Textbook Chap 4 Gravitational force and fields answers
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Student book answers
4.1 What is gravity?
Pages 116–117
Check your learning 4.1
Describe and explain
1
Explain how Aristotle’s explanation for the motion of heavy objects was discredited.
Aristotle argued that every object has a natural place and need to go there, and that the heavier an object is the quicker it will fall. Galileo proved that this is not the case with a series of experiments.
2
Describe how the word ‘gravity’ changed meanings in Newton’s writings
.
Gravity originally meant ‘heavy’ but Newton began using it to refer to a force between objects.
Apply, analyse and interpret
3
In his treatise on motion, Newton compared the gravitational force to the electrostatic and magnetic forces. He said that gravity was an attractive force only. Deduce how this differs from the electrostatic and magnetic forces.
Electrostatic and magnetic forces can also be attractive (for electrostatics a + and – charge will attract; for magnets a N-pole and a S-pole will attract). We say ‘unlike poles/charges attract’, but for mass, like masses attract. (If there was an anti-mass it may repel a mass).
Investigate, evaluate and communicate
4
It is said that gravity was invented by Newton, whereas others say he discovered it. Evaluate these two claims.
Newton gave the name ‘gravity’ to the natural phenomenon, whereby objects are attracted to each other. It could be said that he invented a model or laws that explain falling bodies, and the attraction between planets. Astronomers before Newton had discovered the orbits of planets, but he was the first to use the term ‘the force of gravity’ in his model of the universe.
5
Propose
, with reasons, whether or not this statement is true: Gravitational waves didn’t exist until they were discovered in 2015.
Something seems to happen when massive bodies like black holes collide and this phenomena has always existed. Signals are detected on Earth and these are said to be proof of gravitational waves. It is a good model and has supporting evidence, but the notion of gravitational waves being an explanation for this effect has been around for a 100 years. So, we could say the theory or model of gravitational waves was invented 100 years ago. 1
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Student book answers
4.2 Newton’s law of universal gravitation
Pages 118–124
Check your learning 4.2
Describe and explain
1
Newton’s law of gravitation is said to be an inverse square law. Explain what that means.
The gravitational force is inversely proportional to the square of the distance between the two objects.
2
Determine what happens to the gravitationalforce when:
a
one of the masses is doubled
The force is doubled
b
the distance between the objects is halved.
One-quarter the force
3
Explain why Earth’s radius needs to be taken into account for objects near the surface, whencalculating gravitational forces.
When using Newton’s law of gravitational attraction the objects should be treated as point masses (without size) so if
the object is on the surface of the Earth it is the Earth radius away from the centre.
4
Calculate the force between the Sun (
m = 2.0 × 10
30 kg) and Earth (
m = 5.97 × 10
24 kg) assuming their centres are 1.5 × 10
8 km apart.
5
A 10 kg rock rests on the ground. Calculate the gravitational force acting on it, using Newton’s law of universal gravitation.
2
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Apply, analyse and interpret
6
Black holes are supermassive collapsed stars. The closest anything can get to one and still escape its gravitational force is called the ‘event horizon’. Determine the force acting on a 15 tonne spacecraft at the event horizon of 30 km from a black hole that has a mass equal to 10 times that of the Sun.
Mass of Sun = 2.0 × 10
30 kg; 1 tonne = 1000 kg
7
When a star collapses to form a black hole, the size of the star is greatly reduced although the mass remains the same. Deduce what would happen to the gravitational force of the Sun on our Earth if the Sun unexpectedly collapsed to form a black hole?
Gravity acts between bodies as if they were point masses, so a smaller Sun with the same mass would have no effect on the gravitational force between the Sun and Earth. In reality, if the Sun shrank there would be enormous heating effects from friction and an outpouring of radiation that would affect Earth.
Investigate, evaluate and communicate
8
Jupiter is about 300 times more massive than Earth so it would be easy to deduce that an object on the surface of Jupiter would weigh 300 times more than on the surface of Earth. For example, a rover (spacecraft) with a weight of 9000 N on Earth might be expected to weigh 2 700 000 N on the surface of Jupiter. But this is not the
case. A 9000 N rover on Earth weighs only about 27000 N on the surface of Jupiter. Evaluate this scenario and identify any misunderstandings.
Diameter of Earth = 12742 km, therefore radius of Earth r
E
= 6371 km
Diameter of Jupiter = 139822 km, therefore radius of Jupiter r
J
= 69911
The 9000 N rover weighs 22 422 N on the surface of Jupiter.
