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Active Physics Full Solutions to Textbook Exercises
63 Orbital Motions under Gravity
63 Orbital Motions under Gravity
(b)
F
Note that circular orbits are special cases of
elliptical orbits.
(c)
F
If the time is not expressed in years we
should not use T2 = a3. However, Kepler’s
third law is still valid.
(a)
F
The line only fits the data of the celestial
bodies revolving around the Sun.
(b)
T
Note that T2 is directly proportional to a3.
(c)
F
The planets on the right of the Earth have
larger orbits but are not necessarily more
massive.
Checkpoint
Checkpoint 1 (p. 59)
1.
2.
(a)
F
The Sun is at the focus of the elliptical orbit.
(b)
T
(c)
T
While sweeping out equal areas in equal
time intervals, a planet nearer the Sun has to
travel a longer distance than it does when it
is farther away. In other words, the planet
travels a longer distance for a fixed time
interval when it is nearer the Sun (i.e. higher
speed).
3.
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(a)
p.1
Checkpoint 3 (p. 69)
1.
The orbital period in seconds is
The orbital semi-major axis in metres is
(b)
Perihelion distance
= 3.05/2 − 0.145 = 1.38 AU
Since
, we have
Aphelion distance
= 3.05/2 + 0.145 = 1.67 AU
(c)
3.
(a)
Perihelion. It is nearer the Sun than the
aphelion (and in fact it is the nearest point to
the Sun).
2.
T
Given that the areas shown are equal, the six
time intervals should be equal. Therefore,
the orbital period is six times any one of the
time interval.
(b)
T
(c)
F
In fact, the speed of the planet is the lowest
at aphelion.
(a)
Y
Note that Kepler’s third law is always valid.
(b)
Y
(c)
N
Phobos and Deimos are not planets revolving
around the Sun and this relation cannot be
applied.
Checkpoint 4 (p. 72)
1.
(a)
Checkpoint 2 (p. 63)
1.
Applying
(a)
,
T
The law states that
. When
either mass is halved, the gravitational PE
becomes halved.
(b)
Y
(c)
N
When the distance is halved, the
gravitational PE is doubled.
(a)
The gravitational PE is always negative.
(b)
The gravitational PE is always negative.
This implies a = 2.8 AU.
2.
Y
Recall that
2.
.
Also, it is indirectly proportional to r.
Active Physics Full Solutions to Textbook Exercises
63 Orbital Motions under Gravity
(a)
The escape speed becomes higher.
Checkpoint 5 (p. 75)
(b)
The escape speed remains unchanged.
1.
(c)
The escape speed becomes higher.
(a)
C
Its distance from the star is the longest.
Gravitational PE is always negative and is
zero at infinity. The farther away the two
bodies the larger the gravitational PE
between them.
(b)
(c)
2.
Since mechanical energy is conserved,
1.
Option B is incorrect. The gravitational force
varies with the distance. Since the objects are in an
elliptical orbit, the force must have changed in
magnitude (and also direction) when they travel.
2.
C
2.
(a)
3.
0.5U
3.
Reduced by 0.5U
Since mechanical energy is conserved, the
change of KE is equal to the change in PE i.e.
U – 0.5U = 0.5U.
F
It is the gravitational force that keeps a
spacecraft in the orbit.
(b)
F
Note that the spacecraft and the objects
inside are exerted the same gravitational
force from the massive object.
(c)
F
The astronaut is falling with the same
acceleration as the spacecraft. He feels
weightless because the spacecraft exert no
reaction on him.
Time intervals II and III
Neglecting air resistance, gravitational force is the
only force that acts on the aeroplane and the
astronaut when the engine is shut down.
Recall that
, When
the distance is doubled, the gravitational PE
becomes halved.
(b)
(a)
Moving at the same speed cannot guarantee
weightlessness. For example, a man in a lift
travelling downwards at a uniform speed can
still feel his weight.
Option A is incorrect because as the KE of the
spacecraft has been reduced to decelerate during
the change in orbit.
