1
Which of the following statements about Newton’s law of gravitation is correct?
Newton’s gravitational law explains
A
the origin of gravitational forces.
B
why a falling satellite burns up when it enters the Earth’s atmosphere.
C
why projectiles maintain a uniform horizontal speed.
D
how various factors affect the gravitational force between two
particles.
(Total 1 mark)
2A spacecraft of mass m is at the mid-point between the centres of a planet of mass M1 and its moon of
mass M2. If the distance between the spacecraft and the centre of the planet is d, what is the
magnitude of the resultant gravitational force on the spacecraft?
A
B
C
D
(Total 1 mark)
3
An object on the surface of a planet of radius R and mass M has weight W.
What would be the weight of the same object when on the surface of a planet of radius 2R and
mass 2M?
A
B
C
W
D
2W
(Total 1 mark)
Page 1 of 45
4
Two identical conducting spheres on insulating supports carry charges of magnitude Q and 2Q
respectively. When separated by distance d, the electrostatic repulsive force is F. The spheres
are made to touch and then restored to their original separation d. If there is no loss of charge
what is the new force of repulsion?
A
B
C
D
(Total 1 mark)
5
(a)
State the law that governs the magnitude of the force between two point masses.
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(2)
(b)
The table shows how the gravitational potential varies for three points above the centre of
the Sun.
(i)
distance from centre of Sun/108 m
gravitational potential/1010 J kg–1
7.0 (surface of Sun)
–19
16
–8.3
35
–3.8
Show that the data suggest that the potential is inversely proportional to the distance
from the centre of the Sun.
(2)
Page 2 of 45
(ii)
Use the data to determine the gravitational field strength near the surface of the Sun.
(3)
(iii)
Calculate the change in gravitational potential energy needed for the Earth to escape
from the gravitational attraction of the Sun.
mass of the Earth
=
6.0 × 1024 kg
distance of Earth from centre of Sun
=
1.5 × 1011 m
(3)
(iv)
Calculate the kinetic energy of the Earth due to its orbital speed around the Sun and
hence find the minimum energy that would be needed for the Earth to escape from its
orbit. Assume that the Earth moves in a circular orbit.
(3)
(Total 13 marks)
Page 3 of 45
6
(a)
Define gravitational field strength at a point in a gravitational field.
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(1)
(b)
Tides vary in height with the relative positions of the Earth, the Sun and the moon which
change as the Earth and the Moon move in their orbits. Two possible configurations are
shown in Figure 1.
Configuration A
Configuration B
Figure 1
Consider a 1 kg mass of sea water at position P. This mass experiences forces FE, FM and
FS due to its position in the gravitational fields of the Earth, the Moon and the Sun
respectively.
(i)
Draw labelled arrows on both diagrams in Figure 1 to indicate the three forces
experienced by the mass of sea water at P.
(3)
Page 4 of 45
(ii)
State and explain which configuration, A or B, of the Sun, the Moon and the Earth will
produce the higher tide at position P.
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(2)
(c)
Calculate the magnitude of the gravitational force experienced by 1 kg of sea water on the
Earth’s surface at P, due to the Sun’s gravitational field.
radius of the Earth’s orbit
= 1.5 × 10 11 m
mass of the Sun
= 2.0 × 1030 kg
universal gravitational constant, G
= 6.7 × 10−11 Nm2 kg−2
(3)
(Total 9 marks)
7
When two similar spherical objects of radius R are touching, the gravitational force of attraction
between them is F. When the gravitational force between them is F/4, the distance between the
surfaces of the spheres is
A
R
B
2R
C
4R
D
6R
(Total 1 mark)
8
(a)
State, in words, Newton’s law of gravitation.
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(3)
Page 5 of 45
(b)
By considering the centripetal force which acts on a planet in a circular orbit,
show that T2
R3, where T is the time taken for one orbit around the Sun and R is the
radius of the orbit.
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(3)
(c)
The Earth’s orbit is of mean radius 1.50 × 10 11 m and the Earth’s year is 365 days long.
(i)
The mean radius of the orbit of Mercury is 5.79 × 1010 m. Calculate the length of
Mercury’s year.
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(ii)
Neptune orbits the Sun once every 165 Earth years.
Calculate the ratio
.
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(4)
(Total 10 marks)
Page 6 of 45
9
The Global Positioning System (GPS) is a system of satellites that transmit radio signals which
can be used to locate the position of a receiver anywhere on Earth.
(a)
A receiver at sea level detects a signal from a satellite in a circular orbit when it is passing
directly overhead as shown in the diagram above.
(i)
The microwave signal is received 68 ms after it was transmitted from the satellite.
Calculate the height of the satellite.
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(ii)
Show that the gravitational field strength of the Earth at the position of the satellite is
0.56 N kg–1.
mass of the Earth
mean radius of the Earth
=
=
6.0 × 1024 kg
6400 km
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(4)
Page 7 of 45
(b)
For the satellite in this orbit, calculate
(i)
its speed,
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(ii)
its time period.
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(5)
(Total 9 marks)
10
(a)
State, in words, Newton’s law of gravitation.
