DD Gravitation

Solved examples on Gravitation

Example 4

In a binary star system, two stars (one of mass m and the other of 2m) distance d apart rotate about

their common centre of mass. Deduce an expression for the period of revolution. Show that the ratio of

their angular momenta about the centre of mass is the same as the ratio of their kinetic energies.

Solution:

The centre of mass C will be at a distance d/3 and 2d/3 from the masses 2m and m respectively as

shown in figure 1.10. Both the stars rotate with the same angular velocity ω around C in their

respective orbits. Here the gravitational force acting on each star due to other supplies the necessary

centripetal force.

Gravitational force on each star = G (2m) m/d2.

Centripetal force of star (mass m)

= m r ω2 = m (2d/3)ω2

.·. G(2m)m/d2 = m(2d/3) ω2

ω = √((3Gm/d3 ) )

.·. T = 2Π/ω=2Π√((d3/3Gm) ) Ans.

Ratio of angular momentum

=(Iω)big/(Iω)small = (2m(d/3)2)/(m(2d/3)2) = 1/2 Ans.

This is same as that of their angular momenta.

Solution:

The situation is shown in figure given above.

Following forces act on each ball

(i) Weight of the ball mg

(ii) Tension in thread T

(iii) Force of Gravitational attraction

F = G(mm/r2).

As the system is in equilibrium these three forces can be represented by the sides of a triangle

say ACP in which AC represents the weight mg, CP represents the force F and PA represents the tension

T. Hence

tan = ((r-r')/2)/l = F/mg ...... (2)

But F/mg = (Gm2/r'2)/mg ...... (1)

From equations (1) and (2)

((r-r')/2)/l = (Gm2/r'2)/mg or (r-r')mg/2 = lGm2/r'2

gr'2 (r-r') = 2l Gm. (Proved)

Example 3

A planet of mass m moves in an elliptical orbit around the Sun so that its maximum and minimum

distances from the Sun are equal to r1 and r2 respectively. Find the angular momentum of this planet

relative to the centre of the Sun.

Solution:

As the angular momentum of the planet is constant (no external torque is acting on it), we have

mv1r1 = mv2r2

or v1r1 = v2r2

Further, the total energy of the planet is also constant, hence

-GMm/r1 + 1/2 mv12 = -GMm/r2 + 1/2 mv2

2

where M is the mass of the Sun.

2

3


Example 4




Solution:




centripetal force.



= m r ω2 = m (2d/3)ω2

.·. G(2m)m/d2 = m(2d/3) ω2

ω = √((3Gm/d3 ) )

.·. T = 2Π/ω=2Π√((d3/3Gm) ) Ans.




4

Example 5

Imagine a planet whose diameter and mass are both one half of those of Earth. The day's temperature

of this planet's surface reaches upto 800K. Find whether oxygen molecules are possible in the

atmosphere of this planet.

Solution:

Escape velocity, ve = √2GM/R

Let vp = escape velocity on the planet.

ve = escape velocity on the Earth.

.·. vp/ve =√(MP/RP × RE/ME) = √(1/2×2/1) = 1

.·. vp = ve = 11.2 km/s.

From kinetic theory of gases

vrms = √(3RT/M) = √(3NKT/M) = √(3NKT/Nm)

where N = Avogadro's number

m = mass of oxygen molecule

K = Boltzmann constant

vrms = √3RT/m

= √((3×1.38×10-23×800)/(5.3×10-26)) (m = 5.3 × 10-26 kg).

vr.m.s = 0.79 km/s Ans.

As vrms is very small compared to escape velocity on the planet, molecules cannot escape from the

surface of the planet's atmosphere.

5

Example 6

Halley's comet has a period of 76 years and in 1986 it had a distance of closest approach to the Sun

equal to 8.9 × 10+10 m. What is the comet's farthest distance from the Sun if mass of Sun is 2 × 1030 kg

and G = 6.67 × 10-11 MKS units.

Solution:

Whenever a heavenly body revolves round another body it follows an elliptical path having a distance o

closest and farthest approach.

For these orbits

T2 = (4Π2/GM)a3 , where a and T are average distance of comet from Sun and time period of revolution

respectively.

.·. a = [T2GM/4Π2]1/3

or, a = [((76×3.15×107)2× 6.67×10-11×2×1030)/4Π2 ]1/3

≈ 2.7 × 1012 m.

But for ellpse,

2a = rmin. + rmax.

(where a is the semi major axis of the ellipse)

or, rmax. = 2a - rmin.

rmax = 2 × 2.7 × 1012 - 8.9 × 1010

rmax ≈ 5.3 × 1012 m. Ans.

Example 7

A uniform sphere has a mass M and radius R. Find the pressure P inside the sphere caused by

gravitational compression, as a function of distance r from its centre.

Solution:

Density of the given sphere, ρ = M/(4/3ΠR3)

Let us consider a layer of thickness 'dr' at a distance r from the centre. The part of sphere inside this

spherical shell will attract this with a net force but the outer part would not have any effect on this as

inside a hollow spherical shell force on any particle is zero.

6


Mass of sphere of radius r, M1 = 4/3ΠR3 ρ

.·. M1 = M/R3.r3

Mass of the layer of thickness dr is

dm = 4Πr2.dr.ρ

= 4Πr2.dr.M/(4/3ΠR3 )

= 3M/R3dr . r2

.·. Force on small layer due to M1

dF = (GM1.dm)/r2 = (GM.r3.3M.r2)/(R3.r2.R3) dr

= 3GM2/R6 dr.r3

Thus pressure on the layer of mass 'dm'

dp = dF/4Πr2

= 3GM2/R6 . (r3.dr)/r2 = 3GM2/(4ΠR6).rdr

.·. Total pressure due to mass extending from r to R is

p = ∫rR dp = 3GM2/R6 ∫r

R rdr

= 3GM2/(8ΠR6 ) . [r2/2]rR

= 3GM2/(8ΠR6 ) (R2 - r2)

.·. p = 3GM2/(8ΠR4 ) (1-r2/R2 ) Ans.

7


Example 8

Assuming the Earth to be a sphere of uniform density, calculate the energy needed to completely

disassemble it against gravitational pull amongst its constituent particles given product of mass and

radius of Earth = 2.5 × 1031 kgm.

Solution:

Let us suppose that Earth is made of a infinite number of very thin concentric spherical shells. It can be

completely disassembled by removing these shells one by one.

Let us study Earth when only a sphere of radius x is left and find the energy required to remove a shell

of thickness dx.

Now Potential of this sphere of radius x is u = Gms/x

[ms = mass of sphere = 4/3 Πx3ρ where ρ = density of Earth = M/(4/3ΠR3) where M, R are mass and

radius of Earth respectively].

Now work done (dw) to remove a shell of thickness dx would be equal to the change in potential energy

= Gmsdm/x,

where dm = mass of shell = 4Πx2 dxρ.

=> dW = (4/3 x3 ρ)(4Πx2 dxρ)/x = 16/3 GΠ2ρ2x4 dx..

W = ∫dW = ∫0R 16/3GΠ2 ρ2 x4 dx = 16/15 GΠ2 [m/((4/3)Π R3 )]2 R5..

Which, on substituting the given value, is 3/5 GM2/R = 3/5 g MR

= 1.5 × 1032 J.