Option B is incorrect. The PE does not change
while the spacecraft is at the point it decelerates.
The PE depends on the position of the spacecraft.
C
Option A is incorrect. It is the gravitational force
from the Moon that keeps the astronauts in the
orbit.
Checkpoint 6 (p. 82)
1.
p.2
Checkpoint 7 (p. 86)
A
The speed of the planet is the highest when it
is closest to the star.
The planet has the same mechanical energy
at all positions as the mechanical energy is
conserved.
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You can imagine that the aeroplane is undergoing
projectile motion during the time interval.
Exercise
Exercise 3.1 (p. 63)
1.
The escape speed on the Moon is
A
By T2 = a3, the period should be
The above is valid only when the time is expressed
in years and the length is expressed in AU.
2.
4.
Recall that
, the escape speed is
related to the mass and the radius of the planet
but independent of the mass of the object itself.
A
The orbital period T and the semi-major axis a
must satisfy the relation T2 = a3. See the table
below.
T/y
T2 / y2
a / AU
a3 / AU3
Active Physics Full Solutions to Textbook Exercises
3.
2.83
8.01
2
8
2.83
8.01
4
64
4
16
2
8
4
16
4
64
(a)
(b)
4.
63 Orbital Motions under Gravity
4.
By
5.
By
6.
(a)
By
(b)
(i)
, the orbital speed is
, the period is
Therefore, the ratio is
(a)
The semi-major axis is
(b)
By T2 = a3, the period is
(a)
The aphelion distance is
(ii)
(2)(2.217)−0.33 ≈ 4.10 AU
(c)
(i)
(ii)
, the mass is
It has a shorter period.
The shorter the orbital radius, the
shorter the orbital period
(
).
By T2 = a3, the semi-major axis is
a = T2/3 = 3.32/3 = 2.217 ≈ 2.22 AU
(b)
p.3
The orbital period is not proportional to the
orbital radius. The correct relation should be
T = a3/2 = 1.5253/2 ≈ 1.88 yr
5.
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Y
The aphelion distance of Comet Encke
is shorter than the semi-major axis of
Jupiter.
By
,
OR: By
, the period is
The orbital period of Jupiter is longer.
By T2 = a3, the longer the semi-major
axis, the longer the orbital period.
Note that 1 d = 24 × 60 × 60 s.
Exercise 3.2 (p. 70)
1.
Recall
2.
7
A
, we have
.
B
From
, when the mass of the star is
doubled, the relation becomes
where T is expressed in years and a is expressed in
AU.
The relation is independent of the mass of the
planet.
3.
By
B
By
, we have
Or:
8.
(a)
When the comet travels along its orbit, it
experiences the gravitational pull from the
Sun. As it moves away from the Sun, the pull
always has a component opposite to the
comet’s velocity in direction. Therefore the
comet slows down when it moves away.
Note that the component of the pull
perpendicular to the velocity does no work
on the comet and will not change the speed
of the comet.
Active Physics Full Solutions to Textbook Exercises
(b)
63 Orbital Motions under Gravity
If the mass M of the star is not large enough,
the centre of the circular orbit will not be the
star. The orbital radius does not equal the
distance between the star and the planet. As
a result, we cannot derive from
7.
.
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p.4
(a)
The acceleration is
(b)
(i)
The gravitational PE is
(ii)
Since the spacecraft is moving in a
circular orbit, the mechanical energy is
Exercise 3.3 (p. 82)
1.
A
The PE is
Since the speed is unknown, we cannot find the KE.
Options B and C cannot be correct.
Option D is incorrect (due to conservation of
energy).
2.
(iii) The KE is
E – U = −1.299 × 1010 – (−2.599 × 1010)
≈ 1.30 × 1010 J
B
The mechanical energy is
8.
The KE of the probe just before it hits the surface
of the Moon is (in joules)
where r is the distance of the satellite from the
centre of the Earth.
3.
A
Imagine that a spacecraft is initially in the orbit X.