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(2)
Page 8 of 45
(b)
Some of the earliest attempts to determine the gravitational constant, G, were regarded as
experiments to “weigh” the Earth. By considering the gravitational force acting on a mass at
the surface of the Earth, regarded as a sphere of radius R, show that the mass of the Earth
is given by
where g is the value of the gravitational field strength at the Earth’s surface.
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(2)
(c)
In the following calculation use these data.
radius of the Moon
= 1.74 × 106 m
gravitational field strength at Moon’s surface
= 1.62 N kg
mass of the Earth M
= 6.00 × 1024 kg
gravitational constant G
= 6.67 × 10–11 N m2 kg–2
–1
Calculate the mass of the Moon and express its mass as a percentage of the mass of the
Earth.
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(3)
(Total 7 marks)
11
Which one of the following graphs correctly shows the relationship between the gravitational
force, F, between two masses and the distance, r, between them?
A
B
Page 9 of 45
C
D
(Total 1 mark)
12
(a)
State Newton’s law of gravitation.
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(2)
(b)
In 1798 Cavendish investigated Newton’s law by measuring the gravitational force between
two unequal uniform lead spheres. The radius of the larger sphere was 100 mm and that
of the smaller sphere was 25 mm.
(i)
The mass of the smaller sphere was 0.74 kg. Show that the mass of the larger sphere
was about 47 kg.
density of lead = 11.3 × 103 kg m–3
(2)
(ii)
Calculate the gravitational force between the spheres when their surfaces were in
contact.
answer = ..................................... N
(2)
Page 10 of 45
(c)
Modifications, such as increasing the size of each sphere to produce a greater force
between them, were considered in order to improve the accuracy of Cavendish’s
experiment. Describe and explain the effect on the calculations in part (b) of doubling the
radius of both spheres.
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(4)
(Total 10 marks)
13
Two protons, each of mass m and charge e, are a distance d apart. Which one of the
following expressions correctly gives the ratio
for the forces
acting between them?
A
B
C
D
(Total 1 mark)
Page 11 of 45
14
A projectile moves in a gravitational field. Which one of the following is a correct statement for
the gravitational force acting on the projectile?
A
The force is in the direction of the field.
B
The force is in the opposite direction to that of the field.
C
The force is at right angles to the field.
D
The force is at an angle between 0° and 90° to the field.
(Total 1 mark)
15
Which one of the following graphs correctly shows the relationship between the gravitational
force, F, between two masses and their separation r.
(Total 1 mark)
Page 12 of 45
16
Two identical spheres exert a gravitational force F on each other. What is the gravitational force
between two spheres, each twice the mass of one of the original spheres, when the separation of
their centres is twice the original separation?
A
F
B
2F
C
4F
D
8F
(Total 1 mark)
17
(a)
Give two examples of the techniques used by geologists to obtain values of the strength of
the local gravitational field of the Earth.
In each of your quoted examples, describe the information that the geologists can derive
from their measurements.
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(4)
In 1774, Nevil Maskelyne carried out an experiment near the mountain of Schiehallion in
Scotland to determine the density of the Earth.
Figure 1 shows two positions of a pendulum hung near to, but on opposite sides of, the
mountain. The centre of mass of the mountain is at the same height as the pendulum.
Figure 1
Page 13 of 45
(b)
(i)
Explain why the pendulums do not point towards the centre of the Earth.
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(ii)
Suggest why Maskelyne carried out the experiment on both sides of the mountain.
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(3)
(c)
Figure 2 shows measurements made with the left-hand pendulum in Figure 1.
Figure 2
(i)
The mountain is in the appropriate shape of a cone 0.50 km high and 1.3 km base
radius; it rises from a locally flat plain.
Show that the mass of the mountain is about 2 × 1012 kg.
density of rock = 2.5 × 103 kg m−3
(ii)
Figure 2 shows the left-hand pendulum bob lying on a horizontal line that also
passes through the centre of mass of the mountain. The bob is 1.4 km from the
centre of the mountain and it hangs at an angle of 0.0011° to the vertical.
Calculate the mass of the Earth.
Page 14 of 45
(iii)
The answer Maskelyne obtained for the mass of the Earth was lower than today’s
accepted value even though he had an accurate value for the Earth’s radius.
Suggest one reason why this should be so.
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(7)
(Total 14 marks)
18
(a)
The weight w of an object on the Earth can be represented either as w = mg or
(i)
Explain the meaning of g and G in these equations.
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(3)
(ii)
Use the equations above to show that
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(1)
Page 15 of 45
(iii)
Calculate the mass of the Earth to a precision consistent with the data below.
mean radius of the Earth, = 6.4 × 106 m
G = 6.7 × 10–11 N m2 kg–2
g = 9.8 N kg–1
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mass of the Earth ...................................... kg
(3)
(b)
The figure below shows a satellite in a geostationary orbit around the Earth.
(i)
State the time period for a geostationary satellite.
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(1)
(ii)
The height of a geostationary satellite in orbit is approximately 36 000 km above the
surface of the Earth.
Calculate the radius of a geostationary orbit.
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radius ...................................... m
(1)
Page 16 of 45
(iii)
Calculate the speed, in km s–1, of a satellite in a geostationary orbit.
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speed ...................................... km s–1
(3)
(iv)
State a common use for a geostationary satellite.