Example 9

A satellite revolving close to the surface of Earth from west to east, appears over a certain point at the

equator every 11.6 hours. If radius of Earth is 6400 km. calculate the mass of Earth.

8

Solution:

Since the rotation of Earth is also west to east the apparent angular velocity of the satellite is equal to

ωSE = ωS - ωe,

Where ωS is the angular velocity of the satellite w.r.t. an irrotational frame of reference fixed on the

axis of Earth's rotation and ωe is the angular velocity of Earth w.r.t. the same frame of reference.

.·. ωS = ωSE + ωe

now (GMem)/R2 m Rω2,

where R = radius of Earth as the satellite is rotating close to its surface and M e is the mass of Earth and

m is the mass of satellite.

.·. M = (R3 ω2)/G = R3/G [2Π/TSE + 2Π/TE ]2.

where TSE is the apparent time period of satellites revolution and TE = 24 hrs. is the time period of

Earth's rotation.

Substituting the values

M = 6.0 × 1024 kg.

9

Two questions on gravitation were included in the IIT-JEE 2010 question paper. They are given below with solution. The first question is a ‘single correct choice type’ (single answer type multiple choice) question where as the second question is an ‘integer type’ question which has a single digit integer ranging from 0 to 9 as answer.

(1) A thin uniform annular disc (see figure) of mass M has outer radius 4R and inner radius 3R. The work required to take a unit mass from point P on its axis to infinity is(A) (2GM /7R) (4√2 – 5)(B) – (2GM /7R) (4√2 – 5)(C) GM /4R(D) (2GM /5R) (√2 – 1)The work required to take a unit mass from infinity to point P is the gravitational potential at P and it will be negative. Therefore, the work required to take a unit mass from point P to infinity will be numerically equal to the gravitational potential at P but it will be positive.The annular disc can be considered to be made of a large number of concentric rings of radii ranging from 3R to 4R. Consider one such ring of radius r and small thickness dr. Its mass dm is given bydm = [M/π(16R2 – 9R2)] 2πr dr = 2Mrdr/7R2

If the rim of the ring is at distance x from the point P, the gravitational potential (dV) at P due to the ring is given by dV = – Gdm/x = – (2GM/7R2)(rdr/x) But x = (16R2 + r2)1/2 and so we have dx= (½ )×(16R2 + r2)–1/2 ×2rdrOr, dx = rdr/(16R2 + r2)1/2 = rdr/xTherefore dV = – (2GM/7R2)dxThe gravitational potential at P due to the entire annular disc is given byV = ∫ dV = – ∫(2GM/7R2)dx = – (2GM/7R2)[x] The integration is between the appropriate limits of x, which are (16R2 + 9R2)1/2 and (16R2 + 16R2)1/2, on substituting the values r = 3R and r = 4R respectively in the expression, x = (16R2 + r2)1/2. Therefore we haveV = – (2GM/7R2)[(16R2 + 16R2)1/2 – (16R2 + 9R2)1/2] = – (2GM/7R2)[R√(32) – R√(25)] Thus V = – (2GM/7R) (4√2 – 5)The negative sign is to be dispensed with to obtain the work required to take a unit mass from point P to infinity and the correct choice is (2GM/7R) (4√2 – 5). (2) Gravitational acceleration on the surface of a planet is (√6/11)g, where g is the gravitational acceleration on the surface of the earth. The average mass density of the planet is 2/3 times that of the earth. If the escape speed on the surface of the earth is taken to be 11 kms–1, the escape speed on the surface of the planet in kms–1

will be :[This is is an ‘integer type’ question. You have to work out this question and the single digit integer answer is to be shown in the Objective Response Sheet (ORS) by darkening the appropriate bubble].

10

http://2.bp.blogspot.com/_lRKubY9oOyI/S8XYWj04lhI/AAAAAAAABBg/Zi-SEaGpDBM/s1600/Gravitation-pp14-4-10.jpg

We have vesc = √(2gR) where g is the acceleration due to gravity and R is the radius of the planet (or earth or star). Therefore vesc α √(gR).If gp and ge are the accelerations due to gravity on the surface of the planet and the earth, Rp and Re are their radii, and gp and ge the escape velocities respectively, we havevp/ve = √( gpRp)/ √( geRe) = √[( gp/ge)(Rp/Re)] ………(i)Since the gravitational acceleration on the surface is given by g = GM/R2 where G is the gravitational constant and M is the mass of the planet (or earth or star), we can writeg = G[(4/3)πR3ρ] /R2 where ρ is the average mass density of the planet (or earth or star).Therefore g α Rρ and gp/ge = (Rp/Re)(ρp/ρe).From this Rp/Re = (gp/ge)/(ρp/ρe).But gp/ge = √6/11 and ρp/ρe = 2/3 as given in the question. Therefore we haveRp/Re = (√6/11)×3/2Substituting these in Eq(i),vp/ve = √[(√6/11) (√6/11)×3/2] = (√6/11) ×√(3/2) = 3/11Since ve = 11 kms–1, the escape speed on the surface of the planet in kms–1 will bevp = 11×(3/11) = 3 kms–1 The answer integer is 3.

11

Solved examples of Gravitation

Example 1

The time period of Moon around the Earth is n times that of Earth around the Sun. If the ratio of

the distance of the Earth from the Sun to that of the distance of Moon from the Earth is 392, find the

ratio of mass of the Sun to the mass of the Earth. (Assume that the bodies revolve in circular orbits)

Solution:

The time period Te of Earth around Sun of mass Ms is given by

Te2 = 4Π2/GMe × re

3, ...... (1)

where re is the radius of the orbit of Earth around the Sun.

Similarly, time period Tm of Moon around Earth is given by

Tm2 = 4Π2/GMe × re

3, ...... (2)

where rm is the radius of the orbit of Moon around the Sun.

Dividing equation (1) by (2), we get

(Te/Tm)2 = (Me/Ms) (re/rm)3

.·. (Ms/Me) = (Tm/Te )2× (re/rm)3 ...... (3)

Substituting the given values, we get

(Ms/Me ) = (392)3 n2 Ans.

Example 2

Two balls of mass m each are hung side by side, by two long threads of equal length l. If the distance

between upper ends is r, show that the distance r' between the centres of the ball is given by gr'2 (r-r')

= 2l G M.

12

Solution:

The situation is shown in figure given above.

Following forces act on each ball

(i) Weight of the ball mg

(ii) Tension in thread T

(iii) Force of Gravitational attraction

F = G(mm/r2).

As the system is in equilibrium these three forces can be represented by the sides of a triangle

say ACP in which AC represents the weight mg, CP represents the force F and PA represents the tension

T. Hence

tan = ((r-r')/2)/l = F/mg ...... (2)

But F/mg = (Gm2/r'2)/mg ...... (1)

From equations (1) and (2)

((r-r')/2)/l = (Gm2/r'2)/mg or (r-r')mg/2 = lGm2/r'2

gr'2 (r-r') = 2l Gm. (Proved)

Example 3

A planet of mass m moves in an elliptical orbit around the Sun so that its maximum and minimum

distances from the Sun are equal to r1 and r2 respectively. Find the angular momentum of this planet

relative to the centre of the Sun.