9
A 1 tonne communications satellite is orbiting a planet. A student said that ‘if you double the mass of the satellite or the planet the force will double’. Another student said that doubling the mass of the planet had to have a bigger effect than just going from a 1 tonne satellite to a 2 tonne satellite. Evaluate both claims and decide who is correct.
The formula for the gravitational force between two objects is . It doesn’t matter whether the value of M
is doubled to 2
M
, or m
is doubled to 2
m
, it still has the same effect on the force (doubling it).
3
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Student book answers
4.3 Gravitational fields
Pages 125–131
Check your learning 4.3
Describe and explain
1
Explain how the unit symbol N kg
–1 is equivalent to m s
–2
.
2
Describe a gravitational field.
A gravitational field is a region of space where a gravitational force is experienced.
3
Define ‘gravitational field strength’ and state its units.
Gravitational field strength is the gravitational force per unit of mass. Units: Newtons per kilogram (N kg
-1
)
4
Explain what is meant by ‘the gravitational field strength does not depend on the mass of the object in the field’.
Gravitational field strength is a measure of the force per unit of mass. The force depends on the mass but gravitational field strength is the factor that relates mass to strength.
5
Calculate the gravitational field strength at a point 100 000 km above Earth’s centre.
6
Clarify how the direction of a gravitational field is defined.
It is the direction that a mass would fall freely in the field.
4
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Apply, analyse and interpret
7
Determine the gravitational field strength at a point whose distance from Earth’s surface is equal to
three Earth radii.
8
Determine the altitude above Earth’s surface where the gravitational field strength is one-eighth the value on the surface. 9
Derive a location between the Sun and Earth where the gravitational field between them due to their mass is equal to zero.
Let x be the distance from the point to the Sun.
5
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10
Construct a graph of gravitational field strength versus distance from Earth’s centre, using the data from Table 1 on page 129. Determine the relationship. Construct a second graph to confirm your prediction.
The graph appears to be inverse or inverse square. The formula is an inverse square relationship, so to linearise we plot F
vs 1/
d
2
. This has linearised it.
11
The gravitational field strength 100 km from the Earth’s centre is 3.99 × 10
4 m s
–2
. That’s 40 000 m s
–2
. Deduce why this nonsensical.
The formula doesn’t hold for inside the Earth. Besides, there would be a cancelling out of field strength from mass above this point.
Investigate, evaluate and communicate
12
Jupiter is about 300 times more massive than Earth so it would be easy to deduce that an object on the surface of Jupiter would weigh 300 times more than on the surface of Earth. But this is not the case. It is just three times as heavy. Evaluate this scenario and identify any misunderstandings.
Radius of Earth r
E
= 6371 km
Radius of Jupiter r
J
= 69911
Ratio:
The force acting on an object on the surface of Jupiter is about 2.5 times the weight on the surface of Earth. The misunderstanding is that Jupiter has a bigger radius so the force acting on the surface is less because of distance even
though it is more due to extra mass of Jupiter.
6
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7
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Student book answers
Chapter 4 Review
Pages 132–135
Revision questions
The relative difficulty of these questions is indicated by the number of stars beside each question number: * = low; ** = medium; *** = high.
For questions that follow, use the following data:
Constant
Value
G
(on surface of Earth) = 9.8 ms
–2
r
Mo
(radius of Moon’s orbit around Earth) = 3.8 10
11
m
G
= 6.67 10
–11
Nm
2
kg
–2
r
Eo
(radius of Earth’s orbit around Sun) = 1.5 10
11
m
Body
Mass
Radius
Earth
m
E
= 5.97
10
24
kg
r
E = 6.37 10
6
m
Moon
m
M
= 7.34 10
22
kg
r
M
= 1.74 10
6
m
Sun
m
S
= 2.0 10
30
kg
r
S = 6.96 10
8
m
Multiple choice
1
Select the diagram in Figure 1 that cannot be used as a diagram for a gravitational field.
FIGURE 1
Answer: A. Reason: The field should point towards mass.
2
There are many planets in the universe orbiting a star other than our own Sun. Most of these planets are quite huge – as big as Jupiter. The planet Kepler-7b has a mass 0.433 times that of Jupiter and a radius 1.614 times that of Jupiter. If the gravitational field strength at the surface of Kepler-7b is g
K
, and the gravitational field strength at the surface of Jupiter to be g
J
, the ratio is:
*A
0.17
B
0.27
C
0.69
8
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D
1.1
Reason:
3
Two equal masses, m
, separated by a distance, r
, exert a force, F
, on each other due to their gravitational attraction. Calculate the magnitude of the gravitational force between an object of mass m and an object of
mass 2m separated by the same distance r
.
A
B
*C
2
F
D
4
F
Reason:
4
Select the diagram in Figure 2 that best describes the forces acting on the Moon (M) due to Earth E.