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(1)
(v)
Explain why a geostationary orbit is necessary for this use.
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(1)
(Total 14 marks)
19
The gravitational force between two uniform spheres is 3.1 × 10–9 N when the distance between
their centres is 150 mm. If the mass of one sphere is 2.5 kg, what is the mass of the other?
A
0.043 kg
B
0.42 kg
C
2.8 kg
D
4.1 kg
(Total 1 mark)
Page 17 of 45
20
Two protons are 1.0 × 10–14 m apart. Approximately how many times is the electrostatic force
between them greater than the gravitational force between them?
(Use the Data and Formulae booklet)
A
1023
B
1030
C
1036
D
1042
(Total 1 mark)
21
A projectile moves in a gravitational field. Which one of the following is a correct statement about
the gravitational force acting on the projectile?
A
The force is in the direction of the field.
B
The force is in the opposite direction to that of the field.
C
The force is at right angles to the field.
D
The force is at an angle between 0° and 90° to the field.
(Total 1 mark)
22
Which one of the following statements about Newton’s law of gravitation is correct?
Newton’s law of gravitation explains
A
B
C
D
the origin of gravitational forces.
why a falling satellite burns up when it enters the Earth’s atmosphere.
why projectiles maintain a uniform horizontal speed.
how various factors affect the gravitational force between two particles.
(Total 1 mark)
Page 18 of 45
23
If an electron and proton are separated by a distance of 5 × 10–11 m, what is the approximate
gravitational force of attraction between them?
A
2 × 10–57 N
B
3 × 10–47 N
C
4 × 10–47 N
D
5 × 10–37 N
(Total 1 mark)
24
At the surface of the Earth the gravitational field strength is g, and the gravitational potential is V.
The radius of the Earth is R. An object, whose weight on the surface of the Earth is W, is moved
to a height 3R above the surface. Which line, A to D, in the table gives the weight of the object
and the gravitational potential at this height?
weight
gravitational
potential
A
B
C
D
(Total 1 mark)
Page 19 of 45
25
Masses of M and 2M exert a gravitational force F on each other when the distance between their
centres is r. What is the gravitational force between masses of 2M and 4M when the distance
between their centres is 4r?
A
0.25 F
B
0.50 F
C
0.75 F
D
1.00 F
(Total 1 mark)
26
(a)
(i)
Define gravitational field strength and state whether it is a scalar or vector quantity.
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(2)
(ii)
A mass m is at a height h above the surface of a planet of mass M and radius R.
The gravitational field strength at height h is g. By considering the gravitational force
acting on mass m, derive an equation from Newton’s law of gravitation to express g
in terms of M, R, h and the gravitational constant G.
(2)
(b)
(i)
A satellite of mass 2520 kg is at a height of 1.39 × 107 m above the surface of the
Earth. Calculate the gravitational force of the Earth attracting the satellite.
Give your answer to an appropriate number of significant figures.
force attracting satellite ........................................ N
(3)
Page 20 of 45
(ii)
The satellite in part (i) is in a circular polar orbit. Show that the satellite would travel
around the Earth three times every 24 hours.
(5)
(c)
State and explain one possible use for the satellite travelling in the orbit in part (ii).
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(2)
(Total 14 marks)
27
When a space shuttle is in a low orbit around the Earth it experiences gravitational forces FE due
to the Earth, FM due to the Moon and FS due to the Sun. Which one of the following correctly
shows how the magnitudes of these forces are related to each other?
mass of Sun = 1.99 × 1030 kg
mass of Moon = 7.35 × 1022 kg
mean distance from Earth to Sun = 1.50 × 1011 m
mean distance from Earth to Moon = 3.84 × 108 m
A
FE > FS > FM
B
FS > FE > FM
C
FE > FM > FS
D
FM > FE > FS
(Total 1 mark)
Page 21 of 45
28
Figure 1 shows (not to scale) three students, each of mass 50.0 kg, standing at different points
on the Earth’s surface. Student A is standing at the North Pole and student B is standing at the
equator.
Figure 1
Figure 2
The radius of the Earth is 6370 km.
The mass of the Earth is 5.98 × 1022 kg.
In this question assume that the Earth is a perfect sphere.
(a)
(i)
Use Newton’s gravitational law to calculate the gravitational force exerted by the Earth
on a student.
force ................................................ N
(3)
(ii)
Figure 2 shows a closer view of student A.
Draw, on Figure 2, vector arrows that represent the forces acting on student A.
(2)
Page 22 of 45
(b)
(i)
Show that the linear speed of student B due to the rotation of the Earth is about 460
ms–1.
(3)
(ii)
Calculate the magnitude of the centripetal force required so that student B moves
with the Earth at the rotational speed of 460 ms–1.
magnitude of the force ................................................ N
(2)
(iii)
Show, on Figure 1, an arrow showing the direction of the centripetal force acting on
student C.
(1)
(c)
Student B stands on a bathroom scale calibrated to measure weight in newton (N). If the
Earth were not rotating, the weight recorded would be equal to the force calculated in part
(a)(i).
State and explain how the rotation of the Earth affects the reading on the bathroom scale
for student B.
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(3)
(Total 14 marks)
Page 23 of 45
29
Which one of the following gives a correct unit for
A
N m−2
B
N kg−1
C
Nm
D
N
?
