Download Chapter 5 Center of Mass and Linear Momentum

Chapter 7
Chapter 7
Gravitation
Gravitation
Basic Requirements
1. Concept of the gravitational force;
2. Newton's law of gravitation;
3. Gravitation and the principle of superposition;
4. Gravitation near Earth's surface;
5. Gravitation inside Earth;
6. Gravitational potential energy;
7. Kepler's Laws;
8. Satellites: orbits and energy.
Review and Summary
The Law of Gravitation Any particle in the universe attracts any other particle with
a gravitational force whose magnitude is
F G
m1m2
r2
(Newton’s law of gravitation)
(7-1)
where m1 and m2 are the masses of the particles, r is their separation, and
G (  6.67 1011 N m2 / kg 2 ) is the gravitational constant.
Gravitational Behavior of Uniform Spherical Shells Equation 7-1 holds only for
particles. The gravitational force between extended bodies must generally be found by
adding (integrating) the individual forces on individual particles within the bodies.
However, if either of the bodies is a uniform spherical shell or a spherically symmetric
solid, the net gravitational force it exerts on an external object may be computed as if
all the mass of the shell or body were located at its center.
Superposition Gravitational forces obey the principle of superposition; that is, if

n particles interact, the net force F1, net on a particle labeled as particle 1 is the sum
- 168 -
Chapter 7
Gravitation
of the forces on it from all the other particles taken one at a time:
n 

F1, net   F1i
(7-5)
i2

in which the sum is a vector sum of the forces F1i on particle 1 from

particles 2,3,..., n . The gravitational force F1 on a particle from an extended body is
found by dividing the body into units of differential mass dm, each of which produces

a differential force dF on the particle, and then integrating to find the sum of those
forces:


F1   dF
Gravitational Acceleration
(7-6)
The gravitational acceleration a g of a particle (of
mass m ) is due solely to the gravitational force acting on it. When the particle is at
distance r from the center of a uniform, spherical body of mass M , the magnitude
F of the gravitational force on the particle is given by Eq.7-1. Thus, by Newton’s
second law,
F  mag
(7-8)
GM
r2
(7-9)
which gives
ag 
Free-Fall Acceleration and Weight

The actual free-fall acceleration g of a

particle near Earth differs slightly from the gravitational acceleration a g , and the
particle’s weight (equal to mg) differs from the magnitude of the gravitational force
acting on the particle as computed with Eq.7-1, because Earth is not uniform or
spherical and because Earth rotates.
Gravitational Within a Spherical Shell A uniform shell of matter exerts no net
gravitational force on a particle located inside it. This means that if a particle is located
inside a uniform solid sphere at distance r from its center, the gravitational force
exerted on the particle is due only to the mass M ins that lies inside a sphere of radius
- 169 -
Chapter 7
Gravitation
r . This mass is given by
M ins  
4r 3
3
where  is the density of the sphere.
Gravitational Potential Energy
The gravitational potential energy U ( r ) of a
system of two particles, with masses M and m and separated by a distance r , is the
negative of the work that would be done by the gravitational force of either particle
acting on the other if the separation between the particles were changed from infinite
(very large) to r . This energy is
U 
GMm
r
Potential Energy of a System
(gravitational potential energy)
(7-17)
If a system contains more than two particles, its total
gravitational potential energy U is the sum of terms representing the potential
energies of all the pairs. As an example, for three particles, of masses m1 , m2 , and
m3 ,
 Gm1m2 Gm1m3 Gm2 m3 

