Download Answers to Coursebook questions – Chapter 2.11

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Answers to Coursebook questions – Chapter 2.11
1
Mm
and this plays the role of the centripetal force on
R2
mv 2
mv 2
Mm
M
 G 2 , i.e. v 2  G .
the satellite, i.e.
. Equating the two gives
R
R
R
R
The net force on the satellite is F  G
2
2
GM  2πR 
 2πR 

 .
 and so
R
 T 
 T 
2
For circular motion we have that v  
Simplifying gives the result.
2
From Q1 we know that R 3 
GMT 2
. The angular velocity is the angle swept per unit time, i.e.
4π 2
2π
2π
, i.e. T 
.
T

GM (2π)2 GM
 2 .
Substituting, R3 
4π 2 2


3
mv 2
Mm
M
 G 2 we deduce that v 2  G . Now,
From
R
R
R
R  6400  500  6900 km  6.9  106 m and so
6.67  10 11  6.0  10 24
v
 7.6  10 3 m s 1 .
6
6.9  10
2πR
2πR
The period is then found from v 
T 
 5.7 103 s  95 min .
T
v
GM

R
4
The period has to be one day, i.e. 24 hours. Then v 
deduce that v 2  G
R
5
3
mv 2
Mm
2πR
 G 2 we
and from
T
R
R
M
. Combining the results we get
R
GMT 2 3 6.67 1011  6.0 1024  (24  60  60)2

 4.2 107 m .
2
2
4π
4π
a
EP  
b
V 
c
v
GM earth M moon
6.67 1011  5.98 1024  7.35 1022

 7.64 1028 J .
8
r
3.84 10
GM earth
6.67 1011  5.98 1024

 1.04 106 J kg 1 .
r
3.84 108
GM earth
6.67 1011  5.98 1024

 1.02 103 m s 1 .
8
r
3.84 10
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6
We must plot the function EP  
GM earth m GM moon m
giving the graph in the

r
d r
answers. Here m is the mass of the spacecraft and d the separation of the earth and the
moon (centre-to-centre). Putting numbers in,
6.67  10 11  5.98  10 24  3.0  10 4 6.67  10 11  7.35  10 22  3.0  10 4

r
3.84  10 8  r
1.2  1019
1.5  1017


r
3.84  10 8  r
1.2  1019 / 3.84  10 8 1.5  1017 / 3.84  10 8


r / 3.84  10 8
1  r / 3.84  10 8
3.1  1010 3.9  10 8


x
1 x
r
where x 
. In this way the function can be plotted on a calculator to give the graph
3.84  10 8
EP  
in the answers (see page 798 in Physics for the IB Diploma).
7
a
V 
GM
5Re
6.67  10 11  5.98  10 24
5  6.4  10 6
 1.2  10 7 J kg 1

b
EP  
GMm
5Re
6.67  10 11  5.98  10 24  500
5  6.4  10 6
 6.2  10 9 J

8
The net force is the gravitational force and this must point towards the centre of the earth. This
happens only for orbit 2.
9
As shown in the text, the reaction force from the spacecraft floor is zero, giving the impression
of weightlessness. More simply, both spacecraft and astronaut are in free fall with the same
acceleration.
10
The difference is U  
GMm  GMm  GMm GMm
 R  h  R
 


 GMm 
.

 R(R  h) 
Rh 
R 
R
Rh

h

U  GMm 
. When h is small compared to R this expression is approximately
 R(R  h) 
U 
GMm
GM
h  m 2 h  mgh .
2
R
R
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11
The work done by an external agent in moving an object from r = a to r = b at a small
constant speed.
12
a
The potential at the surface of the planet is
GM
6.67  10 11  M
V 

 4.9  1012 J kg 1 .
5
R
2.0  10
Hence M  1.5  10 28 kg .
b
The escape speed is obtained from
But V  
13
2GM
.
R
GM
and so GM  VR . Substituting, v  2V .
R
c
v  2V  2  4.9  1012  3.1  10 6 m s1
d
The work required is W  mV . This is
W  1500  (1.0 1012  (4.9 1012 ))  5.8 1015 J .
e
We have that
1 2
mv  mV1  mV2  v  2(V2  V1 ) .
2
This gives v 
2(2.2  1012  (4.9  1012 ))  2.3  10 6 m s1 .
a
At r = 0.75, g 
M P (0.75d)2
GM P
GM m

