Gravitation: DPP 1

Gravitation - Daily Practice Problems

1 Summary / Formula Sheet – 1.2. Acceleration due to Gravity and its

Gravitation Variation
• At Earth’s surface:
1.1. Newton’s Law of Gravitation, Gravi-
tational Field and Superposition GME
g= 2
RE
• Force between two point masses:
m1 m2 • At height h above surface (h ≪ RE ):
F = G 2 , G = 6.67 × 10−11 N·m2/kg2
r
 
2h
• Force is along the line joining the masses and gh = g 1 −
obeys superposition: RE
X
F⃗net = F⃗i • At depth d below surface:
i  
d
• Gravitational field (intensity): gd = g 1 −
RE
F⃗ M
⃗g = −∇V = = − G 2 r̂ • At latitude λ (rotation effect):
m r
• For continuous distribution: gλ = g − ω 2 RE cos2 λ
Z
dm
⃗g = −G r̂
r2 • Relation with density:
• Field due to uniform ring on axis: 4
g = πGρRE
GM x 3
g=
(x2 + R2 )3/2
1.3. Gravitational Potential and Potential
• Field due to uniform circular disc on its axis: Energy
 
x • Potential at distance r from mass M :
g = 2πGσ 1 − √
x2 + R 2
GM
M V =−
where σ = . r
πR2
• Field due to uniform spherical shell: • Relation with field:
GM dV
– Outside (r > R): g = g=−
r2 dr
– Inside (r < R): g = 0
• Field due to solid sphere: ⃗g = −∇V
GM ∂V ∂V ∂V
– Outside (r > R): g = or equivalently, gx = − , gy = − , gz = −
r2 ∂x ∂y ∂z
GM r
– Inside (r < R): g = (varies linearly
R3
with r)
Gravitation: DPP 2
• Potential due to uniform spherical shell: 1.5. Kepler’s Laws of Planetary Motion
GM • First Law: Planets move in ellipses with Sun
V =− (r ≤ R) at one focus.
R

• Potential due to solid sphere: • Second Law: Equal areas in equal times (con-
stant areal velocity).
GM
V =− (3R2 − r2 ) (r ≤ R) 1 L
2R3 Ȧ = r2 θ̇ =
2 2m
• Potential energy of two masses:
• Third Law: Square of period ∝ cube of semi-
Gm1 m2 major axis.
U =−
r 4π 2 3
T2 = a
GM
• Relation between field and potential energy:
Z • Vis-viva equation:
∆U = − F⃗ · d⃗r
 
2 2 1
v = GM −
r a
1.4. Escape Velocity and Motion of Satel-
lites • Energy of planet in orbit:

• Escape velocity from planet surface: GM m

E=−
r 2a
2GM p
ve = = 2gR • Ratio of perihelion and aphelion speeds:
R
vp ra
• Orbital velocity for circular orbit: =
va rp
r
GM
v=
r

• Relation between ve and v:

√
ve = 2 v

• Total energy of satellite:

GM m
E =K +U =−
2r
2 Problems
• Time period of revolution:
r (Take G = 6.67 × 10−11 N m2 /kg2 , ME = 6 × 1024
r3 kg, RE = 6.4 × 106 m wherever necessary )
T = 2π
GM
2.1 Newton’s Law of Gravitation, Gravi-
• Geostationary orbit conditions: tational Field & Superposition Principle
T = 24 h, r ≈ 4.216 × 107 m 1. State Newton’s law of universal gravitation and
explain how the constant G is defined. Write its
• Minimum work to shift satellite from r1 to r2 : SI units and dimensions.
 
1 1 2. Two particles of masses m1 and m2 are separated
∆E = GM m −
2r1 2r2 by distance r. Derive the expression for the mu-
tual gravitational force acting between them.