9
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FIGURE 2
Answer B. There is a gravitational force pulling on the Moon that acts towards Earth. This provides the centripetal force that keeps the Moon in orbit. There is only one force (
F
g
) but it is equal in magnitude and direction to F
c
. We can eliminate A as F
c
points the wrong way. We can eliminate D as F
c
points the wrong way and is a different magnitude. Option C has both forces pointing the right way and equal in size but because there is only one force (not
two) it is wrong.
5
Select the statement that best defines gravitational field strength.
A
It is the force acting on a small object in the field.
*B
It is the force acting on a 1 kg object in the field.
C
It is the work done on a 1 kg object to shift it 1.0 m in the field.
D
It is the work done by a field on shifting a 1 kg object 1.0 m in the field.
Reason: The gravitational field strength (
g
) is the net force (
F
) per unit mass (
m
) at a particular point in the gravitational field.
6
On the surface of a planet with a radius r
P
, a mass experiences a gravitational force of attraction of g
. At a height of 4
r
P
above the planet, the mass experiences a force of:
A
B
C
*D
Reason:
10
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11
© Oxford University Press 2019
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7
The force of gravity on a 1 kg object on the surface of Earth is 9.8 N directed downwards. Select the answer that best states the force on a 1 kg object on the surface of Saturn given that Saturn is 100 times the mass of Earth and 10 times the diameter.
A
B
1 N
*C
10 N
D
100 N
Reason:
8
Imagine that the Earth preserved its mass but occupied a sphere only half its current diameter. Determine the weight of a person who normally weighs 600 N on the surface of this alternative Earth.
A
300 N
B
600 N
C
1200 N
*D
2400 N
Reason:
12
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9
Two planets have the same gravitational potential at their surfaces. Select one of the following ratios that must also be the same for the two planets.
A
*B
C
D
radius
10
Planet X has mass M and radius R
. Planet Y has mass 10
M and radius 4
R
. Determine the ratio:
A
0.5
*B
1.6
C
2.0
D
2.5
Reason:
14
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Short answer
Describe and explain
*11 Explain the difference between weight and mass.
Weight is a measure of the local force of gravity on an object.
Mass is a measure of an object’s resistance to change in motion (inertia).
*12
A satellite is in a circular orbit around Earth. Explain why an astronaut aboard the satellite feels weightless if the value of g at that altitude is 6.5 m s
–2
?
Earth-orbiting astronauts feel weightless because there is no external contact force pushing or pulling their body. Gravity is the only force acting on their body, even if it is a reduced value of 6.5 m s
–2
instead of the 9.8 m s
–2
they would experience on the surface of the Earth. The orbiting astronauts still have weight (
F
g
= m × 6.5); that is, there is
a force of gravity acting upon their body that supplies the centripetal force to allow for the orbiting circular motion. The force of gravity is the only force acting upon their body and the astronauts are in free-fall.
Many students believe that orbiting astronauts are weightless because they do not experience a force of gravity. This is wrong as gravity keeps the space station in orbit. Other students say they feel weightless because of the reduced gravity. They say ‘with less gravity, there would be less weight and thus they would feel less than their normal weight.’ While this is partly true, it does not explain their sense of weightlessness. The force of gravity acting upon an astronaut on the space station is certainly less than on Earth's surface. But it is not small enough to account for a significant reduction in weight. In summary, astronauts feel weightless because they are in free fall.
*13 Recall the formula for Newton’s law of universal gravitation and explain the meaning of the symbols.
F
is the gravitational force between the bodies
G
is Gravitational constant = 6.67 x 10
-11
N m
2
kg
-2
M
is the mass of one body (in kg)
m
is the mass of the other body (in kg)
R
is the radial distance between the centres of mass of each body (in metres)
*14 Explain the link between acceleration due to gravity and gravitational field strength.
They are identical quantities but with different (but equivalent) units.
*15 Calculate the gravitational force between:
a
two 60 kg people 1.0 m apart
15
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b
two protons (
m
p = 1.7 × 10
–27
kg) at a distance of 5 × 10
–15
m apart.
.
Apply, analyse and interpret
*16 Clarify the difference between heliocentric and geocentric models of the solar system.
The heliocentric model has the sun (‘helios’) at the centre and all planets revolve around it. The geocentric model has the Earth (‘geo’) at the centre and the sun and planets revolve around it. The heliocentric model is the accepted model.
*17
A satellite is orbiting Earth at a distance of 35 kilometres. The satellite has a mass of 500 kg. Determine the force between the planet and the satellite.