U  


r13
r23 
 r12
(7-18)
Escape Speed An object will escape the gravitational pull of an astronomical body
of mass M and radius R (to reach an infinite distance) if the object’s speed near the
body’s is at least equal to the escape speed, given by
v
2GM
R
(7-24)
Kepler’s Laws Gravitational attraction holds the solar system together and makes
possible orbiting Earth satellites, both natural and artificial. Such motions are
governed by Kepler’s three laws of planetary motion, all of which are direct
consequences of Newton’s laws of motion and gravitation:
- 170 -
Chapter 7
Gravitation
1. The law of orbits. All planets move in elliptical orbits with the Sun at one focus.
2. The law of areas. A line joining any planet to the Sun sweeps out equal areas in
equal times as the planet orbits the Sun. (This statement is equivalent to conservation
of angular momentum.)
3. The law of periods. The square of the period T of any planet about the Sun is
proportional to the cube of the semimajor axis a of the orbit. For circular orbits with
radius r , the semimajor axis a is replaced by r and the law is written as
 4 2  3
r
T 2  
 GM 
(law of periods),
(7-30)
where M is the mass of the attracting body-the Sun in the case of solar system. This
equation is generally valid for elliptical planetary orbits, when the semimajor axis a
is inserted in place of the circular radius r .
Energy in Planetary Motion
When a planet or satellite with mass m moves in a
circular orbit with radius r , its potential energy U and kinetic energy K are given
by
GMm
GMm
and K 
r
2r
The mechanical energy E  K  U is then
GMm
E
2r
For an elliptical orbit of semimajor axis a ,
GMm
E
2a
U 
(7-33)
(7-35)
(7-37)
Einstein’s View of Gravitation Einstein pointed out that gravitation and acceleration
are equivalent. This principle of equivalence led him to a theory of gravitation (the
general theory of relativity) that explains gravitational effects in terms of a curvature
of space.
- 171 -
Chapter 7
Gravitation
Examples
Example 1 Assume a space flyer circles around the Earth in a circular orbit with
speed v0  3.00 104 km / s , the radius of the Earth is R  6.40  103 km , and the
height of the flyer above the Earth is h  800km . Now we want to change the orbit to
an ellipse. In a very short period of time the flyer ejects gases toward the outer side of
the orbit in the direction of the radius vector, then the flyer gains a speed
vr  800km / s pointing toward the Earth. Assume that the mass of the flyer remain
the same before and after the ejection of the gases, after the flyer orbits around the
Earth in the final elliptic orbit, what is the distance between its perigee and the center
of the Earth, and what is the distance between its apogee and the center of the Earth?
Solution: The key idea is that, before and after the ejection of the gases the angular
momentum of the flyer remains the same:
mv0 r0  mvr
(1)
where r0 is the distance between the flyer and the center of the Earth when the flyer
orbited the Earth in a circular orbit, i.e., r0  R  h ; and r is the distance between
the apogee (or perigee) of its elliptic orbit and the center of the Earth, v is the speed at
that point, and m is the mass of the flyer.
For the flyer on the circular orbit, we have
G
mE m
v02

m
r02
r0
(2)
After the alternation of the orbit, we take the Earth and the flyer as a system and we
have the conservation of mechanical energy, therefore:


1
m m 1
m m
m v02  vr2  G E  mv2  G E
2
r0
2
r
Eliminating mE and v from Eqs.(1), (2) and (3), we have
2


1
1
v02 r02    2v02 r0    v02  vr2  0
r
r
- 172 -
(3)
Chapter 7
Gravitation
Solving it, we get two roots r1 and r2 , i.e. ,
r1 
v0 r0
v R  h 
vr
v R  h 
 0
, r2  0 0  0
v0  vr
v0  vr
v0  vr
v0  vr
(Answer)
After substituting the known numbers into it, we have r1  7.0110 km and
3
r2  7.40km . These are the distances between the apogee and the perigee and the
center of the Earth respectively.
Example 2 A geosynchronous satellite is one that stays above the same point on the
Earth, which is possible only if it is above a point on the equator. Such satellites are
used for such purpose as cable TV transmission, for weather forecasting, and as
communication relays. Determine (a) the height above the Earth’s surface such a
satellite must orbit and (b) such a satellite’s speed. (c) Compare to the speed of a
satellite orbiting 200km above Earth’s surface.
Solution: (a) The only force on the satellite is gravity, assuming the satellite moves
in a circle, then
G
mSat mE
v2