3. Calculate the gravitational force between two

spheres of masses 10 kg and 20 kg separated by
0.5 m (take G = 6.67 × 10−11 N m2 /kg2 ).
 Gravitation: DPP 3
4. State the principle of superposition for gravita- 17. A thin disc of radius 0.3 m and uniform surface
tional forces and illustrate it for three masses density σ = 5 kg/m2 has its axis vertical. Find
placed at the corners of an equilateral triangle. the field intensity at a point 0.2 m above its cen-
tre.
5. Three equal masses M are placed at the corners
of an equilateral triangle of side a. Find the net 18. Two identical uniform discs of radius R are placed
gravitational force on one mass due to the other coaxially with their planes parallel and separated
two. by distance x. Show by superposition that the
field at a point midway between them is zero.
6. Define gravitational field intensity at a point. De-
rive an expression for the field due to a point mass 19. Define gravitational flux and state Gauss’s law in
M at distance r. gravitation. Using it, derive g = GM/r2 for a
spherical distribution.
7. A mass M = 2 × 1024 kg creates a field. Find
the gravitational field intensity at a point r = 20. A spherical planet of mass 6 × 1024 kg and radius
6 × 106 m from its centre. 6.4×106 m is surrounded by a concentric spherical
shell of mass 1 × 1024 kg and radius 12.8 × 106 m.
8. Show that for multiple masses m1 , m2 , . . . located Find the gravitational field at a point (a) inside
at position vectors ⃗r1 , ⃗r2 , . . ., the net field at P is the planet, (b) between planet and shell, (c) out-
X mi side the shell.
⃗g = −G (⃗r − ⃗ri ).
|⃗r − ⃗ri |3
i
21. Conceptually explain how the field at any point
9. Find the gravitational field at a point on the per- can be obtained using vector superposition, and
pendicular bisector of two equal masses M sep- describe how symmetry helps simplify the inte-
arated by 2a. Express the result in terms of gration for continuous mass distributions.
G, M, a, and the distance x of the point from the
midpoint. 22. Discuss the physical significance of the inverse-
square nature of gravitational force and how de-
10. Two equal masses M are placed at x = +a and viations from it would alter planetary orbits.
x = −a. At what point on the x-axis between
them is the net field zero if a third small mass m
2.2 Acceleration due to Gravity and its
is placed there?
Variation with Altitude, Depth and Lat-
itude
11. State and prove that inside a uniform spherical
shell of mass M and radius R, the gravitational
23. Define acceleration due to gravity. Derive g =
field is zero. GM
at the surface of Earth.
R2
12. Derive the expression for the gravitational field at
a point outside a uniform spherical shell, showing 24. Show that g decreases inversely as the square
that the shell acts as if all its mass were concen- of distance from Earth’s centre, i.e. g′ =
2
trated at its centre.

R
g .
R+h
13. A uniform spherical shell of radius 1 m has mass
5 kg. Find the field at a point 0.5 m from its cen- 25. Calculate g ′ at an altitude of 1000 km above
tre (inside) and 2 m from its centre (outside). Earth’s surface (R = 6400 km).

14. Obtain an expression for the gravitational field on 26. For small altitude h ≪ R, derive the approximate
the axis of a uniform ring of mass M and radius ∆g 2h
relation =− .
R at a point P distant x from its centre. g R

15. A ring of radius 0.2 m has mass 2 kg. Find the 27. Compute percentage decrease in g at a height of
field at a point on its axis 0.1 m from the centre. 400 m from Earth’s surface.

16. Derive the expression for the gravitational field 28. Derive the expression for variation of g with depth
at a point on the axis of a uniform thin disc of d inside Earth assuming uniform density.
radius R and surface mass density σ.
 Gravitation: DPP 4
29. Find g ′ at a depth of 1200 km from surface 44. Explain why the value of g at mountains is
(R = 6400 km). slightly smaller than at sea level even though the
altitude is small.
30. Show that the rate of change of g with depth
equals that with height near Earth’s surface. 45. A pendulum clock calibrated at equator is moved
to a pole. Will it gain or lose time? Explain
31. Derive the expression for effective g at latitude λ quantitatively using the concept of effective g.
due to Earth’s rotation.
46. Explain how the variation of g with latitude and
32. At what latitude is g maximum and where is it altitude affects weight measurements on Earth.
minimum? Explain physically.
2.3 Gravitational Potential and Poten-
33. Calculate the difference in g at equator and poles tial Energy
using ω = 7.27 × 10−5 rad/s, R = 6.37 × 106 m.
47. Define gravitational potential at a point. Derive
34. Find the value of latitude where g is 0.3% less expression for potential at a distance r from a
than its value at the pole. point mass M .

35. Show that the fractional decrease in g due to ro- 48. Show that gravitational potential V and field g
ω2R dV
tation is at the equator. are related by g = − .
g dr
36. If Earth’s angular velocity doubled, find the per- GM
centage change in g at the equator. 49. If V = − , find gravitational field at a point
r
where r = 2RE .
37. A tunnel is drilled through Earth along a diam-
eter. Derive the expression for acceleration of a 50. Derive an expression for gravitational potential
particle inside Earth as a function of its distance energy of two point masses (m1 , m2 ) separated
from the centre. by distance r.

