m, T=243600s
Then, r = 
m = 42240×103m =42240km
If the satellite at a height h from the earth’s surface then, r = R +h
So, h = (42240 –6380)km = 35860km ≈ 36000km.
Use: 1. It is used in communication system and telecast T.V. programs.
- It is used in weather forecast.
- It is used to explore the upper region of atmosphere.
Binary stars: Two stars orbiting around a common centre of mass are called binary stars.
Let us consider two stars A and B of masses M and m respectively revolving around a common center of mass O with radii of revolution a and b.
From definition of centre of mass Ma = mb or, 
= 
or, + 1 = + 1 or, 
= 
= 
Or, Ma + ma = mx or, a = 
similarly, b = ——-(1)
V1 and V2 are the speed of revolution of stars A and B and T is the common time periodof revolution about the centre of mass O, then V1 = 
and V2 =
——–(2)
The gravitational attraction force between A and B provides the required centripetal force for revolution.
So,
= 
or, 
= 
or, 
= [using equation (2)] or, = 
[using equation (1)]
Or, (M+m) = 
——(3)
From equation (1) and (3) we can calculate M and m.
Weightlessness in space: Let us consider a spaceship of mass MS revolves around earth with orbital radius r with orbital speed v. mass of earth is M and mass of astronaut inside the spaceship is m.
The force of attraction between the earth and spaceship provides the required centripetal force to the spaceship for circular motion. So, 
= 
or, 
=
—–(1)
Let N is the reaction force on astronaut by the floor of spaceship. The force towards the centre of earth is ( – N) provides the centripetal force .
So, – N =
Or, – N = [using equation (1)] so, N = 0
Therefore astronaut is in weightlessness condition in space inside the spaceship.
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