(Total 1 mark)
Page 24 of 45
Mark schemes
1
2
3
4
5
D
[1]
C
[1]
B
[1]
C
[1]
(a)
force is proportional to the product of the two masses
B1
force is inversely proportional to the square of their separation
(condone radius between masses)
or
equation M0 : masses defined A1 separation defined A1
B1
2
(b)
(i)
appreciation that potential x distance from centre of sun =
constant
or calculation of Vr for two sets of values (1.33 × 1020)
or uses distance ratio to calculate new V or r
C1
calculation of all three + conclusion
or uses distance ratios twice+ conclusion
conclusion must be more than ‘numbers are same’
(condone ‘signs’ and no use of powers of 10)
A1
2
Page 25 of 45
(ii)
V = GM/r and g = GM/r2
or
g = V/r (no mark for E or g= V/d or E = V/r )
B1
substitution of one set of data to obtain GM (1.33 × 1020)
or 19 × 1010/7 × 108 seen
B1
271 N kg−1 (m s−2) (J kg−1 m−1)
B1
3
(iii)
potential energy of the Earth = (−)GMm/r
or potential difference formula + r2 =∞
or potential at position of Earth = −8.87 × 108 J kg−1
(from Vr =1.33 × 1020)
C1
correct substitution (allow ecf for GM from (ii))
or
potential energy = potential x mass of Earth
C1
change in PE = 5.32 × 1033 J (cnao)
Fd approach is PE so 0 marks
A1
3
Page 26 of 45
(iv)
speed of Earth round Sun = 2πr/T or
or 3.0 × 104 m s−1
or KE=
B1
KE of Earth = ½ 6 × 1024 × their v2 (2.68 × 1033J)
B1
energy needed = difference between (iii) and orbital KE
(2.64 × 1033 J)
or KE in orbit = half total energy needed to
escape (−1 for AE)
B1
3
[13]
6
(a)
force acting per unit mass or g = F / m or g =
with terms defined
(1)
(b)
(i)
direction of FE correct in each diagram
B1
direction of FM correct in each diagram
B1
direction of FS correct in each diagram
B1
FS must be distinguished from FM
penalty of 1 mark for any missing labelling
(3)
(ii)
sun and moon pulling in same direction / resultant of FM and FS is greatest /
clear response including summation of FM and FS
M1
configuration A
A1
(2)
Page 27 of 45
(c)
F = GMm / R2
C1
correct substitution
C1
(5.95 or 5.96 or 5.9 or 6.0) × 10−3 N kg−1
A1
(3)
[9]
7
8
B
[1]
(a)
attractive force between point masses (1)
proportional to (product of) the masses (1)
inversely proportional to square of separation/distance apart (1)
3
(b)
mω2R = (–)
(use of T =
(1)
gives)
(1)
G and M are constants, hence T2
R3 (1)
3
(c)
(i)
(use of T2
R3 gives)
(1)
Tm = 87(.5) days (1)
(1) (gives RN = 4.52 × 1012 m)
(ii)
ratio =
= 30(.1) (1)
4
[10]
9
(a)
(i)
h (= ct) (= 3.0 × 108 × 68 × 10–3) = 2.0(4) × 107 m (1)
(ii)
g = (–)
(1)
r (= 6.4 × 106 + 2.04 × 107) = 2.68 × 107 (m) (1)
(allow C.E. for value of h from (i) for first two marks, but not 3rd)
g=
(1)
(= 0.56 N kg–1)
4
Page 28 of 45
(b)
(i)
g=
(1)
v = [0.56 × (2.68 × 107)]½ (1)
= 3.9 × 103m s–1 (1)
(3.87 × 103 m s–1)
(allow C.E. for value of r from a(ii)
[or v2 =
= (1)
v=
(1)
= 3.9 × 103 m s–1 (1)]
(ii)
=
(1)
= 4.3(5) × 104s (1)
(12.(1) hours)
3
(use of v = 3.9 × 10 gives T = 4.3(1) × 104 s = 12.0 hours)
(allow C.E. for value of v from (I)
[alternative for (b):
(i)
(1)
= 3.8(6) × 103 m s–1 (1)]
(allow C.E. for value of r from (a)(ii) and value of T)
(ii)
(1)
= (1.90 × 109 (s2) (1)
T = 4.3(6) × 104 s (1)
5
[9]
10
(a)
attractive force between two particles (or point masses) (1)
proportional to product of masses and inversely proportional to
square of separation [or distance] (1)
2
Page 29 of 45
(b)
(for mass, m, at Earth’s surface) mg =
(1)
rearrangement gives result (1)
2
(c)
(1)
= 7.35 × 1022 kg (1)
(= 0.0123) ∴ 1.23%
3
[7]
11
12
D
[1]
(a)
force of attraction between two point masses (or particles) (1)
proportional to product of masses (1)
inversely proportional to square of distance between them (1)
[alternatively