38. Show that the motion of a particle dropped into 51. Compute gravitational potential energy between
a frictionless tunnel through Earth is simple har- Earth and Moon given ME = 6 × 1024 kg, MM =
monic and find its time period. 7.4 × 1022 kg, r = 3.84 × 108 m.

39. Calculate the time period of oscillation of a body 52. A 10 kg mass is taken from infinity to a point
in such a tunnel (assume R = 6400 km, g = 9.8 2 × 106 m from a planet of mass 8 × 1023 kg. Find
m/s2 ). work done by external force.

40. A projectile is launched vertically upward to a 53. Show that the potential at any point outside a
height where acceleration due to gravity becomes uniform spherical shell is the same as that of a
half its surface value. Find that height in terms point mass at its centre.
of radius of Earth.
54. Find potential at a point inside a thin uniform
41. At what depth (in terms of radius of Earth) be- spherical shell of mass M and radius R.
low Earth’s surface will the value of g be reduced
to half of its surface value? 55. Derive the expression for potential at a point in-
side a solid uniform sphere of radius R and mass
42. Compare variation of g with altitude and with M.
depth on the same graph and explain how they
differ. 56. Sketch the graph of potential V (r) for a uniform
sphere and mark continuity at r = R.
43. If a satellite moves very close to Earth’s surface,
what is its orbital period? Show that it equals the 57. Find potential difference between centre and sur-
time period of oscillation in the tunnel of Q38. face of a uniform sphere of mass M and radius
R.
 Gravitation: DPP 5
58. If M = 6 × 1024 kg and R = 6.4 × 106 m, find 2.4 Escape Velocity and Motion of Satel-
Vc − Vs for Earth. lites

59. Show that the energy required to move a mass m 72. Define escape velocity and derive its expression
GME m in terms of G, M , and R of the planet.
from Earth’s surface to infinity equals .
RE
73. Calculate the escape velocity from Earth’s surface
60. A satellite of mass 500 kg is placed in an orbit at (ME = 6 × 1024 kg, RE = 6.4 × 106 m).
RE + 300 km. Calculate the change in potential
energy compared to that at Earth’s surface.
74. Find escape velocity from Moon (MM = 7.4×1022
kg, RM = 1.74 × 106 m).
61. Prove that gravitational potential energy of a uni-
3GM 2
form solid sphere is U = − . 75. Show that escape velocity can be expressed as
5R √
2gR for bodies where g is known.
62. A planet of mass 8×1023 kg and radius 4×106 m.
Compute its self-gravitational potential energy. 76. A body projected at 0.8 times escape velocity
from Earth. Find the maximum height attained
63. Show that escape velocity can also be obtained before it comes momentarily to rest.
using the concept of potential energy difference
between surface and infinity. 77. A satellite is orbiting close to Earth’s surface (M,
R). Derive its orbital velocity and show relation
64. Two point masses 2 kg and 3 kg are 1 m apart. with escape velocity.
Find potential at midpoint and potential energy
of the system. 78. Find ratio of escape velocity to orbital velocity
for a circular satellite orbit around Earth.
65. A 5 kg body is moved from 20 m to 10 m from
another 50 kg body. Find change in potential en- 79. Find orbital velocity of a satellite orbiting 400
ergy. km above Earth’s surface (ME = 6 × 1024 kg,
RE = 6.4 × 106 m).
66. Derive expression for potential due to a circular
ring of mass M and radius R at a point on its 80. A 1000 kg satellite orbits Earth at 300 km alti-
axis. tude. Find (a) orbital speed, (b) total energy, and
(c) kinetic and potential energies.
67. A ring of mass 10 kg and radius 0.5 m. Find po-
tential at its centre and at a point 0.5 m on its 81. Show that total energy of a satellite in a circular
axis. orbit is half of its potential energy and negative.

68. Using superposition, find potential at the centre 82. Derive expression for period of revolution of a
of a square of side a whose corners each carry satellite in circular orbit of radius r.
equal mass M .
83. Compute period of a satellite orbiting 400 km
69. Show that if potential at infinity is zero, then po- above Earth’s surface.
tential everywhere due to positive masses is neg-
ative. 84. Define geostationary satellite. Find its orbital ra-
dius around Earth.
70. Explain why potential inside a uniform shell is
constant though field is zero. 85. Find orbital speed of a geostationary satellite.