m
Sat
r2
r
This equation seems to have two unknowns, r and v . But we know that v must be
such that the satellite revolves around the Earth with the same period that the Earth
rotates on its axis, namely once in 24 hours. Thus the speed of the satellite must be
2r
T
where T  1day  (24h)(3600s / h)  86400 s . We put this into the first equation
above and obtain (after canceling mSat on both sides):
v
mE 2r 

r2
rT 2
2
G
We solve for r:
GmET   6.67 10
r 

4 2
3
11
N  m2 / kg 2  5.98 1024 kg  864000 s 
4 2
- 173 -
 7.54 1022 m3
Chapter 7
Gravitation
and, taking the cube root, r  4.23 10 m , or 42300km from the Earth’s center.
7
We substrate the Earth’s radius of 6380km to find that the satellite must orbit about
36000km (about 6rE ) above the Earth’s surface.
(b) v 
GmE

r
6.67 10
11


N  m2 / kg2 5.98  1024 kg
 3070m / s
4.23  107 m


We get the same result if we use v  2r / T .
(Answer)
(c) Since v  1 / r , for r  rE  h  6380km  200km  6580km , we get
v'  v r / r '  3070m / s  42300km /6580km  7780m / s
(Answer)
Example 3 There is a thin rod with mass m ,
length l and linear density of mass  , if a mass
point with mass m is placed on the extended line
of the rod and distance r away from the end of the
Fig.7-1 Example 3.
rod as shown in Fig.1, what is the gravitational
potential energy and the gravitational force on the mass point by the rod?
Solution: As shown in the figure, take an element of the rod d x and its mass is
dm  dx . The distance between the rod element and the mass point m is x , and
the gravitational potential energy between the mass element dm’ and the mass point
m is:
dE p  
Gmdm'
Gmdx

x
x
Adding up all the gravitational potential energies of mass elements dm dm’ and mass
point m , i.e., integrating from x  r to x  r  l , we have the gravitational
potential energy of mass point m at point P as follow:
Ep  
r l
r
dE  Gm 
p
After the integration, we have
- 174 -
r l
r
dx
x
Chapter 7
Gravitation
E p  Gm ln
r
rl
(Answer)
The gravitational force on mass point m can be directly calculated by the universal
gravitational law. In Fig.1, mass element dm  dx has the following gravitational
force on mass point m :

mdm' 
dF  G 2 i
x

The direction of dF is pointing to the positive ox axis, then the gravitational force of
the rod on mass point m is:


r  l dx 
1 
1
F   dF  Gm 
i

G

m


i
r
x2
r lr
i.e.,

mm' 
F G
i
r l  r 
(Answer)
Problem Solving
1 (2) In Fig. 2a, particle A is fixed in
place at x  0.20m on the x axis and
particle B, with a mass of 1.0kg , is
fixed in place at the origin. Particle C
(not shown) can be moved along the
x axis, between particle B and x   .
Figure 2b shows the x component
Fig.7-2 Problem 1.
Fnet , x of the net gravitational force on
particle B due to particle A and C, as a function of position x of particle C. The plot
actually extends to the right, approaching an asymptote of 4.7 1010 N as x  .
What are the masses of (a) particle A and (b) particle C?
Solution: (a) The component Fnet , x of the net gravitational force on particle B is:
- 175 -
Chapter 7
Fnet , x  G
Gravitation
mB mC
mAmB
G
2
x
0.20m2
The key idea is that as x   , the net gravitational force on particle B is only due to
particle A:
 4.17  1010 N  G
Substituting mB  1.0kg and G  6.67 10
11
mAmB
0.20m 2
N  m2 kg 2 into the above equation,
yields
mA  0.25kg
(Answer)
(b) The key idea is that from the Fnet , x ( x ) curve, as x  0.4m , Fnet , x  0 , so
0G
mB mC
 0.4m 
2
G
mA mB
 0.2m 
2
Substituting mA  0.25kg , mB  1.0kg into the above equation, and
mC  1.0kg
(Answer)
2 (5) Two dimensions. In Fig.3, three point particles are fixed in place in xy plane.
Particle A has mass mA , particle B has mass 2.00m A , and
particle C has mass 3.00m A . A fourth particle D, with
mass 4.00m A , is to be placed near the other three
particles. In terms of distance d , at what (a)
x coordinate and (b) y coordinate should particle D be
placed so that the net gravitational force on particle A
Fig.7-3 Problem 2.
from particle B, C, and D is zero?
Solution: The key idea is that, as the net gravitational force on particle B from the
other three particles is zero, the gravitational force on particle A from D is the same in
- 176 -
Chapter 7
Gravitation
magnitude and opposite in direction to the composite gravitational force on article A
from B and C.
As shown in the diagram, there is:
2
2
FAD  FAB
 FAC
y
That is,
mm
 mm   mm 
G 2 A D2   G A 2 B   G A C2 
x y
d   1.5d  