quoting an equation, F =
with all terms defined (1)
reference to point masses (or particles) or r is distance between
centres (1)
F identified as an attractive force (1)]
max 2
(b)
(i)
mass of larger sphere ML (=
= 47(.3) (kg) (1)
πr3ρ) =
π × (0.100)3 × 11.3 × 103 (1)
[alternatively
use of M µ r3 gives
(1) (= 64)
and ML = 64 × 0.74 = 47(.4) (kg) (1)]
2
(ii)
gravitational force F
(1)
= 1.5 × 10−7 (N) (1)
2
Page 30 of 45
(c)
for the spheres, mass µ volume (or µ r3, or M =
πr3ρ) (1)
mass of either sphere would be 8 × greater (378 kg, 5.91 kg) (1)
this would make the force 64 × greater (1)
but separation would be doubled causing force to be 4 × smaller (1)
net effect would be to make the force (64/4) = 16 × greater (1)
(ie 2.38 × 10−6 N)
max 4
[10]
13
14
15
16
17
D
[1]
A
[1]
D
[1]
A
[1]
(a)
technique one (1)
information derived from it (1)
technique two (1)
information derived from it (1)
4
Page 31 of 45
(b)
(i)
gravitational attraction t o …(1)
…centre of gravity(mass) of mountain (1)
(ii)
cancellation of some systematic errors (1)
3
(c)
(i)
calculates volume of cone (1)
mass = density × volume seen (1)
2.2 × 1012 kg (1)
(ii)
sideways force/mg = tan (0.0011) (1)
sideways force = Gmsch 0.5/(1400)2 subst seen (1)
2.4 × 1024 kg (1)
(iii)
his density estimate was too low (1)
or mean density of the Earth is higher than that of the mountain (1)
7
[14]
18
(a)
(i)
g gravitational field strength, G gravitational constant
C1
g force on 1 kg (on or close to) Earth’s surface
A1
G universal constant relating attraction of any two masses
to their separation/constant in Newton’s law of gravitation
A1
3
(ii)
equates w and cancels m
B1
1
Page 32 of 45
(iii)
substitutes values into equation
B1
correct calculation 5.99 × 1024
C1
answer to two significant figures 6.0 × 1024 (kg)
A1
3
(b)
(i)
1 day/24 hours/86400 (s)
B1
1
(ii)
4.24 × 107 (m)
B1
1
(iii)
v = 2πr/T or equivalent
C1
conversion of period to seconds (allow in (b)(i))
C1
3.08 (cao)
A1
3
(iv)
communication/specific example of communication (eg
satellite TV/weather)
B1
1
(v)
avoids dish having to track/stationary footprint
B1
1
[14]
19
B
[1]
Page 33 of 45
20
21
22
23
24
25
26
C
[1]
A
[1]
D
[1]
C
[1]
A
[1]
A
[1]
(a)
(i)
force per unit mass ✓
a vector quantity ✓
Accept force on 1 kg (or a unit mass).
2
(ii)
✓
force on body of mass m is given by
gravitational field strength
✓
For both marks to be awarded, correct symbols must be used for M
and m.
2
Page 34 of 45
(b)
✓
(i)
= 2.45 × 103 (N) ✓
to 3SF ✓
1st mark: all substituted numbers must be to at least 3SF.
If 1.39 × 107 is used as the complete denominator, treat as AE with
ECF available.
3rd mark: SF mark is independent.
3
(ii)
✓
F = mω2 (R + h) gives ω2 =
from which ω = 2.19 × 10–4 (rad s–1) ✓
or = 2.87 ✓ 104 s ✓
time period
✓
gives v2
[or F =
from which v = 4.40 ✓ 103 (m s–1) ✓
or = 2.87 × 104 s ✓ ]
time period T
[or T2 =
✓
✓
=
gives time period T = 2.87 × 104s ✓ ]
=
= 7.97 (hours) ✓
number of transits in 1 day =
= 3.01 ( ≈ 3) ✓
Allow ECF from wrong F value in (i) but mark to max 4 (because
final answer won’t agree with value to be shown).
First 3 marks are for determining time period (or frequency). Last 2
marks are for relating this to the number of transits.
Determination of f = 3.46 × 10–5 (s–1) is equivalent to finding T by
any of the methods.
5
Page 35 of 45
(c)
acceptable use ✓
satisfactory explanation ✓
e.g. monitoring weather or surveillance:
whole Earth may be scanned or Earth rotates under orbit
or information can be updated regularly
or communications: limited by intermittent contact
or gps: several satellites needed to fix position on Earth
Any reference to equatorial satellite should be awarded 0 marks.