71. A body is moved slowly from r1 to r2 in the field of 86. A 500 kg satellite is raised from Earth’s surface
mass M . Derive work done W = GM m( r12 − r11 ). to a circular orbit of radius 2RE . Find work done
by external force.

87. Find change in total energy of a satellite when its

orbit radius increases from R to 2R.
 Gravitation: DPP 6
88. Show that for elliptical orbit, total energy E = 103. Derive mathematically that T 2 ∝ r3 for circular
GM m orbits using Newton’s law of gravitation.
− , where a is semi-major axis.
2a
4π 2 3
89. A satellite moves in an elliptical orbit with perigee 104. Given T 2 = r , find the value of GM⊙ using
GM
distance RE and apogee distance 4RE . Find ratio Earth’s data (r = 1.5 × 1011 m, T = 1 year).
of its speeds at perigee and apogee.
105. A planet is 4 times farther from the Sun than
90. Explain qualitatively why satellites remain in or- Earth. Find its orbital period in years.
bit without continuous thrust.
106. State how Kepler’s third law can be modified to
91. Find altitude for a satellite of period T = 2 hours apply for planets around any star of mass Ms .
around Earth.
107. A satellite orbits a planet at r1 = 2 × 107 m with
92. Show that T2
∝ r3
for circular orbits and deter- period T1 = 6 h. Find period at r2 = 4 × 107 m.
mine the proportionality constant in terms of G
and ME . 108. Show that the ratio of squares of orbital periods of
any two satellites around the same planet equals
93. If a satellite is to orbit Mars (M = 6.4 × 1023 kg) the ratio of cubes of their orbital radii.
in 3 hours, find orbital radius.
109. The planet Mars has a mean orbital radius of
94. A satellite is fired from Earth with a speed 1.2 2.28 × 1011 m. Find its period of revolution
times orbital speed. Find the nature of its trajec- around the Sun.
tory and discuss its energy.
110. Explain why planets move faster at perihelion and
95. Show that minimum energy required to launch a slower at aphelion.
satellite of mass m from Earth’s surface
 into a cir-

GM m R
cular orbit at altitude h is 1− . 111. A comet travels in an elliptical orbit with perihe-
2R R+h
lion q = 0.5 AU and aphelion Q = 4.5 AU. Find
its eccentricity and semi-major axis.
96. Find the fractional change in kinetic energy if a
satellite is moved from orbit radius R to 1.5R.
112. Using conservation of angular momentum, derive
the ratio of comet’s speeds at perihelion and aphe-
97. Derive escape speed from the Moon’s orbit
lion.
around Earth for a spacecraft intending to leave
the Earth–Moon system.
113. Compute ratio of perihelion to aphelion speed for
the comet in above Q.
2.5 Kepler’s Laws of Planetary Motion
114. A planet’s angular momentum per unit mass is h.
98. State Kepler’s three laws of planetary motion and h
Show that its areal velocity is .
explain the physical meaning of each. 2

99. Using Newton’s law of gravitation, derive that 115. A planet of mass m moves under Sun’s force. De-
planetary motion under an inverse-square force rive expression for total energy in terms of semi-
leads to Kepler’s second law. major axis a.

100. Show that areal velocity of a planet around the 116. For Earth, find its total11orbital energy per unit
Sun is constant. mass using a = 1.5 × 10 m and GM⊙ = 1.33 ×
1020 m3 /s2 .
101. Derive the expression for areal velocity in terms
GM m
of angular momentum per unit mass. 117. Show that for elliptical orbit, E = − and
2a
that E < 0 implies a bound orbit.
102. For Earth, calculate its areal velocity in its nearly
circular orbit around the Sun (r = 1.5 × 1011 m, 118. A newly discovered planet has period 8 years.
v = 3 × 104 m/s). Find its mean distance from the Sun in AU.
 Gravitation: DPP 7
119. If a planet’s orbit becomes slightly elliptical, how 6. Field from point mass M :
will its period change compared to a circular orbit
of same a? M
⃗g = − G r̂
r2
120. A planet of period T moves in an orbit of ec-
centricity e. Show that its mean orbital speed 7. g = GM/r2 = 6.67×10−11 · 2×1024 /(6×106 )2 .
2πa
vm = .
T g ≈ 3.70 m/s2
121. Derive expression for kinetic and potential energy
8. By superposition of point-mass fields:
of a planet at any point in its elliptical orbit.
X mi
122. At perihelion, a planet’s speed is 3.2 × 104
m/s at ⃗g = −G (⃗r − ⃗ri )
|⃗r − ⃗ri |3
i
distance 1.47×1011 m. Find its speed at aphelion
1.52 × 1011 m.
9. Components cancel horizontally; vertical compo-
nents add:
123. Describe briefly how Kepler’s laws can be deduced
from Newton’s law of gravitation. 2GM x
g(x) = toward midpoint
(a2 + x2 )3/2