*18
An astronaut is on the Moon.
a
Calculate the acceleration due to gravity on its surface.
16
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b
Determine the time it would it take a spanner to fall from rest from a height of 1.5 m to the surface
of the Moon.
*19
A 15 kg object has a weight of 8000 N. Calculate is the gravitational field strength at this point.
*20 Determine the gravitational field strength of the Sun (mass 2 × 10
30 kg) at a distance of 15 × 10
10 m
from its centre.
17
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*21 Calculate the distance between two objects of mass 5 × 10
4 kg and 2.5 × 10
4 kg respectively, if the gravitational
force between them is 2.00 × 10
–8 N.
**22
Determine the gravitational field strength at a point:
a
on Earth’s surface (see data table at start of questions)
b
1.5 Earth radii above Earth’s surface
c
3.0 Earth radii above Earth’s surface
18
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d
1000 m above the surface of the Moon
e
on the surface of the Sun.
**23
A hydrogen atom consists of an electron of mass 9 × 10
–31 kg and a proton of mass1.9 × 10
–27 kg separated by an
average distance of 6 × 10
–11 m. Determine
the gravitational force between them.
19
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**24
A satellite is orbiting Earth in a circular orbit at a distance of two Earth radii from Earth’s surface.
a
Calculate the acceleration due to gravity aboard the satellite.
b
Calculate the weight of a 70 kg astronaut aboard the satellite.
**25
Determine the distance between two objects of mass 6 × 10
5 kg and 3 × 10
5 kg, if the gravitational force between them is 1.5 × 10
8
N.
20
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**26
Deduce the gravitational field strength in the region of a satellite orbiting 8000 km above Earth’s surface.
**27
Determine the height above the surface of Earth at which the gravitational field strength is equal to 2.0 N kg
−1
.
21
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**28
Determine the altitude above Earth’s surface at which the gravitational field strength is one twenty-fifth the value on the surface of the Earth.
Let a be the altitude of the location of A, where g
A
is the value on the surface of the Earth (g).
The radial distance of point A (
r
A
) is the radius of the Earth (
r
E
) plus the altitude (
a
): r
A
= r
E
+
a
Note: T
his can also be done by using the accepted values in the equation for the radius of the Earth, g and G.
***29
Determine the gravitational force of attraction between an electron and a nucleus 1.5 μm apart.
The mass of the electron is 9.11 × 10
–31 kg. Consider the nucleus to be made up of a proton and neutron,
each of mass 1.67 × 10
–27 kg.
22
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***30
Deduce the acceleration due to gravity at the surface of the Moon if its mass is times the mass of Earth and its diameter times that of Earth.
***31
Earth’s gravitational field strength at its surface is g
, and Earth’s radius is r
. Determine the gravitational field strength at a distance of 33
r from the centre of Earth.
***32
Figure 3 shows a satellite orbiting Earth. The satellite is part of the network of global positioning satellites that transmit radio signals used to locate the position of receivers on Earth.
When the satellite is directly overhead, the signal reaches the receiver 67 ms after it leaves the satellite. Radio waves travel at the speed of light (
c = 3 × 10
8 m s
–1
).
24
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FIGURE 3
a
Calculate the height of the satellite above the surface of Earth.
b
Explain why the satellite is accelerating towards the centre of Earth even though its orbital speed is constant.
According to Newton’s 2nd law, the direction of the acceleration is given by the direction of the net force. The only force acting on a satellite is the gravitational force between it and Earth so the satellite has a net force towards the Earth. There is no other force. Hence, the acceleration is also towards the centre of the Earth.
c
Determine the gravitational field strength due to Earth at the position of the satellite.
Investigate, evaluate and communicate
**33
Predict how far would you have to travel upwards from the Earth’s surface to notice a 1 N kg
−1 difference in gravitational field strength. (Earth has a radius of 6400 km.)
Let A
be the position about the Earth’s surface where the field strength is 1 N kg
-1
lower than on the surface of the Earth. This question can be done in two ways:
25
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Alternatively:
**34
Predict the height at which the force of gravity on a satellite will be half that on it at ground level.
26
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**35
Predict the gravitational field strength at a point whose distance from Earth’s surface is equal
to four Earth radii.
Field strength follows an inverse square relationship with radial distance.
If g
= 9.8 N kg
-1
at 1 r
E
,
then at (1 + 4) r
E
the field strength will be:
9.8 = 0.392 N kg
-1
Alternatively:
***36
Evaluate whether the gravitational field strength of Earth at height h above the surface is given by
, where g
s
is the gravitational field strength at the surface and R is the radius of the Earth.
So yes, the gravitational field strength of Earth at height h above the surface is given by
.
27
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