2
[14]
27
A
[1]
Page 36 of 45
28
(a)
(i)
Use of F – GMm/r2
C1
Allow 1 for
-correct formula quoted but forgetting
square in substitution
Correct substitution of data
M1
-missing m in substitution
491 (490)N
A1
-substutution with incorrect powers of 10
Condone 492 N,
(ii)
Up and down vectors shown (arrows at end) with labels
B1
allow W, mg (not gravity); R
allow if slightly out of line / two vectors
shown at feet
up and down arrows of equal lengths
B1
condone if colinear but not shown acting on body
In relation to surface W ≤ R (by eye) to allow for weight vector
starting in middle of the body
Must be colinear unless two arrows shown in which case R vectors
½ W vector(by eye)
(b)
(i)
Speed = 2πr / T
B1
Max 2 if not easy to follow
2π6370000 / (24 × 60 × 60)
B1
463 m s−1
B1
Must be 3sf or more
(ii)
Use of F = mv2/r
C1
Allow 1 for use of F = mrω with ω= 460
2
1.7 (1.66 – 1.68) N
Page 37 of 45
A1
(iii)
Correct direction shown
(Perpendicular to and toward the axis of rotation)
NB – not towards the centre of the earth
B1
(c)
Force on scales decreases / apparent weight decreases
Appreciates scale reading = reaction force
C1
The reading would become 489 (489.3)N or reduced by 1.7 N)
A1
Some of the gravitational force provides the necessary centripetal force
B1
2
or R = mg – mv /r
[14]
29
A
[1]
Page 38 of 45
Examiner reports
2
3
5
6
Direct application of Newton’s law of gravitation easily gave the answer in this question, which
had a facility of 78%. A very small number of incorrect responses came from assuming that the
law gives F ∝ (1 / r) – represented by distractors A and D. Rather more (14%) chose distractor B;
these students probably added the two component forces acting on the spacecraft instead of
subtracting them.
This question which had a facility of 70%, was an algebraic test of the relationship between the
weight of an object at the surface of a planet and the mass and radius of that planet. This
question discriminated well and had no particularly strong distractor.
(a)
This was done well by the majority of candidates. A common error was to state that the
force is inversely proportional to the square of the radius.
(b)
(i)
Most candidates knew a method of showing the inverse proportionality.However,
many used only two of the sets of data or provided only a series of numbers without
any explanation of what they were doing or providing any conclusion. In the worst
cases, answers were set out poorly and any reasoning was hard to follow.
(ii)
Although many arrived at the correct answer, there were many dubious equations to
justify the final result. To gain full credit, candidates were expected to write down an
appropriate gravitational field equation from which to proceed. Some recalled the
value for G although the questions asked them to ‘use the data’.
(iii)
There were relatively few correct answers to this part. Many candidates could not
identify an appropriate equation to use or did not realise that they had the value for
GM from earlier parts. Some determined the energy needed for the Earth to move
from the surface of the Sun to the position of the Earth’s orbit. Those who recalled G,
having no value for the mass of the Sun, determined the energy required for the
Earth to escape from the Earth.
(iv)
Most were able to gain some credit for this part, gaining marks for calculating the
speed of the Earth in its orbit and/or for use of the KE formula. Many either ignored
the last part or added the KE in orbit to their answer to part (iii).
(a)
Most candidates managed to give an acceptable definition of gravitational field strength.
Those who did not usually failed because they omitted to mention unit mass or because
they confused field strength with potential or potential energy.
(b)
(i)
This part was also well done. Some candidates gave confused labelling, showed their
forces in the wrong direction, or omitted to show the forces on both of the diagrams.
(ii)
Explanations were often not clear: some candidates created a difficulty by referring to
the resultant force when they probably were thinking of the resultant force of only FM
and FS. A few candidates sought to give explanations relating to the distances
between the Earth and the Sun or Moon, highlighting the need to advise candidates
not to rely on judgements of distance from diagrams which are not to scale.
Page 39 of 45
(c)
8
This calculation was done well by most of the candidates. A few tried to use an equation for
potential rather than force and some made processing errors, often forgetting to square the
orbital radius even though they had shown it as being squared in their equation.
It was rare for all three marks to be awarded in part (a). Most answers made at least some
reference to the proportionality and inverse proportionality involved in Newton’s law, but
references to point masses or to the attractive nature of the force were scarce.
The essential starting point in part (b) was a correct statement equating the gravitational force
with mω2R; the more able candidates had little difficulty in then applying T = 2π/ω to derive the
required result, and three marks were usually obtained by them.
R3 result in part (b), and the candidates
Both halves of part (c) followed directly from the T2
who realised this usually made excellent progress. Unfortunately, a large proportion tried to go
back to first principles and tied themselves in knots with the algebra and/or arithmetic, often
getting nowhere. Confusion over which unit of time to employ in the different parts caused much
difficulty, especially for candidates who had calculated a constant of proportionality in part (i).
Some very elegant solutions to part (ii) were seen, where the result emerged swiftly from (165)2/3.
The most absurd efforts came from candidates who made the implicit assumption that the Earth,
Mercury and Neptune all travel at the same speed in their orbits, leading to wrong answers of
141 days and 165 respectively.
9
Most candidates scored the mark in part (a) (i) and went to use their answer correctly in part (ii).
A small number of candidates however, failed to add the height calculated in part (i) to the Earth’s
radius or added the radius in km to the height in m. They were usually able to gain some credit
for knowing the correct equation to use.
In part (b) (i), many candidates gave a clear and correct expression, using either the expressions
for centripetal acceleration or the speed in terms of the mass of the Earth. Weaker candidates
confused the symbols for speed and gravitational potential on the data sheet and attempted to
calculate the speed using the expression for gravitational potential. Most candidates who
completed part (i) went on to complete part (ii) successfully, although some lost the final mark as
a result of giving the answer to too many significant figures. Some candidates in part (ii)
successfully related the time period to the radius of orbit and thus gained full credit. A small
minority of candidates gained no credit as a result of misreading part (b), attempting to provide
answers based on a time period of 24 hours.