Solution:

13

As the angular momentum of the planet is constant (no external torque is acting on it), we have

mv1r1 = mv2r2

or v1r1 = v2r2

Further, the total energy of the planet is also constant, hence

-GMm/r1 + 1/2 mv12 = -GMm/r2 + 1/2 mv2

2

where M is the mass of the Sun.

Example 4




Solution:




centripetal force.

14



= m r ω2 = m (2d/3)ω2

.·. G(2m)m/d2 = m(2d/3) ω2

ω = √((3Gm/d3 ) )

.·. T = 2Π/ω=2Π√((d3/3Gm) ) Ans.




Example 5

Imagine a planet whose diameter and mass are both one half of those of Earth. The day's temperature

of this planet's surface reaches upto 800K. Find whether oxygen molecules are possible in the

atmosphere of this planet.

Solution:

Escape velocity, ve = √2GM/R

Let vp = escape velocity on the planet.

ve = escape velocity on the Earth.

.·. vp/ve =√(MP/RP × RE/ME) = √(1/2×2/1) = 1

.·. vp = ve = 11.2 km/s.

From kinetic theory of gases

vrms = √(3RT/M) = √(3NKT/M) = √(3NKT/Nm)

where N = Avogadro's number

15

m = mass of oxygen molecule

K = Boltzmann constant

vrms = √3RT/m

= √((3×1.38×10-23×800)/(5.3×10-26)) (m = 5.3 × 10-26 kg).

vr.m.s = 0.79 km/s Ans.

As vrms is very small compared to escape velocity on the planet, molecules cannot escape from the

surface of the planet's atmosphere.

Example 6

Halley's comet has a period of 76 years and in 1986 it had a distance of closest approach to the Sun

equal to 8.9 × 10+10 m. What is the comet's farthest distance from the Sun if mass of Sun is 2 × 1030 kg

and G = 6.67 × 10-11 MKS units.

Solution:

Whenever a heavenly body revolves round another body it follows an elliptical path having a distance o

closest and farthest approach.

For these orbits

T2 = (4Π2/GM)a3 , where a and T are average distance of comet from Sun and time period of revolution

respectively.

.·. a = [T2GM/4Π2]1/3

or, a = [((76×3.15×107)2× 6.67×10-11×2×1030)/4Π2 ]1/3

≈ 2.7 × 1012 m.

But for ellpse,

2a = rmin. + rmax.

(where a is the semi major axis of the ellipse)

or, rmax. = 2a - rmin.

rmax = 2 × 2.7 × 1012 - 8.9 × 1010

rmax ≈ 5.3 × 1012 m. Ans.

Example 7

16

A uniform sphere has a mass M and radius R. Find the pressure P inside the sphere caused by

gravitational compression, as a function of distance r from its centre.

Solution:

Density of the given sphere, ρ = M/(4/3ΠR3)

Let us consider a layer of thickness 'dr' at a distance r from the centre. The part of sphere inside this

spherical shell will attract this with a net force but the outer part would not have any effect on this as

inside a hollow spherical shell force on any particle is zero.


Mass of sphere of radius r, M1 = 4/3ΠR3 ρ

.·. M1 = M/R3.r3

Mass of the layer of thickness dr is

dm = 4Πr2.dr.ρ

= 4Πr2.dr.M/(4/3ΠR3 )

= 3M/R3dr . r2

.·. Force on small layer due to M1

dF = (GM1.dm)/r2 = (GM.r3.3M.r2)/(R3.r2.R3) dr

= 3GM2/R6 dr.r3

Thus pressure on the layer of mass 'dm'

dp = dF/4Πr2

= 3GM2/R6 . (r3.dr)/r2 = 3GM2/(4ΠR6).rdr

.·. Total pressure due to mass extending from r to R is

p = ∫rR dp = 3GM2/R6 ∫r

R rdr

17

= 3GM2/(8ΠR6 ) . [r2/2]rR

= 3GM2/(8ΠR6 ) (R2 - r2)

.·. p = 3GM2/(8ΠR4 ) (1-r2/R2 )

Example 8

Assuming the Earth to be a sphere of uniform density, calculate the energy needed to completely

disassemble it against gravitational pull amongst its constituent particles given product of mass and

radius of Earth = 2.5 × 1031 kgm.

Solution:

Let us suppose that Earth is made of a infinite number of very thin concentric spherical shells. It can be

completely disassembled by removing these shells one by one.

Let us study Earth when only a sphere of radius x is left and find the energy required to remove a shell

of thickness dx.

Now Potential of this sphere of radius x is u = Gms/x

[ms = mass of sphere = 4/3 Πx3ρ where ρ = density of Earth = M/(4/3ΠR3) where M, R are mass and

radius of Earth respectively].

Now work done (dw) to remove a shell of thickness dx would be equal to the change in potential energy

= Gmsdm/x,

where dm = mass of shell = 4Πx2 dxρ.

=> dW = (4/3 x3 ρ)(4Πx2 dxρ)/x = 16/3 GΠ2ρ2x4 dx..

W = ∫dW = ∫0R 16/3GΠ2 ρ2 x4 dx = 16/15 GΠ2 [m/((4/3)Π R3 )]2 R5..

Which, on substituting the given value, is 3/5 GM2/R = 3/5 g MR

= 1.5 × 1032 J.

Example 9

A satellite revolving close to the surface of Earth from west to east, appears over a certain point at the

equator every 11.6 hours. If radius of Earth is 6400 km. calculate the mass of Earth

Solution:

Since the rotation of Earth is also west to east the apparent angular velocity of the satellite is equal to

ωSE = ωS - ωe,

18

Where ωS is the angular velocity of the satellite w.r.t. an irrotational frame of reference fixed on the

axis of Earth's rotation and ωe is the angular velocity of Earth w.r.t. the same frame of reference.

.·. ωS = ωSE + ωe

now (GMem)/R2 m Rω2,

where R = radius of Earth as the satellite is rotating close to its surface and M e is the mass of Earth and

m is the mass of satellite.

.·. M = (R3 ω2)/G = R3/G [2Π/TSE + 2Π/TE ]2.

where TSE is the apparent time period of satellites revolution and TE = 24 hrs. is the time period of

Earth's rotation.

Substituting the values

M = 6.0 × 1024 kg.

19

1. A solid sphere of uniform density and radius 4 units is located with its centre at origin O of coordinates. Two units of equal radii 1 unit, with their centre at A

(-2,0,0) and B (2,0,0) respectively are taken out of the solid leaving behind spherical cavities as shown in Fig. Choose the incorrect statement from the following.

a)The gravitational force due to this object at origin is zero.

b)The gravitational force at point B (2,0,0)

c)The gravitational potential is the same at all points of the circle Y2 + Z2 = 36

d)The gravitational potential is the same at all points of the circle Y2 + Z2 = 4

Ans. (b)

The distance of each cavity from the centre O is the same. The two cavities are symmetrical with respect to the centre O and the mass of the sphere being concentrated at the centre O, the gravitational force due to the sphere is zero at the centre. Hence choice (a) is correct. Similarly, the gravitational potential is the same at all points of the circle Y2+Z2 = 36 whose radius is 6 units and at all points of the circle Y2+Z2 = 4. So choices (c) and (d) are also correct.