It has to accelerate (e.g. by firing a rocket) at the
leftmost point in order to enter orbit Y. In other
words, the spacecraft has gained KE during the
orbital transition. In contrast, a spacecraft initially
in orbit Z has to decelerate at the rightmost point
in order to enter orbit Y. Therefore, a body should
have the lowest mechanical energy in orbit X and
the highest in orbit Z.
4.
6.
9.
(a)
Applying
, the escape speed is
(b)
By the conservation of energy, the speed
should be also 60.2 km s−1.
C
Recall that
5.
The velocity is
.
(a)
They have the same acceleration of
.
(b)
The object in orbit 2 has less mechanical
energy.
(c)
They have the same PE of
(d)
The object in orbit 2 has less KE.
(e)
The object in orbit 2 is slower.
Substitute:
(f)
[Deleted]
(a)
The gravitational PE is
v1 = 4538 m s−1, G = 6.67 × 10−11N m2 kg−2,
M = 6.42 × 1023 kg,
r1 = 1.71 × 105 + 3390 × 103 = 3.561 × 106 m, and
r2 = 1.78 × 107 + 3390 × 103 = 2.119 × 107 m.
10.
By the conservation of energy,
.
Now we have
(b)
The PE is smaller (more negative).
Active Physics Full Solutions to Textbook Exercises
63 Orbital Motions under Gravity
1.
Statement (3) is incorrect. According to Kepler’s
third law, T2 ∝ a3.Therefore, it should be the
period squared that is proportional to AC cubed.
2.
B
C
3.
The contents are in weightless condition and
appear to remain at the original position.
3.
(a)
4.
5.
(b)
Recall
4.
The astronaut and the spacecraft are free
falling together and hence there is no
reaction from the spacecraft on the
astronaut. As a result, the astronaut feels
weightless.
(i)
(ii)
5.
Statement (1) is incorrect. The planet moves with
the lowest speed at the aphelion.
aphelion / AU
a / AU
1.24
5.68
3.46
19P/Borelly
1.35
5.83
3.59
9P/Tempel
1.51
4.74
3.125
81P/Wild
1.59
5.31
3.45
C
By
, the ratio should be
13/2 : 23/2 : 33/2 = 1 : 23/2 : 33/2
6.
C
Since
, the nearer to the Earth, the
faster the satellite travels. Therefore, Y should
have more KE.
Since
, the farther away from the
Earth, the more the PE the satellite have. (Note the
minus sign!) Therefore, X should have more PE.
7.
C
Since the satellites are moving in circular orbits,
we have E = U/2. Options A and B are incorrect as
E must be changing with r similar to U. Also, E
must be more ‘positive’ than U and option C
should be the best representation. (Note the minus
sign!)
Chapter Exercise: Multiple-choice Questions (p. 91)
D
perihelion / AU
67P/C-G
The longer the semi-major axis, the longer the
orbital period. Therefore, 19P/Borelly should have
the longest orbital period.
The magnitude of acceleration
gradually decreases as the gravitational
force acting on the spacecraft
decreases with the distance between it
and the Earth.
The astronaut feels weightless
throughout the whole journey when he
is moving in the orbit.
B
comet
Chapter Exercise
1.
. Therefore, the period ratio is
The semi-major axes of the comets a are as shown.
Since no net force acts on the marble after
the throw, by Newton’s first law of motion, it
moves in a straight line at a uniform velocity.
The normal reaction from the balance on the block
is zero when they fall freely at the same rate. As a
result, the balance reads zero.
(a)
D
Yes.
Since the spacecraft is travelling in space at a
uniform velocity, the only possible scenario
for her to experience weightlessness is that
the spacecraft is so far away from all celestial
bodies so that the total gravitational force
acting on the spacecraft is too small to be
influential.
(b)
D
The centripetal force is provided by the
gravitational force:
The acceleration is
2.
p.5
Statement (2) is incorrect. The time is not
proportional to the angular separation. According
to Kepler’s second law, the imaginary line joining
the Sun and the planet sweeps out equal areas in
equal time intervals.
The spacecraft has a speed is 765 m s−1 when it is
farthest away from Mars.