10
Missing from most attempted statements in part (a) were the expected references to point
masses and to an attractive force. Many candidates simply tried to put the well-known formula
into words, whilst others referred to the sum of the masses rather than the product of them.
The equation g = – GM / r2 is given in the Data booklet and mechanical rearrangement of it
leads directly to the expression in part (b). However, this was not what was required by the
wording of the question, and the many candidates who tried this approach were not given any
marks. The acceptable starting point was to equate the gravitational force with mg.
Answers to part (c) were frequently completely successful, making an interesting contrast with
the earlier parts of this question. The main problems here were omission of kg after the mass of
the moon, significant figure penalties, and arithmetical slips – typically forgetting to square the
denominator.
Page 40 of 45
12
Many correct statements of Newton’s law of gravitation were seen in part (a). Some candidates
1/r2, not both together) and only received
referred to just one aspect of the law ( M1M2, or
one mark. A reference to point masses – which helps when explaining the meaning of r – was not
common. In fact a clear understanding of the meaning of r was expected in satisfactory answers.
The common inadequate responses, when neither was more fully explained, were ‘radius’ and
‘istance’ Candidates who tried to rely simply on quoting F = GM1M2/r2 were awarded a mark only
when the terms in the equation were correctly identified; a further mark was available to them if
they gave a clear definition of r or referred to the nature of the force as attractive.
r3’. The first
Part (b) (i) could be approached using either ‘mass = volume × density’ or ‘mass
method was far more common, and most answers were satisfactory. On this paper, this was the
first example of a question requiring candidates to ‘show that…’ Convincing answers to this type
of question should include the fullest possible working, in which the final answer is quoted to one
more significant figure than the value given in the question. Here, for example, a value of 47.3 kg
was convincing. Part (b) (ii) also proved to be very rewarding for most candidates.
Common errors here were failing to square the denominator, or to assume that surfaces in
contact meant that r = 0 (whilst still arriving at a finite numerical answer!).
Whilst many correct and well argued answers were seen in part (c), it was clear that some
candidates had not read the question with sufficient care. Two requirements for a satisfactory
answer ought to be clear from the wording of the question: the need to give a quantitative
answer, and to confine the answer to the effect on the calculations in part (b). ‘Calculations’
(plural) was a strong hint that the mass of both spheres would be affected, but there were many
answers in which it was assumed that the masses would not be changed. This meant that a
maximum mark of 1 out of 4 could be awarded, for the 1/r2 relationship alone. The incorrect use
of language sometimes also limited the mark that could be awarded for the answers here:
candidates who stated that doubling the separation would reduce the force ‘by one quarter’ could
not be credited with a mark.
13
This question raised similar demands to the previous question, in that they required more than
one concept to be combined to give an algebraic result. In this question, electrostatic and
gravitational forces had to be considered together. The facility was 64%, and this question was
the best discriminator in the test. Over one-fifth of the responses were for distractor A, which has
(4π
14
) in the numerator instead of the denominator of the required expression.
The direction of forces in gravitational, electric and magnetic fields continues to be an area of
misunderstanding, as illustrated by the responses in this question, which had a facility of 55%.
Despite the fact that this question was about gravitational fields, just over a quarter of the
candidates selected distractor C, where the force is supposed to be at right angles to the field.
This confusion with a magnetic field is no more understandable than that of the 11% who chose
distractor B, where the force would be in the opposite direction to the field. Perhaps this latter
group were thinking of electrons in an electric field. Such incorrect responses suggest that
candidates were not always reading the questions with sufficient care.
Page 41 of 45
15
This question was a graphical test of inverse proportionality, as represented by the universal law
of gravitation. The facility of the question was 65%, but one quarter of the candidates were
tempted into choosing distractor C.
16
This question was about gravitational forces. Application of the inverse square law was
completed successfully by 70% of the candidates in the former question.
18
In part (a) (i), nearly all candidates correctly identified g and G; few were rigorous in their
explanations of what the quantities mean.
Few candidates did not equate the two equations in part (a) (ii), cancel m and rearrange into the
form shown.
The vast majority of candidate performed the calculation in part (a) (iii) correctly, but a significant
number quoted the final answer to either one or three significant figures (instead of the correct
two). A small minority of candidates forgot to square the radius of the Earth.
In part (b) (i), most candidates recognised that the period would be 24 hours.
Difficulty was had by some candidates in part (b) (ii) who struggled to add the quantities written in
different forms.
Part (b) (iii) was done well either by candidates dividing the circumference of the obit by the
period in seconds or else using the mass of the Earth calculated in part (a) (iii).
Most candidates gave an appropriate use for geostationary satellites in part (b) (iv), however
GPS and ‘mobile phones’ were not accepted.
In part (b) (v) few candidates were able to discuss the avoidance of dishes tracking by having
geostationary satellites.
19
20
This question, involving a rearrangement of the force equation from Newton’s law, had a facility
of 77%.
Another reused question combined Coulomb’s law with Newton’s law of gravitation and needed
candidates to take data from the Data and Formulae Booklet. The incorrect responses were
distributed fairly evenly across the three remaining distractors.
Page 42 of 45
21
22
23
24
25
The candidates in 2010 found this question to be slightly easier than their predecessors, with the
facility advancing from 55% to 59%. One in four candidates demonstrated their confusion with
magnetic fields by opting for distractor C, where the force was perpendicular to the field.