2. If the distance between the earth and the sun were half its present value, the number of days in a year would have been

20

a)64.5 (b) 129

b)182.5 (d) 730

Ans. (b)

According to Kepler’s law of periods,

T1/T2 = (R1/R2)3/2 = [ R1/ (R1/2)]3/2

= (2)3/2 = 2√2

T2 = T1/2√2 = 365 days/2√2 = 129 days

3. An artificial satellite moving in a circular orbit around earth has a total (kinetic+potential) energy E0. Its potential energy is

(a) –E0 (b) 1.5E0

(c) 2E0 (d) E0

Ans. (c)

For a satellite we know

Kinetic energy = GmM/2r

Potential energy = -GmM/r

Total energy E0 = KE + PE

= (GmM/2r) – (GmM/r) = -GmM/2r = PE/2 ---------- ( -GmM/r = PE)

or PE = 2E0

4. A satellite S is moving in an elliptical orbit around the earth. The mass of the satellite is very small compared to the mass of the earth. Which of the following statement is correct ?

21

(a) The acceleration of S is always directed towards the centre of the earth.

(b) The angular momentum of S about the centre of the earth changes in direction, but its magnitude remains constant.

(c) Total mechanical energy of S remains constant.

(d) The linear momentum of S remains constant in magnitude.

Ans. (a)

For elliptical orbit, the earth is at one focus of the ellipse. For spherical bodies, the gravitational force is central. Hence statement (a) is correct.

5. Two objects of masses m and 4m are at rest at infinite separation. They move towards each other under mutual gravitational attraction. Then at a separation r, which of the following is true?

(a) The total energy of the system is not zero.

(b) The force between them is not zero.

(c) The centre of mass of the system is at rest.

(d) All the above are true

Ans. (d)

At a finite separation the total kinetic energy of the system of masses and the force between them are both finite. Since the two masses are at rest initially and there is no external force, the centre of mass cannot move. Hence the correct choice is (d)

6. A satellite is launched in to a circular orbit of radius R around the earth. A second satellite is launched into an orbit of radius 1.01R. The period of the second satellite is longer than that of the first by approximately

0.5% (b) 1.0%

1.5% (d) 3.0%

22

Ans. (c)

We know according to Kepler’s law of periods, T2 = KR3 where k is a constant. Taking logarithm of both sides, we have

2 log T = log k + 3 log R

Differenting, we get

2 (dT / T) = 0 + (3 dR / R)

= 3/2 x [(1.01R – R) / R] x 100 = 1.5%

so correct choice is (c)

7. A simple pendulum has a time period T1 when on the earth surface, and T2 when taken to a height R above the earth’s surface, where R is the radius of the earth. The value of T2/T1 is

(a) 1 (b) √2

(c) 4 (d) 2

Ans. (d)

The acceleration due to gravity at a height h above the surface of the earth is given by

g2 = g1 [R / R+h]2

where g1 is the value at the surface of the earth

T2 = 2π √l/g2 and T1 = 2π √l/g1

T2 / T1 = √g1/g2 = R+h / R = R+R / R = 2 (h=R)

8. An ideal spring with spring constant is k is hung from the ceiling and a block and a block of mass M is attached to its lower end. The mass is released with the spring initially unstretched. Then the maximum extension in the spring is

23

(a) 4 Mg/k (b) 2 Mg/k

(c) Mg/k (d) Mg/2k

Ans. (b)

Let x be the extension in the spring when it is loaded with mass M. The change in gravitational potential energy = Mgx. This must be the energy stored in the spring which is given by ½ kx2. Thus ½ kx2 = Mgx or x = 2mg / k

Hence the correct choice is (b)

24

Gravitation Practice Problems - Solutions

1. All planets orbit the sun in ellipses, not perfect circles which means that there is a point in each planet’s orbit when it is closest to the sun and when it is farthest from the sun. Comets orbit the sun in much the same way, but the distance differences are much more drastic. As Halley’s Comet orbits the sun, its closest distance is 5*1010m and its farthest distance is 4.5*1012m.

a. Without doing any calculations, compare the gravitational forces between the comet and the sun at both the closest and farthest points in the orbit?

The farthest distance is 90 (4.5*1012/5*1010 = 90) times greater than the closest distance. Therefore, the force becomes (1/90)2 [or 0.000123]

times the original size.

b. If we assume the comet has a mass of 3*1014kg, calculate the two gravitational forces.FG closest = G*3*1014*2*1030/ (5*1010)2 = 1.6*1013 N

FG farthest = G*3*1014*2*1030/ (4.5*1012)2 = 1.98*109 N

2. How do we know that Jupiter has a force exerted on it?It moves in a circular orbit

3. What exerts the force on Jupiter?The Sun exerts a gravitational force on Jupiter

4. What do we mean when we say that Jupiter falls?Jupiter falls from its straight line or tangential path onto the curved path of its orbit

5. How do gravitational force and distance compare?Gravitational force is proportional to the inverse square of the distance

6. At what height above the earth would a 300kg satellite have to orbit in order to experience a gravitational force half as strong as it is on the Earth’s surface?

FG on the Earth’s surface = G*300*6*1024/(6.4*106)2 = 2931 N½ * 2931 = 1465.5 = G*300*6*1024/d2

d2 = G*300*6*1024/1465.5 = 8.19*1013 d = 9.05*106m = x +REARTH

x = 9.05*106 – 6.4*106 = 2.65*106m above the surface of the Earth

7. The acceleration due to gravity on Venus is 0.89 that of Earth. If the radius of Venus is 6.05*106m, what is Venus’ mass?

FG = 0.89*9.8 = 8.72= G*1*MVENUS/(6.05*106)2

MVENUS = 8.72*(6.05*106)2/(G*1) = 4.79*1024 kg

8. Which exerts a greater force, the Sun on Venus or Venus on the Sun?They both exert the same force on one another.

By Newton’s Third Law, we know this must be true

25

Newton’s Law of Gravitation

Newton's Law of Gravitation states that "Every particle in the universe attracts every other particle with

a force that is directly proportional to the product of their masses and inversely proportional to the

square of the distance between them".

Consider two bodies A and B of masses mA and mB, attracting each other with forces AB (force on A due

to B) and BA (force on B due to A), respectively.

Then, AB = - BA

and

AB = | AB| = magnitude of the attractive force

= G(mA mB)/d2

where d is the distance between them. G is a universal constant known as Universal Gravitational

constant. Its value was first measured by Cavendish and is now known to be:

G = 6.62726 × 10-11 N-m2/kg2

"The space around a body within which its force of gravitational attraction is perceptible (by any other

body in this space) is called its gravitational field."

The intensity E, , of the gravitational field of a mass 'm' at a point at distance 'r' from it is the force

experienced by a unit mass placed at this point in the field. (Assuming that the presence of unit mass

does not affect the gravitational field of the mass m)

Thus, E = (-m/r2)G (the negative sign is because the intensity of the field is directed towards the

mass and away from the unit mass).

If 1, 2, 3 ...... are the gravitational forces acting on the particle A due to particles P1, P2, P3, ......,

respectively, then net gravitational force on a particle A due to particles P1, P2, ......, Pn is given by:

= 1 + 2 +........+ n

Illustration:

Two Lead balls of radius 10 cm and 1 cm are placed with their centres 1 metre apart. Calculate the

force of attraction between them. The density of Lead is 5.51 × 103 kg/m3.