Exercise 3.4 (p. 87)
|
8.
A
Note that
9.
A
.
Active Physics Full Solutions to Textbook Exercises
63 Orbital Motions under Gravity
Statement (2) is correct only when both the
astronaut and the spacecraft remain still relatively
to each other.
10.
(b)
|
p.6
Applying T2 = a3, the period is
T = 1.4553/2 = 1.755 ≈ 1.76 y (1M+1A)
Note that the distance must be in AU and the
period is in years.
D
Option A is incorrect. The Moon is orbiting
around the Earth due to the gravitational pull from
the Earth. It is not logical to deduce that the
gravitational field produced by the Earth is zero at
a position nearer than the Moon.
(c)
(i)
The change in PE is
Option B is incorrect because an astronaut
orbiting the Earth is not always between the two
bodies.
Option C is incorrect. Note that centripetal force is
not a real force.
11.
(1M+1A)
A
(ii)
Since
, the closer the comet to the Sun,
the smaller the PE (the more negative the PE).
12.
Therefore, the KE change is
−1.91 × 1024 J. (1A)
A
When the engine is on, it produces a constant
thrust. At the same time, the gravitational force
acting on the astronaut gradually decreases (with
the distance between the rocket and the Earth).
Hence the g-force rises with an increasing trend
during the first 10 s.
By conservation of energy, the KE
decrease should be equal to the PE
increase in value. (1M)
17.
(d)
The Sun is not at the centre of the orbit. (1A)
(a)
The semi-major axis is 205 + 6370 × 2 +
50 930 = 31937.5 ≈ 3.19 × 104 km. (1A)
(b)
Applying
, the period is
When the engine is switched off, the rocket and
the astronaut inside are only subjected to the
gravitational force. Therefore, the astronaut will
feel weightless and the g-force is hence zero.
13.
(1M+1A)
B
Also accept 56 830 / (60 × 60) ≈ 15.8 h.
The change in PE is −5.8 × 109 – (−3 × 109)
= −2.8 × 109 J.
(c)
By conservation of energy,
Options C and D must be incorrect because the PE
has to be decreasing when the distance becomes
smaller.
14.
Substitute:
A
The original PE is
is
r1 = (205 + 6370) × 103 = 6.575 × 106 m, and
r2 = (50 930 + 6370) × 103 =5.73 × 107 m.
and the original KE
. When the spacecraft
Now we have
is very far away, the PE becomes zero and the KE is
. The speed is thus
15.
B
(1M+1A)
Note that the orbit of the Earth is not completely
circular.
(d)
(i)
The radius of the orbit is
(1740+200) × 103 = 1.94 × 106 m
Applying
Chapter Exercise: Structured Questions (p. 93)
16.
(a)
The semi-major axis is (1.13+1.78)/2
= 1.455 AU. (1A)
Moon is
, the mass of the
Active Physics Full Solutions to Textbook Exercises
63 Orbital Motions under Gravity
|
p.7
(1A)
The mechanical energy E is the sum of PE
and KE, i.e.
(1M+1A)
(ii)
Applying
.
(1A)
(b)
, the speed is
(i)
The mechanical energy E decreases. (1A)
As
and r decreases, E also
decreases (more ‘negative’). (1A)
OR: Part of the mechanical energy has
been converted to heat and internal
energy when the satellite moves across
air.
(1M+1A)
(e)
(f)
If the total mechanical energy is conserved,
Chang’e 1 would have travelled in a fixed
orbit but would not change its orbit. (1A)
In the circular orbit (1A)
At the point of transition, Chang’e 1
decelerates to transit from the elliptical orbit
to the less elliptical one and finally to the
circular one. In other words, the PE of
Chang’e 1 remains unchanged at that point
but the KE decrease. Hence we can deduce
that Chang’e 1 had the least mechanical
energy in the circular orbit. (1A)
18.
(a)
(ii)
The speed increases. (1A)
As
and E takes a more
negative value, we can deduce that v
increases. (1A)
20.