10. Equal masses at ±a give zero field at the mid-

point.
x=0

11. By shell theorem, net internal field cancels every-

where.
⃗g = 0

3 Answers 12. For r > R, field equals that of a point mass at

centre.
M
3.1 Newton’s Law of Gravitation, Gravi- ⃗g = − G 2 r̂
r
tational Field & Superposition Principle
1 m1 m2 13. Inside g = 0, outside g = GM/r2 = 6.67×10−11×
1. F ∝ m1 m2 and F ∝ 2
⇒F =G 2 .
r r 5/4.

[G] = N m2 kg−2 = m3 kg−1 s−2 g(0.5) = 0, g(2) = 8.34 × 10−11 m/s2

2. Inverse-square central force along the line joining 14. Project ring elements on axis; tangential parts
the masses: cancel:
m1 m2 GM x
F⃗12 = − G 2 r̂ g(x) =
r (R + x2 )3/2
2

(10)(20) GM x
3. F = G = 6.67×10−11 ×800. 15. g = .
(0.5)2 + x2 )3/2
(R2

F = 5.34 × 10−8 N g ≈ 1.19 × 10−9 m/s2

4. By vector addition for 60◦ angle between forces: 16. Disc as ring superposition:
√  
Fnet = 3F x
g(x) = 2πGσ 1 − √
x2 + R 2
5. Each force F = GM 2 /a2 ; resultant at 60◦ .
17. Substitute R = 0.3, x = 0.2, σ = 5.
√ GM 2
Fnet = 3 toward centre g ≈ 9.3 × 10−10 m/s2
a2
 Gravitation: DPP 8
18. Fields from identical discs oppose each other ex- 28. Uniform-density Earth ⇒ M (r) ∝ r3 ⇒ g(r) ∝ r:
actly.
⃗gmid = 0
 d
gd = g 1 −
R
⃗ = −4πGM , derive g = GM/r2 .
H
19. From ⃗g ·dA
29. d/R = 1200/6400 = 0.1875 ⇒ g ′ = 0.8125 g (≈
g=
GM 7.96 m/s2 ).
r2
g ′ ≈ 0.8125 g (≈ 7.96 m/s2 )
20. Inside planet: g = GMp r/Rp3 ;
between shells:
GMp /r2 ; outside: G(Mp +Ms )/r2 .
30. Near surface: g(h) ≈ g(1 − 2h/R), g(d) =
GMp r GMp g(1 − d/R). For equal magnitude h and d/2, first
(a) g = , (b) g = , derivatives match:
Rp3 r2
G(Mp + Ms ) dg 2g dg g
(c) g = . = , =
r2 dh 0 R dd 0 R

R dm
21. ⃗g = −G r̂, reduced by symmetry. 31. Effective g: ⃗geff = ⃗g − ω
⃗ × (⃗
ω ×⃗r) with magnitude:
r2
Superposition + symmetry simplify integration. g(λ) = g0 − ω 2 R cos2 λ

22. If F ∝ 1/rn , n ̸= 2, orbits not closed; Kepler’s 32. ω 2 R cos2 λ is max at λ = 0◦ and zero at 90◦ :
laws fail.
gmax at poles, gmin at equator
Inverse–square law ensures closed conic orbits.
33. ∆g = gpole − geq = ω2R ≈
3.2 Acceleration due to Gravity and its (7.27×10−5 )2 (6.37×106 ):
Variation with Altitude, Depth and Lat-
itude ∆g ≈ 3.37 × 10−2 m/s2 (∼ 0.34%)