Newton’s Law of Gravitation

26

Solution:

Density of balls, ρ = 5.51× 103 kg/m3

Force of attraction between them, F = Gm1m2/r2

where m1 = 4/3Πr23 × ρ = 5.51 × 103 × 4/3 Π10-3 ≈ 23 kg

and m2 = 4/3Πr23 × ρ ≈ 23 × 10-3 kg

G = 6.67 × 10-11 Nm2 kg-2

.·. F = 6.67 × 10-11 × 21 × 21 × 10-3N = 3.5 × 10-11 N Ans.

Important points

← The gravitational force is an attractive force.

← The gravitational force between two particles does not depend on the medium.

← The gravitational force between two particles is along the straight line joining the particles

(called line of centers).

Illustration:

Three identical particles, each of mass m, are placed at the three corners of an equilateral triangle of

side 'a'. Find the force exerted by this system on another particle of mass m placed at

(a) the mid-point of a side

(b) the centre of the triangle

Solution:

(a) Using the principle of superposition

= A + B + C

When the particle is placed at the mid point of a side (at P), C = - B, and they cancel each other.

Hence, force experienced by the particle, = A

| | = | FA | = Gmm/(AP)2 = Gm2/(a sin 600)2

27

(b) If the particle is placed at the centre of the triangle, the net force on the particle P due to particles

placed at the corners A, B and C will be zero.

Hence, = A + B + C= 0.

Gravitational Field and Intensity

The space around a body where the gravitational force exerted by it can be experienced by any other

particle is known as the gravitational field of the body. The strength of this gravitational field is referred

to as intensity, and it varies from point to point.

Consider the gravitational field of a particle of mass m located at the origin (O).

Suppose that a test particle of mass m0 is placed at the point P(x, y, z). The force of gravitational

attraction exerted on the test particle is given by,

g = (GMmo/r2)

where the position vector = r,

r = OP = | | = | | = r

and the unit vector, = / r

The intensity of this gravitational field at a point (P) is given by the force per unit mass on a test

particle kept at P, i.e.

= g/mo

28

where is the gravitational intensity and g is the gravitational force acting on the mass m0. The

gravitational field is, therefore, a vector field.

The gravitational field at P due to a particle of mass m kept at the point O (origin) is given by

= g/mo = {-(Gmmo/r2) } * 1/mo = Gm/r2

where = xi + yj + zk represents the position vector of the point P with respect to the source at the

origin and = / r represents the unit vector along the radial direction.

The superposition principle extends to gravitational field (intensities) as well:

= 1 + 2 + 3 +....+ n

where 1, 2,.... n are the gravitational field intensities at a point due to particle 1, 2, ......, n respectively.

For a continuously distributed mass, the formula changes to =∫d , where d gravitational field

intensity due to an elementary mass dm.

Gravitational Field and Intensity

The gravitational field of a ring on its axis

29

Let us consider a ring of mass M in the plane perpendicular to the plane of the paper. We want to find

the gravitational field on its axis at a distance x.

Consider a differential length of the ring of mass dm.

dE = Fdm/z2

The Y-components of the fields due to diametrically opposite elements cancel each other. Thus, the X-

components add up.

E = ∫Gdm/z2 cosα = Gcosα/z2 ∫dm = GMcosα/z2 = GMx/(a2+x2)3/2 ←

Field due to a uniform thin spherical shell

Consider a thin spherical shell of radius 'a', mass M and of negligible thickness. Out of the spherical

shell we consider a small ring of thickness (R dθ). The shaded ring has mass dm = (M/2) sin θ dθ. The

field at P due to thing ring is

dE = Gdm/z2 cosα = GM/2 (sinθ dθ cosα/z2)←

From ΔOAP,

z2 = a2 + r2 - 2ar cos θ

or 2z dz = 2ar sin θ dθ

or sin θ dθ = z dz/ar

Also, from ΔOAP,

30

a2 = z2 + r2 - 2zr . cos α ; cos α = (z2+r2-a2)/zr

Thus, dE = GM/(4ar2)[1-(a2-r2)/z2]dz

or ∫dE = GM/(4ar2)[z+(a2-r2)/z2]

Case I:

P is outside the shell, r > a

E = GM/(4ar2 ) [z+(a2-r2)/z2 ](r-a)(r+a) = GM/r2

We see that the shell may be treated as a point particle of the same mass placed at its centre to

calculate the gravitational field at an external point.

Case II:

P is outside the shell, r > a

E = GM/(4ar2)[z+(a2-r2)/z2 ](a-r)(a+r) = 0

We see that the field inside a uniform spherical shell is zero.

Gravitational field outside a solid sphere

The sphere can be thought of as composed of many shells from radius = 0 to radius = a.

The point P is at a distance r from the centre of all these concentric shells.

=> E = G/r2[ΔM1 + ΔM2 +...]

E = GM/r2

Gravitational field inside a uniform solid sphere of radius 'R'

To find the field at a point P inside the sphere at a distance r < R form the centre, let us consider a

sphere of radius r.

31

Consider a point P on the surface of the shaded sphere. Since this point is inside the shells having radii

larger than r, they do not contribute to the field at P. Shells that are less than radius 'r', contribute to

the gravitational field at P.

The mass of the sphere of radius r is

M' = (M.4/3 Π r3)/(4/3 Π R3 ) = (Mr3)/R3

.·. EP = (GM')/r2 = (GM r)/R3

The adjacent graph shows the variation of E due to a solid sphere of radius R with the distance r from

its centre.

E = GM/r2 (r > R)

E = (GM/R3) r (r < R)

← This result holds good for the earth if it is assumed to be a uniform solid sphere.

← As by definition, g = Fg/m and also E = Fg/m , so g = E, i.e. acceleration due to gravity and

gravitational intensity E at a point are synonymous.

Illustration:

Two concentric shells of masses M1 and M2 are present. Calculate the gravitational force on 'm' due to

M1 and M2 at points P, Q and R.

32

Solution:

Field at P, EP = 0 => F = 0

Field at Q, due to M2 will be zero but there will be field due to m1,

EQ = (GM1)/b2

=> F = (GM1m)/b2

Field at R, is the sum of fields due to M1 and M2, ER = G(M1+M2 )/c2

Acceleration due to Gravity (g)

The Earth attracts a mass m on its surface by a force F given as:

F = GMe m/Re2 ,

where Me is the mass of the Earth and Re its radius.

This force imparts an acceleration to the mass m, which is known as acceleration due to gravity

(g).

By Newton's second law, acceleration = Force/ Mass ,

=> g = F/m = GMe/Re2

Illustration: Find the value of g at the surface of Earth?

Radius of Earth = 6.37 × 106 meter.

Mass of Earth = 6 × 1024 kg.

Solution:

As g = GM/R2 ...... (1)

G = 6.67 × 10-11 Nm2 kg-2

33

M = 6 × 1024 kg

R = 6.37 × 104 m

Put all values in (1), we get g = 9.8 m/s2 Ans.

Variation of Acceleration due to Gravity (g)

i) Due to altitude

Consider a mass m at a height h from the surface of the earth. Now, the force acting on the mass due

to gravity is F = G , where M is the mass of the earth and R is the radius of the earth.