2
α
2
(1)
C
A
B
α
And
y
d

(2)
x
1.5d
Substituting mB  2.00mA , mC  3.00mA , and mD  4.00mA into Eqs.(1) and (2)
and solving for x and y :
tan  
x  0.99d , y  0.66d
(Answer)
3 (10) A projectile is shot directly away from Earth’s surface. Neglect the rotation of
Earth. What multiple of Earth’s radius RE gives the radial distance a projectile reaches
if (a) its initial speed is 0.500 of the escape speed from Earth and (b) its initial kinetic
energy is 0.500 of the kinetic energy required to escape Earth? (c) What is the least
initial mechanical energy required at launch if the projectile is to escape Earth?
Solution:
The escape speed is:
v
2GM
RE
(a) As the projectile’s initial speed is 0.500v , the key idea is that when it reaches the
maximum radial distance R1 , its kinetic energy is zero and potential energy
U1  
GMm
, then
R1
- 177 -
Chapter 7
Gravitation
 GMm 
1
GMm
2
  
m0.500v    
2
RE 
R1

yields R1 
4
RE
3
(Answer)
(b) As the projectile’s initial kinetic energy is 0.500 of
1 2
mv , the maximum
2
radial distance it reaches is R2 , so
GMm
1
 GMm
0.500 mv2  

RE
R2
2

yields R2  2RE
(Answer)
(c) When the projectile escapes Earth, its least required kinetic energy is zero and the
potential energy is zero, too. As the mechanical energy is conserved, the least initial
mechanical energy required at launch is zero.
4 (12) Zero, a hypothetical planet, has a mass of 5.0 1023 kg , a radius of 3.0 106 m ,
and no atmosphere. A 10kg space probe is to be launched vertically from its surface.
(a) If the probe is launched with an initial energy of 5.0 107 J , what will be its
kinetic energy when it is 4.0 106 m from the center of Zero? (b) If the probe is to
achieve a maximum distance of 8.0 106 m from the center of Zero, with what initial
kinetic energy must it be launched from the surface of Zero?
Solution:
(a) The key idea is that, the mechanical energy of the space probe is conserved, then:
K f  U f  Ki  U i
 GMm 
 GMm 

  Ki   
K f   
R
R
1
0 



R0  3.0  106 m is the radius of Zero and R1  4.0  106 m ,
K i  5.0  107 J . Substituting G  6.67 1011 N  m2 kg 2 , M  5.0 1023 kg
where
- 178 -
Chapter 7
Gravitation
and m  10kg into the above equation, yields:
K f  2.33 107 J
(Answer)
(b) The key idea is that, as the probe achieves the maximum distance
R2  8.0  106 m , its minimum final kinetic is zero, we have:
 GMm 
 GMm 

 
  Ki ' 
R
R
2 
0 


Solving for Ki ' , yields K i '  6.94  10 J
7
(Answer)
So it must be launched with at least 6.94 10 J kinetic energy.
7
5 (16) A 20kg satellite has a circular orbit with a period a 2.4h and a radius of
8.0 106 m around a planet of unknown mass. If the magnitude of the gravitational
acceleration on the surface on the planet is 8.0m / s 2 , what is the radius of the planet?
Solution: The gravitational acceleration of the satellite:
v 2 2r / T 
4 2 r