23. From F = GM m/R2 and F = mg at surface:

34. Set ω 2 R cos2 λ = 0.003 gp :
GM
g= 0.003 gp
R2 cos2 λ = ≈ 0.872 ⇒ λ ≈ 21◦
ω2R
GM  R 2
24. At distance R+h: g ′ = = g .
(R + h)2 R+h ∆g ω2R
35. At equator: geq = g − ω 2 R ⇒ = .
 2 g g
′ R
g =g
R+h ∆g ω2R
=
g g
 6400 2
25. g ′ = g ≈ 0.748 g (≈ 7.33 m/s2 if g =
7400 36. If ω → 2ω, reduction becomes 4ω 2 R; extra de-
9.8).
crease is 3ω 2 R:
g ′ ≈ 0.748 g (≈ 7.33 m/s2 )
3ω 2 R
% ↓ g ≈ 100 × ≈ 1.03%
26. Binomial for h ≪ R: g′ ≈ g(1 − 2h/R) so g
∆g/g ≈ −2h/R.
37. Inside uniform Earth: g(r) = GM (r)/r2 =
∆g 2h
≈− GM r/R3 toward centre:
g R
g
2h 800 a(r) = − r
27. % drop ≈ 100× = 100× ≈ 0.0125%. R
R 6.4 × 106

≈ 0.0125%
 Gravitation: DPP 9
38. a(r) = − Rg r is SHM with ω = g/R:

p
3.3 Gravitational Potential and Poten-
tial Energy
s
R R r GM
T = 2π 47. V (r) = − ∞ ′2 dr′
g r
GM
39. T = 2π
p p
R/g = 2π 6.4×106 /9.8: V (r) = −
r

T ≈ 5.07 × 103 s ≈ 84.6 min 48. For central V (r): ⃗g = −∇V = −

dV
r̂
dr
g
2 √
40. Set g ′ = : R
= 1
⇒ h = ( 2 − 1)R: dV
2 R+h 2 g= −
dr
h ≈ 0.414 R (≈ 2.65 × 103 km) GM GM
49. g(r) = 2
⇒ g(2RE ) = 2
r 4RE
1 R
41. Set gd = g/2: 1 − d/R = 2 ⇒d= 2: g0
g(2RE ) =
4
d= R
2 (≈ 3.2 × 103 km)
R r Gm1 m2 ′
50. U (r) = − ∞ dr
42. Altitude: g(h) ∼ g(1 − 2h/R) initially, then r′2
∝ 1/(R + h)2 ; depth: g(d) = g(1 − d/R) linear to Gm1 m2
zero at centre. U (r) = −
r
Altitude: nonlinear (↓ as 1/r2 ); Depth: linear decrease GME MM
51. U = −
p r
43. Circular orbit just above surface: T = 2π R/g,
U ≈ −7.7 × 1028 J
same as tunnel-SHM period.
GM m
52. Wext = ∆U = U (r) − U (∞) =
s
R r
Torbit, near surface = 2π
g
Wext ≈ 2.67 × 108 J

44. Reduced g from altitude (h) and local mass deficit 53. By shell theorem / integration, outside acts like
(rock density) lowers weight slightly at moun- point mass at centre
tains.
GM
V (r) = − (r ≥ R)
Smaller g ⇒ slightly smaller weight at high altitude r

p 54. Inside thin shell, g = 0 ⇒ V = const = V (R)

45. geq < gpole ; pendulum T = 2π L/g. Moving to
pole: g ↑, T ↓: GM
V (r) = − (r ≤ R)
R
∆T ∆g
Clock gains time; ≈ − 12 ≈ −0.17% 55. Match V and dV /dr at r = R for uniform density
T g
GM
3R2 − r2


46. Apparent weight W = mgeff ; geff decreases with V (r) = − 3
(r ≤ R)
2R
altitude and toward equator, hence weights vary
geographically. GM
56. Vout = −GM/r, Vin = − 3 (3R2 − r2 ) are con-
2R
W ∝ geff (latitude, altitude) tinuous at r = R
V continuous at r = R, dV /dr (slope) changes