This question involving statements about Newton’s law of gravitation, had a facility of 85%.
When pre-tested, this question had been found appreciably harder but was more discriminating
than on this occasion.
Data for the gravitational constant and the masses of the electron and proton had to be extracted
from the Data Sheet (see Reference Material) for use in this question where the topic was the
gravitational force between two particles. Over four-fifths of the students succeeded with this.
This question continued the theme of gravitation. At first sight, it should be easy. In fact it was the
most demanding question in the test, with a facility of only 33%. Marginally more candidates
chose the incorrect distractor D than the correct answer. This was a fairly simple test of inverse
square proportion for force and inverse proportion for potential. Candidates made matters difficult
by confusing the distance from an external point to the centre of the Earth with the distance to the
surface of the Earth.
This was a fairly demanding calculation on the inverse square law of gravitation, in which
candidates had to consider the effect of changing both the size of the attracting masses and their
separation. Just over half reached the correct conclusion. No doubt it was errors in rearranging
the arithmetic and/or algebra that caused 34% of candidates to opt for distractor B, where the
new force was double what it ought to be.
Page 43 of 45
26
The definition in part (a)(i) was well known. Because the quantity concerned is called
gravitational field strength, there was frequent confusion as to whether it is a vector or a scalar,
with many answers being crossed out and changed. Part (a)(ii) was also generally very
rewarding. The main problem was a failure to show how the terms from the data booklet
equations (m1, m2 and r) translated into the terms in the question (m, M, R and h). In the
derivation, some students cancelled M instead of m. However, others had so little confidence in
their use of algebra that they could make little progress even in a simple derivation such as this.
Part (b)(i) caused few problems and marks were generally high. Sometimes incorrect values had
been extracted from the data booklet for the mass and radius of the Earth. Three significant
figures were expected in the answer; therefore a minimum of three significant figures should also
have been used in the substitution and working. When h = 1.39 × 107 was used as the radius of
the orbit one mark was lost and the value of the force thus obtained was carried forward to make
most marks available in part (b)(ii). Part (b)(ii) offered a very wide range of approaches to enable
students to show that the satellite would make three transits of Earth in every 24 hours. Apart
from the three alternatives given in the mark scheme (all of which were frequently seen) a very
concise calculation showed that a satellite with an angular speed of 2.19 × 10−4 rad s−1 would
move through an angle of 18.9 rad in one day, equating to (18.9 / 2π =) 3.01 transits.
Use of polar orbiting satellites for monitoring the Earth (weather forecasting, spying, surveying,
etc.) were well known in part (c), although some students confused the application with an
equatorial geosynchronous satellite. Explanations of the application were often less satisfactory:
reference to the rotation of Earth beneath the orbit, allowing the whole surface to be scanned,
was the key here. The ability to provide regular updates of the information obtained was also an
acceptable explanation. Students who mentioned the use of the polar satellite for
communications gained the first mark but were usually unable to point out its limitations, caused
by intermittent contact.
27
In this question the candidates had to decide about the relative magnitudes of the forces from the
Earth, the Moon and the Sun acting on a spacecraft when close to the Earth. Values for the
relevant masses and distances were provided in case candidates needed to perform a
calculation, or to carry out a check on their intuition. Obviously the spacecraft would not be in
orbit around Earth if FE was smaller than either of the other two forces. Hence FE must be the
largest of the three forces. The relative sizes of FM and FS then comes down to the ratio M / R2,
because the local gravitational field strength caused by each of the masses is GM / R2. The
facility of the question was 56%. 29% of the candidates chose distractor C; they appreciated that
FE is largest but thought that FM would be greater than FS.
Page 44 of 45
28
(a)
(b)
(i)
Most candidates were able to make good progress with this calculation and there
were many correct answers.
(ii)
Many attempts were unconvincing and frequently carelessly drawn. Weight and
reaction forces were often shown as not being collinear. Some showed a reaction
force at one of the feet but not the other. That the length of a vector should represent
magnitude was not realised by many candidates.
(i)
A good proportion of correct approaches were seen but many candidates seemed
unsure what equation to use so quoted some that were not relevant. Good structure
in a mathematical argument is an important skill in all problems but even more so in
‘show that’ type questions were marks are awarded for each step.
(ii)
Again there was a good proportion of correct response. Some candidates used F =
mrω2; and 460 m s-1 for ω.
(iii)
(c)
Misunderstanding about centripetal force was common here and there were relatively
few correct answers. The majority showed the force acting toward the centre of the
Earth. Whilst a component of this force provides the centripetal force, the direction of
the centripetal force is toward the centre of rotation which in the diagram is
perpendicular and toward the axis of rotation of the Earth.
There were very good answers from candidates who understood that the scales read the
reaction force.There were many who knew the formula mg – R = mv2 / r but thought that
the scales would record mg and assumed R to remain constant so that the centripetal force
increased the scale reading.
29
This question was on gravitational effects. Rearrangement of possible units to obtain the ratio of
the quantities g2 / G was required; almost 70% of the candidates could do this correctly but 20%
chose distractor B (N kg-1 instead of N m-2).
Page 45 of 45