 2
r
r
T
2
ag 
Substituting T  2.4h and r  8.0 106 m into the above equation and yielding:
ag  4.2m / s 2
The gravitational acceleration at the surface:
ag 0 
GM
R2
(1)
where R is the radius of the planet, and
ag 
GM
r2
(2)
From Eqs. (1) and (2) we have:
r
 
ag  R 
ag 0
2
(3)
Substituting ag 0  8.0m / s , ag  4.2m / s , r  8.0  10 m into Eq.(3) ,yields
2
2
- 179 -
6
Chapter 7
Gravitation
R  5.8  106 m
(Answer)
6 (18) Three identical stars of mass M form an equilateral triangle that rotates
around the triangle’s center as the stars move in a common circle about the center. The
triangle has edge length L . What is the speed of the stars?
Solution: Analyze the free-body diagram of one of the stars as shown.
The net force is:
Fnet  2 F cos 30  3F  3
F
F
GMM
L2
(1)
As the star is rotating around the triangle’s center, then:
30°
Fnet 
Mv 2
r
where v is the speed of the star and r 
(2)
3
L is the distance
3
between the star and the triangle’s center.
Solving Eqs. (1), (2) for v , and
v
The speed of the three identical stars is
GM
L
(Answer)
GM
.
L
7 (19) One way to attack a satellite in Earth orbit is to launch a swam of pellets in
the same orbit as the satellite but in the opposite direction. Suppose a satellite in a
circular orbit 500km above Earth’s surface collides with a pellet having mass 4.0g .
(a) What is the kinetic energy of the pellet in the reference frame of the satellite just
before the collision? (b) What is the ratio of this kinetic energy to the kinetic energy of
a 4.0g bullet from a modern army rifle with a muzzle speed of 950m / s ?
Solution: (a) In the reference frame of Earth, since the pellet is in the circular orbit
with the speed of v , there is:
- 180 -
Chapter 7
m
Gravitation
v2
Mm
G 2
r
r
(1)
Substituting the mass of Earth M  5.98 10 kg , G  6.67 10
24
11
N  m2 kg 2
and r  R  500km (the radius of Earth R  6.37 10 m ) into Eq. (1), yields
6
v  7.6  103 m / s
In the reference frame of the satellite, since the pellet moves in the opposite direction
to the satellite, the speed of the pellet is then 2v before the collision. Then the kinetic
energy of the pellet:
K


2
1
2
m2v   2mv2  2  4.0  10 3 kg  7.6  103 m / s  4.6  105 J
2
(Answer)
(b) The kinetic energy of the bullet:
Kb 
1
1
2
2
mb vb   4.0  10 3 kg950m / s   1.8  103 J
2
2
So the ratio is:
K 4.6  105

 256
K b 1.8  103
(Answer)
8 (24) A typical neutron star may have a mass equal to that of the Sun but a radius of
only 10km . (a) What is the gravitational acceleration at the surface of such a star? (b)
How fast would an object be moving to the surface if it fell from rest through a
distance of 1.0m on such a star? (Assume the star does not rotate.)
Solution:
(a) The gravitational acceleration is
ag 
11
GM
R2
Substituting G  6.67 10 N  m kg , the mass of the sum M  1.99 10 kg ,
and R  10km into the above equation and yielding:
2
2
ag  1.33 1012 m / s 2
- 181 -
30
(Answer)
Chapter 7
(b) v 
Gravitation
2ag h  2  1.33  1012 m / s 2  1.0m  1.63  106 m / s
(Answer)
9 (25) A very early, simple satellite consisted of an inflated spherical aluminum
balloon 30m in diameter and of mass 20kg . Suppose a meteor having a mass of
7.0kg passes within 3.0m of the surface of the satellite. What is the magnitude of
the gravitational force on the meteor from the satellite at the closest approach?
Solution:
The gravitational force:
F G
Mm
r2
(1)
Since the radius of the satellite is 3.0m , then
r
30m
 3.0m  18m
2
Substituting Eq.(2) and G  6.67 10
11
(2)
N  m2 kg 2 , M  20kg , m  7.0kg into
Eq.(1), then
F  2.88  1011 N
(Answer)
10 (28) The orbit of Earth around the sun is almost circular: The closest and farthest
distances are 1.47 108 km and 1.52 108 km respectively. Determine the
corresponding variations in (a) total energy, (b) gravitational potential energy, (c)
kinetic energy, and (d) orbital speed. (Hint: Use conservation of energy and
conservation of angular momentum.)
r2
r1
Solution:
(a) Since the only force acting on Earth is the gravitational
force, the total energy of Earth is conserved, so
E  0
(Answer)
(b)
Fig.7-4 Problem 10.
 GMm   GMm 
1 1
   