(a)
By
, the orbital speed is
(1M+1A)
(b)
The acceleration is
The gravitational PE is
(1M+1A)
(b)
(c)
W = mg = (100)(9.8049 ×
≈ 0.980 N (1M+1A)
10−3)
(1M+1A)
The mechanical energy is E = U/2
= −7.612 × 1010 / 2
= −3.806 × 1010 ≈ −3.81 × 1010 J (1M+1A)
The escape speed is
(c)
The minimum extra KE needed is
3.806 × 1010 J. The speed is
(2A)
Therefore, the ball can escape because its
speed is higher than the escape speed. (1A)
(d)
For the spacecraft to just escape from the
Earth forever, its mechanical energy after the
turning off the thrusters has to be zero. (1M)
The speed should be
(1A)
(d)
No. (1A)
The spacecraft is not on the Earth’s surface
initially. The KE needed and thus the value of
v are smaller. (1A)
(1M+1A)
19.
(a)
The PE is
21.
(a)
(i)
At infinity, all KE has changed to PE.
Since the PE at infinity is zero, KEmin is
+GMm/r. (1A)
(ii)
This is only the minimum because
there may be work done against other
resistive forces. (1A)
. (1A)
The centripetal force is provided by the
gravitational force. Therefore, the KE can be
deduced by
Active Physics Full Solutions to Textbook Exercises
63 Orbital Motions under Gravity
(iii) The initial KE is
. (1A)
Rearranging
p.8
The satellite actually keeps ‘falling’ but do
not fall to the surface of the Earth.
, we have
(1A)
(b)
(i)
The centripetal force need to keep the
satellite in orbit is
(iv) Substitute
|
. (1A)
This is provided by the gravitational
force, i.e.
into the above. (1A)
(1A)
(1A)
(v)
(1A)
Since v is indirectly proportional to √r,
v decreases as r increases. (1A)
(b)
(i)
The KE is
. The velocity is thus
(ii)
Subsitute v = 2πr/T into the above.
(1A)
(1M+1A)
22.
(a)
(ii)
Its speed is far lower than the escape
speed required. (1A)
(i)
The length of the orbit is 2 π (1.8 × 108)
= 1.13 × 109 m. (1A)
r3/2,
Since T ∝
T increases as r
increases. From (i) we can deduce that
T increases as v decreases. (1A)
(c)
The period is
The speed is thus 1.13 × 109/(21 × 60 ×
60) ≈ 1.5 × 104 m s−1. (1A)
(ii)
The centripetal force is
. (1A)
(1M+1A)
The value of the force is
(d)
The gravitational PE has been converted to
internal energy (heat) and KE. (2A)
(a)
A satellite in a geostationary orbit appears to
be stationary as viewed from the ground
because its orbital period is exactly the same
as the self-rotation period of the Earth. (1A)
. (1A)
(iii) The gravitational force is
24.
. (1A)
Therefore, the mass is
The transmitter and receiver on the ground
do not need to track such satellite as time
passes by. (1A)
(1A)
(b)
(i)
By
, the PE is
(b)
The centripetal force is provided by the
gravitational force.
(1M+1A)
(ii)
The KE is
.
(1M+1A)
(1A)
(iii) The total mechanical energy (PE+KE) is
negative. The rock does not have
sufficient KE to escape. (1A+1C)
23.
(a)
(c)
(i)
.
The KE is
. (1A)
The phrase ‘gravity is diminishing’ means
that the gravitational force is getting weaker
as the distance increases (
The PE is
The total mechanical energy is
). (1A)
There is no such a force pointing away from
the Earth acting on a satellite orbiting
around the Earth. (1A)
Also, the forces on the satellite are actually
unbalanced. (1A)
(1A)
(ii)
The change in energy is
Active Physics Full Solutions to Textbook Exercises
63 Orbital Motions under Gravity
(1M+1A)
(iii) Since
T2
∝
a 3,
we have
(1M)
The time is 10.57/2 ≈ 5.29 h. (1A)
You can also find T using
.
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p.9