If the acceleration due to gravity at the given height is g', then mg' = G ,

=> g' = G

(Expanding binomially and neglecting the higher order terms).

(ii) Due to depth:

If a particle of mass m is kept at a depth 'd' form the surface of earth, then gravitational force exerted

on the particle of mass 'm'.

(GM' m)/(R-d)2 = GM(R-d)/R3

where M' = mass of earth within radius of (R - d)

.·. M' = M/R3(R-d)3 = (GM/R3) R(1-d/R) = g(1-d/R)

(iii) Due to rotation of the earth:

Consider a body at a point with altitude θ, on the surface of the earth.

Let R = radius of the earth and ω = angular velocity of the earth about its own axis.

F.B.D. of the body with respect to an inertial frame is shown in the adjacent figure.

34

Acceleration of the body with respect to the earth's centre O is (Rcos θ)ω2 directed towards the axis of

rotation (i.e. the centripetal acceleration).

From Newton's second law in the radial direction

mg - Fn = m(R cos θ) ω2 cos θ

or Fn = m[g - Rω2 cos2 θ]

or Fn/m = g' = g - Rω2 cos2 θ

where g' is the apparent value of the acceleration

At poles, θ = 90o => g' = g

At the equator, θ = 0o => g' = g - Rω2

Figure given below has illustrated the variation of g with the distance of separation from Earth's centre.

Illustration:

Find the value of g at a height equal to the radius of Earth.

Solution:

g = GM/(R+h)2 and g0 = GM/R2 (At the surface of Earth).

g/g0 = R2/(R+h)2 = R2/4R2 ·.·h = R

=> g = g0/4 = 9.8/4 = 2.45 m/s2. Ans.

Caution: Here h = R, so we cannot apply g' = g(1-2h/Re)

35

Illustration:

At what angular velocity Earth should rotate, for the weight of an object at the equator to be zero?

What would be the duration of a day in this case?

Radius of Earth = 6.4 × 106 m

g0 = 9.8 m/s2

Solution:

For the weight to be zero, the value of g should be zero. That is

Here, g' = g0 - Reω2 = 0

or ω = √(g0/Re ) = √(9.8/(6.4×106 )) = 1.2 × 10-3 rad/s

The duration of one day will be equal to the time period of rotation

T = 2Π/ω = 2Π/1.2*10-3 sec.

Orbital Velocity

When a satellite revolves in an orbit around a planet, it requires centripetal force to do so. This

centripetal force is provided by the gravitational force between the planet and the satellite. If the mass

of the satellite is m and that of the planet is M, then the gravitational force between them at a height h

above the surface of the planet is

F = GMm/(R+h)2', where R is the radius of the earth.

If the speed of the satellite in its orbit is (R + h), then the required centripetal force is mv2/(R+h).

.·. mv2/(R+h) = (G Mm)/(R+h)2

or v = √(GM/(R+h))=√(GM/R(1+h/R)) = √(gR/(1+h/R))

If the height is very small compared to the radius of the earth, then

v = √gR

If the time period of the satellite is 24 hrs rotating in the same sense as the rotation of the earth and if

the plane of the orbit is at right angle to the polar axis of the planet (earth), then the satellite will

always be above a certain place of the earth.

This kind of a satellite is called geostationary satellite.

Illustration:

36

An artificial satellite of mass 100 kg is in circular orbit at 500 km above the earth's surface. Take the

radius of the earth as 6.5 × 106 m.

(a) Find the acceleration due to gravity at any point along the satellite path.

(b) What is the centripetal acceleration of the satellite?

Solution:

Here, h = 500 km = 0.5 × 106 m

R = 6.5 × 106 m

r = R + h = 6.5 × 106 + 0.5 × 106 = 7.0 × 106 m

(a) g' = g(R/R+h)2 = 9.8((6.5 * 106)/(7.0 * 106)) = 8.45 m/s2

(b) Centripetal force on the satellite, F = mv2/R

.·. Centripetal acceleration, a = F/m = v2/r = (√(gR2/r))2/r

= (gR2)/r2 =g R2/(R+h)2 = 8.45 m/s2

Gravitational Potential Energy

Change in gravitational potential energy of a system is defined as the -ve of the work done by the

gravitational force as the configuration of the system is changed.

Uf - Ui = Wgr

Change in gravitational potential energy of two point masses m1 and m2 as their separation is changed

from r1 to r2 is given by

U(r2) - U(r1) = Gm1m2 [1/r1 - 1/r2]

If, at infinite separation, gravitational potential energy is assumed to be zero, then the gravitational

potential energy of the above two point mass system at separation r,

U(r) = -Gm1m2/r

Illustration:

What is the gravitational Potential energy of the Moon-Earth system, relative to the potential

energy at infinite separation?

37

Solution:

M = Mass of Earth = 5.98 × 1024 kg

m = mass of Moon = 7.36 = 1022 k

d = mean separation between Earth and moon = 3.82 × 108 m.

.·. Gravitational potential energy of the Moon-Earth system (U)

= -GMm/d

Put all values U = -7.68 × 1028 J Ans.

Illustration:

A projectile is fired vertically from the Earth's surface with an initial velocity of 10 km/s. Neglecting

atmospheric retardation, how far above the surface of the Earth would it go? Take the earth's radius as

6400 km.

Solution:

Let the projectile go up to a height h. Then the law of conservation of mechanical energy gives

1/2 mv2 - GMm/R = -GMm/R+h

h = v2R2/2GM-v2R

= v2R2/2gR2-v2R

where g is the acceleration due to gravity on the surface of the Earth.

= ((104)2 (6.4×106)2)/(2(9.8)(6.4×106)2-(104)2 (6.4×106))

Gravitational Potential

Gravitational field around a material body can be described not only by gravitational intensity vector

but also by a scalar function, the gravitational potential V. The gravitational at any point may be

defined as the potential energy per unit mass of a test mass placed at that point.

V = U/m (where U is the gravitational potential energy of the test mass m).

Thus, if the reference point is taken at infinite distance, the potential at a point in the gravitational field

is equal to amount of work done by the external agent per unit mass in bringing a test mass from

infinite distance to that point. The expression for the potential is given by

V = ∫P∞ . d

With the above definition, the gravitational potential due to a point mass M at a distance r from it is

38

V = -∫r∞ GM/r2 . d = ∫r

∞GM/r2 dr = -GM/r

Potential is a scalar quantity. Therefore, at a point in the gravitational field of a number of material

particles, the resultant potential is the arithmetic sum of the potentials due to all the particles at that

point. If masses m1, m2, ......, mn are at distances r1, r2, r3, ......, rn then potential at the given point is

V = -G(m1/r1 + m2/r2 + m3/r3 +..........)

The field and the potential are related as, E = -dV/dr

Gravitational potential due to a shell

(i) at a point outside the shell is: -GM/r (r>R)

(ii) at a point on the surface of the shell is: -GM/R

(iii) at a point inside the shell is: -GM/R

Gravitational potential (V) due to a uniform solid sphere

(i) Outside of the sphere at a distance r form the centre, V = -GM/r.