  GMm  
U  U 2  U1   
r2  
r1 

 r1 r2 
where the subscripts 1 and 2 refer to the parameters of the closest and farthest points.
- 182 -
Chapter 7
Substituting G  6.67 10
11
Gravitation
N  m kg , the mass of the Sun M  1.99 1030 kg ,
8
8
the mass of Earth m  5.98 1024 kg , and r1  1.47 10 km , r2  1.52  10 km
2
2
into the above equation , yields:
U  1.78  1032 J
(Answer)
(c) Since E  K  U  0
K  U  1.78  1032 J
(Answer)
(d) Because the gravitational force acting on the Earth passes through the center of the
Sun, there is no external torque acting on the Earth. The angular momentum is
conserved, so we can get
I11  I 22
mr12
That is
v1
v
 mr22 2
r1
r2
So
v2 r1 1.47
 
v1 r2 1.52
From
m
So
And
v1 
v12
Mm
G 2
r1
r1
GM
6.67  1011  1.99  1030

 3.0  104 m / s
r1
1.47  1011
v  v2  v1 
1.47
v1  v1  990m / s
1.52
(Answer)
11 (32) Several planets (Jupiter, Saturn, Uranus) are
encircled by rings, perhaps composed of material that
failed to form a satellite. In addition, many galaxies
contain ring-like structures. Consider a homogeneous
thin ring of mass M and outer radius R (Fig. 5). (a)
Fig.7-5 Problem 11.
- 183 -
Chapter 7
Gravitation
What gravitational attraction does it exert on a particle of mass n located on the ring’s
central axis a distance x from the ring center? (b) Suppose the particle falls from rest
as a result of the attraction of the ring of matter. What is the speed with which it passes
through the center of the ring?
Solution:
(a) The key idea is to find the net gravitational force of the ring on the particle. Take
an element of the ring dr as shown, and its mass is dM 
M
dr , the gravitational
2R
force on the particle is:
dF  
GmdM
Gm M
 2
dr
2
2
x R
x  R 2 2R
The component of the force along x and y axis is respectively:
GmM
x
dr
2
2
x  R 2R
x  R2
GmM
R
dFy   2
dr
2
2
x  R 2R
x  R2
dFx  


2


For arbitrary dr , there is always an element dr  on the symmetric place to dr , as
shown, and the component of the gravitational force of dr  on the particle is same in
magnitude and opposite in direction to that of dr , so the net force of the thin ring on
the particle along the x axis is:
F   dFx  
2R
0


GmM
x
GmMx

dr
 2
 2
3/ 2
2
2
2
x  R2
 x  R 2R x  R




The gravitational attraction on the particle from the ring is

x
(Answer)
GmMx
2
 R2

3/ 2
in
magnitude.
(b) The change in the kinetic energy of the particle comes from the work done by the
gravitational force
0
1 2
mv  0   Fdx
x
2
Substituting F into the integral for v , yields
- 184 -
Chapter 7
Gravitation
1
v  2GM  
R