 9.
.
Hence


0
M m (0.25d)2
(0.75d)2 (0.25d)2
b
The probe must have enough energy to get to the maximum of the graph. From then on
the moon will pull it in. Then
W
14
1 2 GMm
mv 
 0 , i.e. v 
2
R
1 2
mv  mV  v  2V  2(0.2  1012  (6.42  1012 ))  3.5  106 m s 1.
2
We deduced many times that
v2 
GM
r
and so
ET 
1 2 GMm GMm GMm
GMm
mv 



2
r
2r
r
2r .
ET  
6.67 1011  2.0 1030  6.0 1024
 2.7 1033 J.
11
2 1.5 10
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15
Using EK 
GMm
GMm
GMm
, EP  
and ET  
we deduce that
2r
r
2r
a
B has the larger kinetic energy
b
A has the larger potential energy and
c
A has the larger total energy.
mv 2
Mm
M
 G 2 to be v 2  G .
r
r
r
1 2 1 GMm
The kinetic energy is then EK  mv 
.
2
2 r
GMm
The potential energy is EP  
r
1 GMm GMm
GMm
and so the total energy is ET 
.


2 r
r
2r
GMm
Since r  5R , ET  
.
10R
16
The speed in orbit is given from
17
From Q16, ET  
18
The space station is in an orbit with orbit radius 2Re and so has speed v 
GMm
GMm
. Since we are told that ET  
and energy is conserved,
2r
5R
GMm
GMm
5R
.


r
2r
5R
2
GM
with respect
2Re
to the earth.
Let the satellite be launched with speed u with respect to the space station.
Then the speed with respect to the earth is u  v .
Its total energy is therefore
1
GMm
.
m(u  v)2 
2
2Re
At the escape speed this energy must be zero and so
u
GM
GM
GM
v

 2.3  10 3 ms1 .
Re
Re
2Re
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19
a
The engines do positive work increasing the total energy of the satellite. Since
ET  
b
GMm
it follows that the orbit radius will increase.
2r
Since the kinetic energy is given by EK 
GMm
and the orbit radius has increased, the
2r
speed in the new circular orbit will decrease.
c
The firing of the rockets when the satellite is in the lower orbit make the satellite move
on an elliptical orbit.
After half a revolution the satellite will be at A and further from the earth than in the
original position at P.
As the satellite gets to A its kinetic energy is reduced and the potential energy increases.
At A the speed is too low for the new circular orbit and the engines must again be fired to
increase the speed to that appropriate to the new orbit. (If the engines are not fired at A
then the satellite will remain in the elliptical orbit and will return to P.)
new circular or bit
A
F
old circular
orbit
P
20
The tangential component at A is in the direction of velocity and so the planet increases its
speed. At B it is opposite to the velocity and so the speed decreases.
The normal component does zero work since the angle between force and displacement is a
right angle and cos 90o  0 .
21
The potential energy is given by EP  
GMm
.
r
This is least when the distance to the sun r is the smallest (remember EP is negative).
Therefore since the total energy is conserved, the kinetic energy and hence the speed are
greatest at P.
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22
a
The ratio is
F F
F F
B
moon
B
sun
A
moon
A
sun
GM moon m
GM moon m
M moon
M moon


2
2
2
(d
 Re ) (d moon  Re )
(d
 Re ) (d moon  Re ) 2
 moon
 moon
GM sun m
GM sun m
M sun
M sun


2
2
2
(dsun  Re ) (dsun  Re )
(dsun  Re ) (dsun  Re ) 2
.
Numerically this gives
B
A
Fmoon
 Fmoon
B
A
Fsun
 Fsun
b
23
This indicates the relative importance of the moon.
The escape speed is obtained from
But g 
24
7.35 1022
7.35 1022