3GM GM
57. Vc = − , Vs = −
2R R
GM
Vc − V s = −
2R
 Gravitation: DPP 10
58. Use GM/R = gR Gm
69. V (∞) = 0 and dV = − 2 dr < 0 toward mass
r
gR ⇒ V decreases
Vc − V s = − ≈ −3.14 × 107 J kg−1
2
V ≤ 0 everywhere (zero only at ∞)
59. ∆U = U (∞) − U (R) = 0 − (−GM m/R) R
70. Inside shell, g = 0 ⇒ ∆V = g dr = 0 along any
GM m path
Ereq =
R
GM
  V = constant inside shell = −
1 1 R
60. ∆U = U (R+h) − U (R) = GM m −
R R+h
R r2 GM m
9 71. W = r1 dr
∆U ≈ 1.4 × 10 J r2
 
m(r) dm 1 1
61. Integrate −G over the sphere W = GM m −
r r2 r1

3GM 2
Uself = − 3.4 Escape Velocity and Motion of Satel-
5R
lites
3GM 2 GM
62. U = − 72. Set total specific energy zero: 1 2
−
5R 2 ve R
=0

U ≈ −6.4 × 1030 J r
2GM
ve =
R
63. Set 12 mve2 + (−GM m/R) = 0
√
r
2GM 73. ve = 2gR with g = 9.8, R = 6.4 × 106
ve =
R
ve ≈ 1.12 × 104 m/s
 
2 3 (2)(3)
64. Vmid = −G + , U = −G r
2GMM
0.5 0.5 1 74. ve =
RM
Vmid = −10G J kg−1 , U = −6G J
ve ≈ 2.38 × 103 m/s
 
1 1
65. ∆U = −G(5)(50) −
10 20 GM p
75. Use g = 2
in ve = 2GM/R
R
∆U = −12.5 G ≈ −8.34 × 10−10 J p
ve = 2gR
66. Sum equal
√ contributions of ring elements at dis-
tance R2 + x2 GM GM R
1 2 2
76. 2 (0.8 ve ) − =− ⇒ = 0.36
GM R R+h R+h
V (x) = − √
R 2 + x2  1 
h= − 1 R ≈ 1.78 R
0.36
√ Vc = −GM/R; on axis at x: V (x) =
67. At centre:
−GM/ R2 + x2
v2 GM p
77. Centripetal: = ⇒ v = GM/R and
Vc ≈ −1.33 × 10−9 J kg−1 , V (0.5) ≈ −9.43 × 10−10 J kg−1 √ R R2
ve = 2 v
√
 
GM r
√
68. r = a/ 2 to each corner; V = 4 − √ GM
a/ 2 vorb, surface = , ve = 2 v
R
√
4 2 GM
Vcentre =−
a
 Gravitation: DPP 11
78. rp + ra vp
√ 89. a = = 2.5RE , vis-viva ⇒ =
ve 2 va
= 2 s
vorb 2/rp − 1/a
p 2/ra − 1/a
79. r = RE + 400 km, v = GME /r vp
=4
va
v ≈ 7.67 km/s
90. Gravity provides centripetal force; tangential in-
ertia sustains motion
p
80. r = RE + 300 km, v = GME /r, K =
GME m GME m GME m
, U =− , E=− Bound orbit with no continuous thrust (ideal, no drag)
2r r 2r

v ≈ 7.73 km/s, K ≈ 2.97 × 1010 J, U ≈ −5.95 × 1010 J, E ≈GM

−2.97
T 2
× 1010 J
1/3
91. r = , alt = r − RE for T = 7200 s
4π 2
GM 1 GM GM m
81. With v 2 = : E = K +U = m − alt ≈ 1.65 × 106 m (≈ 1650 km)
r 2 r r

GM m p
E=− = 12 U (< 0) 92. From T = 2π r3 /GM
2r
4π 2 3
2πr p T2 = r
82. T = with v = GM/r GM
v
GM T 2
1/3
r
r3 93. r = for M = 6.4 × 1023 kg, T =
T = 2π 4π 2
GM 10800 s

r ≈ 5.0 × 106 m (alt ≈ 1.6 × 106 m)

p
83. r = RE + 400 km in T = 2π r3 /GME

T ≈ 5.56 × 103 s ≈ 92.6 min 1 2 GM

94. If v = 1.2 vc at radius r: ϵ = 2v − =
r
GM
2π (0.72 − 1) <0
84. ω = , r = (GM/ω 2 )1/3 ; altitude = r − RE r
86400
√
r ≈ 4.216 × 107 m, alt ≈ 3.5786 × 104 km Elliptic, bound; not escape (escape needs v ≥ 2 vc )

p GM m
85. v = GM/r at r from prev Q 95. E = − ⇒ ∆E = E(R+h) − E(R)
2r
v ≈ 3.07 km/s GM m