x R 
1
2
2
(Answer)
That is the speed of the particle when it passes through the center of the ring.
12 (33) The fastest possible rate of rotation of a planet is that for which the
gravitational force on material at the equator just barely provides the centripetal force
needed for the rotation. (Why?) (a)Show that the corresponding shortest period of
rotation is
3
,
G
T
where  is the uniform density (mss per unit volume) of the spherical planet. (b)
Calculate the rotation period assuming a density of 3.0g / cm3 , typical of many
planets, satellites, and asteroids. No astronomical object has ever been found to be
spinning with a period shorter than that determined by this analysis.
Solution: (a) Assume a body of mass m is at the equator, the mass and the radius of
the planet is M and R respectively, the gravitational force is:
Fg  G
Mm
R2
(1)
For the rotating body, its centripetal force is:
Fc  m
v2
R
(2)
where v is the linear speed of the body. Since Fg  Fc , we have:
G
Mm
v2

m
R2
R
(3)
And the shortest period is :
T
2R
v
The uniform density of the spherical planet is:
- 185 -
(4)
Chapter 7
Gravitation

M
(5)
4 3
R
3
Combining Eqs.(3), (4), (5) we have
T
3
G
(6)
(b) Substituting G  6.67 1011 N  m2 kg 2 ,   3.0 g / cm3 into Eq.(6), yields
T  6.9  103 s
(Answer)
13 (34) The gravitational force between two particles with masses m and M , initially
at rest at great separation, pulls them together. Show that at any instant the speed of
2G ( M  m) / d , where d is their separation at
that instant. (Hint: Use the laws of conservation of energy and conservation of linear
momentum.)
Solution: Assume the velocities of the two particles are v1 and v2 , respectively, and
the linear momentum of the two-particle system is conserved:
either particle relative to the other is
mv1  mv2  0
(1)
The particles are pulled together by the gravitational force and the mechanical energy
of the two-particle system is also conserved:
K f  U f  Ki  U i
i.e.
1 2 1 2 
Mm 
mv1  mv2    G
0
2
2
d 

(2)
From Eqs.(1) and (2) we have:
v2 
2G
M
M  m d
and
v1  
And the speed of either particle relative to the other is then:
- 186 -
mv2
M
Chapter 7
v2  v1  v2 
mv2 M  m

M
M
Gravitation
2G
M
M  md
2G M  m 
d
(Answer)
14 (37) Some people believe that the positions of the planets at the time of birth
influence the newborn. Others deride this belief and claim that the gravitational force
exerted on a baby by the obstetrician is greater than the force exerted by the planets.
To check this claim, calculate the magnitude of the gravitational force exerted on a
3kg baby (a) by a 70kg obstetrician who is 1m away and roughly approximated as
a point mass, (b) by the massive planet Jupiter ( m  2 1027 kg ) at its closest approach
to Earth (  6 1011 m ), and (c) by Jupiter at its greatest distance from Earth
(  9 1011 m ). (d) Is the claim correct?
Solution:
(a) The gravitational force exerted on the 3kg baby by the obstetrician is
F1  G
Substituting G  6.67 10
11
Mm0
r12
(1)
N  m2 kg 2 , M  70kg , m0  3kg , r1  1m into
Eq.(1), and
F1  1.4  108 N
(Answer)
(b) The gravitational force exerted on the 3kg baby by the Jupiter planet at its closest
approach to Earth is
F2  G
Substituting G  6.67 10
11
Mm0
r22
(2)
N  m2 kg 2 , m  2 1027 kg , m0  3kg , r2  6 1011 m
into Eq.(2), and
F2  1.1  106 N
(Answer)
(c) The gravitational force exerted on the 3kg baby by the Jupiter planet at its greatest
- 187 -
Chapter 7
Gravitation
distance from Earth is
F3  G
Substituting G  6.67 10
11
mm0
r32
(3)
N  m2 kg 2 , m  2 1027 kg , m0  3kg , r3  9 1011 m
into Eq.(3), and
F3  4.9  107 N
(d) Since F2  F1 and even F3  F1 , the claim is not correct.
- 188 -
(Answer)
(Answer)