(3.84 108  6.4 106 ) 2 (3.84 108  6.4 106 ) 2

 2.2 .
1.99 1030
1.99 1030

(1.5 1011  6.4 106 )2 (1.5 1011  6.4 106 ) 2
a
1 2 GMm
mv 
 0 , i.e. v 
2
R
2GM
.
R
GM
and so GM  gR2 . Substituting, v  2gR .
R2
The net gravitational field strength at the indicated position has magnitude
g
G16m
Gm
G25m G25m



 0.
2
2
(4d / 5) (d / 5)
d2
d2
b
V 
25
G16m
Gm
G20m G5m
25Gm




(4d / 5) (d / 5)
d
d
d .
a
See Q1.
b
T2 
4π 2 r 3
.
GM
Now r  R and  
M 4

3
3 .
Hence, R
Substituting, T 
c
Tplanet
Tearth
M
3M
.

3
4πR
4πR3
3
4π 2 3
3π
.

G 4π
G
earth

 169 

 earth  
 3.95  4 .
planet
planet  85 
2
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26
a
We must use the formula T 2 
4π 2 R3
that we have derived many times already.
GM
GM
 GM  gR 2 .
R2
4π 2 R3 4π 2 R
Substituting, T 2 
.

gR 2
g
Now g 
Hence T  2π
R
.
g
b
3.4 106
T  2π
 5.5 103 s  91 min .
4.5
c
From T 2 
T 2 R3
4 2 R 3
we deduce that 12  13 .
T2
R2
GM
912
(3.4  10 6 )3

Hence
and so R2  4.5  10 6 m .
2
3
140
R2
The height is therefore h  4.5 106  3.4 106  1.1106 m .
27
a
F
GM 2
4R 2
b
GM 2 Mv 2
GM
 2
v2 
2
4R .
4R
R and so
2
2
GM  2πR 
 2πR 

But v  
 .
 and so
4R  T 
 T 
16π 2 R 3
Hence T 2 
.
GM
2
c
16π 2 (1.0 109 )3
T
 2.8 104 s  7.8 h .
11
30
6.67 10 1.5  2.0 10
d
ET 
1
1
GM 2
Mv 2  Mv2 
2
2
2R .
Since
v2 
GM
4R
we have that
ET 
1 GM
GM 2 GM 2 GM 2
GM 2
M
2



2
4R
2R
4R
2R
4R .
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e
f
Since energy is being lost, the total energy will decrease. This implies that the
distance R will decrease. (From the period formula in b the period will decrease
as well.)
GM 2
2 3
ET  
4R and the period is T 2  16π R . Combining
The total energy is
GM
2
GM
the two gives ET  
3/2
 GMT 2 
4
2 
 16π 
i
or ET  cT 3/2 , where c is a constant. Working as we do with propagation of
uncertainties (or using calculus) we have that
ET
T
3
ET 3 T
t .
or t 

ET
2 T
ET
2 T
ii
28
29
ET
T
ET
3 T
3 72 106 s y 1

 
 3.9 109 y 1
t
2 t 2
2.8 104 s
1
 2.6 108 y.
9
1
3.9 10 y
g
The lifetime is therefore
a
The pattern is not symmetrical and so the masses must be different. The spherical
equipotential surfaces of the right mass are much less distorted and so this is the larger
mass.
b
The gravitational field lines are normal to the equipotential surfaces.
c
From far away it looks like we have a single mass of magnitude equal to the sum of the
two individual masses. The equipotential surfaces of a single point mass are spherical.
a
The magnitude of the gravitational field strength is the slope of a potential–distance
graph. Drawing a tangent to the curve at
r
 0.20 we find a slope of approximately
d
4.7 N kg1 .
b
At
r
 0.75 the slope of the graph is zero. But the slope of a potential–distance graph is
d
the magnitude of the gravitational field strength. Hence g is zero at this point.
c
Since g  0 
GM
Gm
GM
Gm



it follows that
2
2
2
r
(d  r)
(0.75d) (d  0.75d)2
M (0.75d)2

 9.
m (0.25d)2
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