R

∆E = 1−
2R R+h
GM m GM m
86. Ei = − , Ef = − ⇒ ∆E = Ef − Ei
2R 4R
K′ R 2
96. K ∝ 1/r ⇒ = =
GM m K 1.5R 3
Wext,min = ∆E =
4R
∆K K′ − K 1
= = − (−33.3%)
K K 3
87. Same computation as prev Q
r
GM m 2GME
∆E = 97. At r ≃ rMoon : vesc = ; for a circular
4R p r
lunar orbit
√ vc = GME /r, minimum tangential
88. From vis-viva v 2 = GM 2/r − a1 and orbit aver- ∆v = ( 2 − 1)vc

age energy
GM m vesc ≈ 1.44 km/s, ∆vmin ≈ 0.414 vc ≈ 0.42 km/s
E=−
2a
 Gravitation: DPP 12
3.5 Kepler’s Laws of Planetary Motion 108. From T 2 = 4π 2 r3 for same M :
GM
98. (1) Orbits are ellipses with Sun at a focus; (2)
equal areas in equal times (⃗τext = 0 ⇒ L const); T12 r13
=
(3) T 2 ∝ a3 . T22 r23

Kepler I, II, III as stated above 109. T = (r/1 AU)3/2 yr, with r = 2.28×1011 m and
1 AU = 1.5×1011 m

GM m
 ⃗
⃗
99. ⃗τ = ⃗r × F = ⃗r × − ⃗
r = ⃗0 ⇒ dL = 0 T ≈ 1.88 years
r3 dt
110. Ȧ const ⇒ v larger where r smaller (perihelion),
Areal velocity constant ⇒ Kepler II smaller where r larger (aphelion).

1 L vp > va by Kepler II
100. Ȧ = r2 θ̇ = = const
2 2m
q+Q 0.5 + 4.5 Q−q
L 111. a = = ,e=
Ȧ = = constant 2 2 Q+q
2m
a = 2.5 AU, e = 0.8
101. With L = mr2 θ̇: vp ra
112. L = mvr = const ⇒ =
L va rp
Ȧ =
2m vp Q
=
va q
102. Ȧ = 12 rv = 12 (1.5×1011 )(3×104 )
113. With Q = 4.5 AU, q = 0.5 AU:
Ȧ = 2.25 × 1015 m2/s
vp
=9
va
mv 2 GM m GM 2πr
103. = 2
⇒ v2 = , T =
r r r v 114. h = L/m = r2 θ̇ ⇒ Ȧ = 21 r2 θ̇
4π 2 3 h
T2 = r Ȧ =
GM 2

115. Vis-viva: v 2 = GM 2/r − a1 , average over orbit

4π 2 r3 11
104. GM = with r = 1.5×10 m, T = 1 yr =
T2
7
3.156×10 s GM m
E=−
2a
20 3 2
GM⊙ ≈ 1.33 × 10 m /s
GM⊙ 1.33×1020
116. ϵ = E/m = − =−
105. T ∝ r3/2 ⇒ T = (4)3/2 yr 2a 2(1.5×1011 )

T = 8 years ϵ ≈ −4.43 × 108 J/kg

117. From central-force two-body result:

106. Replace M⊙ by star mass Ms :
GM m
4π 2 3 E=− < 0 ⇒ bound ellipse
T2 = a 2a
GMs
118. a = (T /yr)2/3 AU with T = 8 yr
T2  r2 3/2
107. = = 23/2 a = 4 AU
T1 r1
p
T2 ≈ 17.0 h 119. For given a, T = 2π a3 /GM unchanged by
small e
T same as circular with same a
 Gravitation: DPP 13
120. Mean speed over one period: vm = 122. L/m = vr const ⇒ va = vp rp = 3.2×104 1.47
path length per cycle 2πa ra 1.52
=
T T
va ≈ 3.09 × 104 m/s
2πa
vm =
T 123. Inverse-square central force ⇒ τ = 0 ⇒ Kepler
II; radial equation ⇒ conic sections (Kepler I);
121. Vis-viva: v 2 = GM 2/r − a1 and U = −GM m/r


energy/orbit geometry ⇒ T 2 ∝ a3 (Kepler III).

GM m
 
1 2 1 1 GM m
K= − U =− Kepler I–III follow from F =
2 mv = GM m
r 2a
,
r r2