(ii) Inside the sphere at a distance r from the centre, V = -GM/R3 (R2/2 - r2/6)

Relation between Gravitational Potential and The Gravitational Field Strength

Gravitational field strength at a point is given by

E = GM/r2

We can also describe the gravitational field of a body by a scalar function called the potential. Let

us begin with a test particle of mass m0 at an infinite separation from the body and move the test

particle toward the body until their separation is r, where the potential energy is U(r). Now the

gravitational potential V at that point is defined as V(r) = U(r)/m0.

As we can calculate the radial component of gravitational force from U(r) according to the

relation

F = -dU/dr => m0E = -m0 dv(r)/dr

Hence, we can find the radial component of the gravitational field intensity from V(r)

According to the relation E = -dV/dr.

Binding Energy

39

Binding energy of a system of two bodies is the amount of minimum energy needed to separate the

bodies to a large distance.

If two particles of masses m1 and m2 are separated by a distance r, then the gravitational potential

energy of the system is given by

U = Gm1m2/r

Let T amount of energy is supplied to the system to separate the bodies by a large distance. When the

bodies are separated by a large distance, gravitational potential energy of the system is zero. For

minimum T, conserving energy for initial and final positions,

T + U = 0

=> T - Gm1m2/r = 0 or T = Gm1m2/r

Hence, binding energy of a system of two particles separated by a distance r is equal to

T = Gm1m2/r, where m1 and m2 are the masses of the particles.

Illustration:

Two spherical bodies of masses 2M and M and of radii 3R and R, respectively, are held at a distance

16R from each other in free space. When they are released, they start approaching each other due to

the gravitational force of attraction. Then, find:

(a) the ratio of their acceleration during their motion.

(b) their velocities at the time of impact.

Solution:

(a) Due to the mutual attraction, the masses attract each other.

If the acceleration are a1 and a2, the net external force on the system = 0

=> cm = 0 => m1a1 - m2a2 = 0

or a1/a2 = m2/m1 = 1/2

40

Gravitational Potential Energy

(b) Taking both the bodies as a system, from conserving momentum of the system,

m1v1 - m2v2 = 0 => m1/m2 = v2/v1 = 2

Now, conserving the total mechanical energy, we have

1/2 (2M) V12 + 1/2MV2

2 - G(2m)M/4R = - G(2m)M/16R and solving it we get.

v1 = √GM/8R and v2 = 2√GM/8R

Note: The velocities and acceleration are w.r.t. the inertial reference frame (i.e. the centre of

mass of the system).

Illustration:

At a point above the surface of earth, the gravitational potential is -5.12 × 107 J/kg and the

acceleration due to gravity is 6.4 m/s2. Assuming the mean radius of the earth to be 6400 km, calculate

the height of this point above the earth's surface.

Solution:

Let r be the distance of the given point from the centre of the earth. Then,

Gravitational potential = -GM/r = -5.12 × 107 J/kg ...... (1)

and acceleration due to gravity,

g = GM/r2 = 6.4 m/s2 ...... (2)

Dividing (1) by (2), we get

r = 5.12*107/6.4 = 8 × 106 m = 8000km

.·. HEscape Velocity

41

Escape velocity on the surface of earth is the minimum velocity given to a body to make it free from

the gravitational field, i.e. it can reach an infinite distance from the earth.

Let ve be the escape velocity of the body on the surface of earth and the mass of the body to be

projected be m. Now, conserving energy at the surface of the earth and infinity,

1/2 mve2 - GM/R = 0 => ve = √2GM/R.

Illustration:

The mass of Jupiter is 318 times that of earth, and its radius is 11.2 times the earth's radius. Estimate

the escape velocity of a body from Jupiter's surface. [Given: The escape velocity from the earth's

surface is 11.2 km/s.]

Solution:

Hence, MJ = 318 Me; RJ = 11.2 Re; ve = 11.2 km/s

We know, vJ = √2GMj/Rj and ve = √2GMe/Re

.·. vJ/ve = √(Mj/Me χ Re/Rj)

=> vJ = ve √(Mj/Me χ Re/Rj)

vJ = 11.2{318Me/Me χ Re/11.2Re}1/2 = 11.2(318/11.2)1/2 = 59.7 km/s

Illustration:

Find the escape speed from a point at a height of R/2 above the surface of earth. Assuming mass of

earth as M and its radius as R.

Solution:

Conserving mechanical energy of a point mass m which is to escape, and earth system we have,

-GMm/(R+R/2) + 1/2mv2 = 0

=> v = √(4GM/3R)

height of the point from earth's surface = r - R = 8000 - 6400 = 1600 km

Illustration:

The escape velocity of a body on the surface of the Earth is 11.2 km/s. A body is projected away

with twice this speed. What is the speed of the body at infinity? Ignore the presence of other heavenly

bodies.

42

Solution:

If v is the velocity of projected and v' is the velocity at infinity, then we have by energy

conservation principle.

1/2mv2 - GMm/R = 1/2mv'2 + 0

Here v = 2ve

.·. (1/2) . 4ve2 - GM/R = 1/2v'2

=> 2ve2 - GM/R = 1/2v'2

Now ve = √2GM/R

=> 2ve2 - v2e/2 = 1/2v'2

or, v'2 = 3 ve2

or, v' = √3 ve = √3 × 11.2 km/s = 19.4 km/s. Ans.

Gravitational Self Energy of a Uniform Sphere

Consider a sphere of radius R and mass M uniformly distributed. Consider a stage of formation at which

the radius of the spherical core is r. Its mass will be 4/3 Πr3ρ at that time, where r is the density of the

sphere.

Let an additional mass be added so that the radius of the core be increased by dr in the form of a

spherical layer over the core. The mass of this layer will be 4Πr2dr. ρ

The mutual gravitational potential energy of the above mass and the spherical core of radius r,

dU = (-G(4/3Πr3ρ)(4Πr2dr.ρ))/r = 16/3 Π2ρ2Gr4dr

Hence, total energy evolved in the formation of the spherical body of radius R i.e., self energy.

Satellites and Planetary Motion

Earth and its Satellite

43

Consider a satellite of mass m revolving in a circular orbit around the Earth, 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 r = R e

+ h, where Re is the radius of the Earth. The gravitational force between Me & m provides the

centripetal force necessary for circular motion, i.e.,

GMem/(Re+h)2 = mv2/(Re+h)

Or v2 = GMe/(Re+h) or v = √GMe/(Re+h)

Hence orbital velocity depends on the height of the satellite above Earth's surface. Time period T

of the satellite is the time taken to complete one revolution.

Therefore T = 2Πr/v = 2Π(Re + h)√(Re+h)/GMe

or T2 = 4Π2(Re+h)3/GMe where r = Re + h

If time period of a satellite is 24 hrs. Then

r = [GMeT2/4Π2]1/3 = 42400 km and h = 36000 km.

This gives the height of a satellite above the Earth's surface whose time period is same as that of

Earth's. Such a satellite appears to be stationary when observed from the Earth's surface and is hence

known as Geostationary satellite.

For a satellite very close to the surface of Earth i.e. h << Re then

r ≈ Re

vorbital = √GMe/Re = √gRe

Satellite in circular orbit

For different velocities, the trajectory of the satellite would be different. Let us consider these

cases.

If v is the velocity given to a satellite and v0 represents the velocity of a circular orbit and ve the

escape velocity.

i.e. v0 = √GMe/(Re+h)

ve = √2GMe/(Re+h)

Where, h is the distance of the satellite from the surface of the Earth.

(1) When v < v0, the satellite follows an elliptical path with centre of the Earth as the further focus. In

this case, if satellite is projected from near surface of the Earth, it will hit the Earth's surface without

completing the orbit.

(2) If v = v0, obviously the satellite follows a circular orbit with centre of Earth as the centre of the

orbit.

44

(3) If v0 < v < ve, then the satellite follows an elliptical orbit with centre of the Earth as the nearer

focus.

(4) If v = ve, the satellite escapes the gravitational field of the Earth along a parabolic trajectory.

(5) If v > ve, the satellite escapes the gravitational field of Earth along a hyperbolic trajectory.

Note: The derivation of the above cases is beyond the scope of JEE.

Enquiry: What should be the energy required to shift a satellite orbiting around the Earth to infinity?

At infinity the Potential Energy of the satellite would be zero and if we want to supply minimum

Energy then its kinetic energy would also be zero. Let us first find the total Energy of the satellite.

Total Energy = Kinetic Energy + Potential Energy

= 1/2 mv2 + (-GMem/r)

= 1/2 m.GMe/r - GMem/r

= -GMem/2r

Now, Binding Energy would be equal to - (Total Energy) as it is the energy needed to shift the

satellite from its orbit to infinity.

So, the energy required = GMem/2r.

Here, r = Re + h.

Enquiry: If we see a satellite from Earth, how long will it take for one revolution?

Let us consider a satellite in circular orbit with a time period Ts. The Earth also rotates with the

time period Te = 24 hrs. If an observer on Earth sees this satellite, the angular velocity of the satellite

will be SE = S - e. Hence, the time period of revolution will seem different from Ts and will be

observed as TSE.

Two cases arise for calculation of TSE.

(i) If the satellite and Earth are revolving and rotating respectively in the same direction.

Hence, 2Π/TSE = |2Π/Ts - 2Π/Te|

=> TSE = TsTe/|Te - Ts|

(ii) If the satellite and Earth are revolving and rotating respectively in the opposite direction.

Hence, 2Π/TSE = 2Π/Ts + 2Π/Te

45

or TSE = TsTe/|Ts - Te|.

Kepler's Laws- Elliptical Motion of Planets and Satellites

One of the greatest ideas proposed in human history is the fact that the earth is a planet, among the

other planets, that orbits the sun. The precise determination of these planetary orbits was carried out

by Jhannes Kepler, using the data compiled by his teacher, the astronomer Tycho Brahe. Johannes

Kelper discovered three empirical laws by using the data on planetary motion.

1. Each planet moves in an elliptical orbit, with the sun at one foci of the ellipse.

2. A line from the sun to a given planet sweeps out equal areas in equal intervals of time.

3. The square of the periods of the planets are proportional to cube of their mean distance (or semi-

major axis) from the sun.

These laws go by the name 'Kepler's laws of planetary motion'. It was in order to explain the origin of

these laws, among other phenomena, that Newton proposed the theory of gravitation.

In our discussion, we are not going to derive the complete laws of planetary motion from Newton's law

of gravitation. Since most of the planets actually revolve in near circular orbits, we're going to assume

that the planets revolve in circular orbits.

Consider a planet of mass m rotating around the sun (mass M >> m) in a circular orbit of radius r with

velocity v. Then, by applying Newton's law of gravitation and the second law of motion, we can write

Gravitational force = mass × centripetal acceleration

i.e. GMm/r2 = m(v2/r) ... (1)

or, v2 = GMm/r ... (2)

As the moment of the gravitational force about S is zero, the angular momentum of the planet about

the sun remains constant. This is the meaning of Kepler's 2nd law of motion, as will be shown later.

The time period of rotation, T, of the planet around the sun is given by,

T = 2Πr/v = -2Πr/√GM/r = 2Π/√GM r3/2 ... (3)

Squaring both sides,

T2 = (4Π2/GM)r3 ... (4)

46

which is Kepler's 3rd law of motion.

Note: The constant of proportionality in the above equation depends only on the mass of the sun (M)

but not on the mass of the planet.

Kepler's Laws are also valid for the motion satellites around the earth.

Kepler's Second Law

Consider a planet P that moves in an elliptical orbit around the sun, and let P and P' be the positions of

the planet at time t and t + Δt (where Δt is a very small time interval). If the angular displacement of

the planet is Δθ, then the are swept out by the line joining the planet and sun (SP) in time Δt is:

ΔA = area of the section SPP'

= 1/2 r2.Δθ; where r = the length SP.

The area velocity, vA = ΔA/Δt = 1/2r2Δθ/Δt = 1/2r2ω = constant .... (5)

In other words,

m × (2vA) = constant as well (m = mass of the planet)

Areal velocity = dA/dt = L/2m ... (6)

This is the expression for the angular momentum of the planet,

L = Iω = mr2ω

= mr2 (dθ/dt) perpendicular to the plane of its orbit.

The gravitational force,

= -GMm/r2 is centripetal, and the torque on the planet is zero,

i.e. = * = * (-GMm/r2 )= 0 ... (7)

47

Hence, the angular momentum of the planet does not vary, i.e. the areal velocity of the planet remains

constant. At its aphelion (farthest point from the sun, r is large), the planet moves slowly and at its

perihelion (nearest point from the sun, r is small) the planet moves fastest.

Illustration:

Calculate the mass of the Sun from the following data; distance between the Sun and the Earth = 1.49

× 1011 m, G = 6.67 × 10-11 SI units and one year = 365 days.

Solution:

Force of attraction between the sun and the earth = GmsmE/dSE2

Considering the orbit of the earth as nearly circular, the centripetal force acting on the earth is m E

dSEω2.

=> mE dSE ω2 = GmsmE/dSE2

mS = dSE2.ω2/G = 4Π2dSE

2/GT2

=(4×(3.14)2×(1.49×1011)2)/(6.67×10-11×(365×24×60×60)2)

=1.32 × 1019 kg.

Illustration:

A Saturn year is 29.5 times the earth year. How far is Saturn from the sun (M) if the earth is 1.5 × 10 8

km away from the sun?

Solution:

It is given that

TS = 29.5 Te; Re = 1.5 × 1011 m

Now, according to kepler's third law

TS2/Te

2 = Rs3/Re

3

RS=Re(TS/Te)2/3=1.5×1011 ((29.5Te)/Te)2⁄3=1.43×1012 m =1.43×109 km

Illustration:

A planet of mass m moves along an ellipse around the sun so that its maximum and minimum

distances form the sun are equal to R and r, respectively. Find the angular momentum of this planet

relative to the centre of the sun.

Solution:

48

According to Kepler's second law, the angular momentum of the planet is constant.

.·. mv1R = mv2r or v1R = v2r

If the mass of the Sun is M, conserving total mechanical energy of the system at two given positions we

have,

-GMm/R + 1/2 mv12 = -(G M m)/r + 1/2 mv2

2

.·. GM[1/R - 1/r] = v12/2 + v2

2/2 or GM[(r-R)/Rr] = v12/2 + v2

2R2/2r2

.·. v12= (2GM(R-r) r2)/Rr(R2-r2 ) = 2GMr/R(R+r)

Now, angular momentum = mv1R = m√2GMR/(R+r)

49

Documents

DD Gravitation