Download A satellite X is in a circular orbit of radius r about the centre

1
A satellite X is in a circular orbit of radius r about the centre of a spherical planet of mass M.
Which line, A to D, in the table gives correct expressions for the centripetal acceleration a and
the speed v of the satellite?
Centripetal acceleration a
Speed v
A
B
C
D
(Total 1 mark)
2
The Earth moves around the Sun in a circular orbit with a radius of 1.5 × 108 km.
What is the Earth’s approximate speed?
A
1.5 × 103ms–1
B
5.0 × 103ms–1
C
1.0 × 104ms–1
D
3.0 × 104ms–1
(Total 1 mark)
Page 1 of 52
3
Two satellites, P and Q, of the same mass, are in circular orbits around the Earth. The radius of
the orbit of Q is three times that of P. Which one of the following statements is correct?
A
The kinetic energy of P is greater than that of Q.
B
The weight of P is three times that of Q.
C
The time period of P is greater than that of Q.
D
The speed of P is three times that of Q.
(Total 1 mark)
4
(a)
(i)
State the relationship between the gravitational potential energy, Ep, and the
gravitational potential, V, for a body of mass m placed in a gravitational field.
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(1)
(ii)
What is the effect, if any, on the values of Ep and V if the mass m is doubled?
value of Ep ...........................................................................................
value of V … … …
................................................................................
.
(2)
Page 2 of 52
(b)
The diagram above shows two of the orbits, A and B, that could be occupied by a satellite
in circular orbit around the Earth, E.
The gravitational potential due to the Earth of each of these orbits is:
orbit A
orbit B
(i)
– 12.0 MJ kg–1
– 36.0 MJ kg–1.
Calculate the radius, from the centre of the Earth, of orbit A.
answer = .......................... m
(2)
(ii)
Show that the radius of orbit B is approximately 1.1 × 104 km.
(1)
(iii)
Calculate the centripetal acceleration of a satellite in orbit B.
answer = ..................... m s–2
(2)
Page 3 of 52
(iv)
Show that the gravitational potential energy of a 330 kg satellite decreases by about 8
GJ when it moves from orbit A to orbit B.
(1)
(c)
Explain why it is not possible to use the equation ∆Ep = mg∆h when determining the
change in the gravitational potential energy of a satellite as it moves between these orbits.
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(1)
(Total 10 marks)
5
A satellite is in orbit at a height h above the surface of a planet of mass M and radius R. What is
the velocity of the satellite?
A
B
C
D
(Total 1 mark)
Page 4 of 52
6
Satellites N and F have the same mass and move in circular orbits about the same planet. N is
the nearer satellite and F is the more distant. Which one of the following is smaller for N than for
F?
A
gravitational force on the satellite
B
speed
C
kinetic energy
D
time for one orbit
(Total 1 mark)
7
The diagram below (not to scale) shows the planet Neptune (N) with its two largest moons,
Triton (T) and Proteus (P). Triton has an orbital radius of 3.55 × 108 m and that of Proteus is 1.18
× 108 m. The orbits are assumed to be circular.
(a)
Explain why the velocity of each moon varies whilst its orbital speed remains constant.
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(1)
Page 5 of 52
(b)
Write down an equation that shows how Neptune’s gravitational attraction provides the
centripetal force required to hold Triton in its orbit. Hence show that it is unnecessary to
know the mass of Triton in order to find its angular speed.
(3)
(c)
Show that
is approximately 5.2.
(4)
(Total 8 marks)
Page 6 of 52
8
(a)
State, in words, Newton’s law of gravitation.
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(3)
(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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Page 7 of 52
(ii)
Neptune orbits the Sun once every 165 Earth years.
Calculate the ratio
.
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(4)
(Total 10 marks)
9
Communications satellites are usually placed in a geo-synchronous orbit.
(a)
State two features of a geo-synchronous orbit.
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(2)
(b)
Given that the mass of the Earth is 6.00 × 1024 kg and its mean radius is 6.40 × 106 m,
(i)
show that the radius of a geo-synchronous orbit must be 4.23 × 107 m,
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Page 8 of 52
(ii)
calculate the increase in potential energy of a satellite of mass 750 kg when it is
raised from the Earth’s surface into a geo-synchronous orbit.
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(6)
(Total 8 marks)
10
What is the angular speed of a satellite in a geo-synchronous orbit around the Earth?
A
7.3 × 10–5 rad s–1
B
2.6 × 10–1 rad s–1
C
24 rad s–1
D
5.0 × 106 rad s–1
(Total 1 mark)
Page 9 of 52
11
A planet of mass M and radius R rotates so rapidly that loose material at the equator only just
remains on the surface. What is the period of rotation of the planet?
G is the universal gravitational constant.
A
B
C
D
(Total 1 mark)
12
A satellite orbiting the Earth moves to an orbit which is closer to the Earth.
Which line, A to D, in the table shows correctly what happens to the speed of the satellite and to
the time it takes for one orbit of the Earth?
Speed of satellite
Time For One Orbit Of Earth
A
decreases
decreases
B
decreases
increases
C
increases
decreases
D
increases
increases
(Total 1 mark)
Page 10 of 52
13
As a comet orbits the Sun the distance between the comet and the Sun continually changes. As
the comet moves towards the Sun this distance reaches a minimum value.
Which one of the following statements is incorrect as the comet approaches this minimum
distance?
A
The potential energy of the comet increases.
B
The gravitational force acting on the comet increases.
C
The direction of the gravitational force acting on the comet changes.
D
The kinetic energy of the comet increases.
(Total 1 mark)
14
The Hubble space telescope was launched in 1990 into a circular orbit near to the Earth.
It travels around the Earth once every 97 minutes.
(a)
Calculate the angular speed of the Hubble telescope, stating an appropriate unit.
answer = .....................................
(3)
(b)
(i)
Calculate the radius of the orbit of the Hubble telescope.
answer = ................................ m
(3)
Page 11 of 52
(ii)
The mass of the Hubble telescope is 1.1 × 104 kg. Calculate the magnitude of the
centripetal force that acts on it.
answer = ................................ N
(2)
(Total 8 marks)
15
(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 12 of 52
(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)
Page 13 of 52
(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)
(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)
Page 14 of 52
16
An artificial satellite of mass m is in a stable circular orbit of radius r around a planet of mass M.
Which one of the following expressions gives the speed of the satellite?
G is the universal gravitational constant.
A
B
C
D
(Total 1 mark)
Page 15 of 52
17
The figure below shows the variation of gravitational potential, V, with distance from the centre of
the Earth, r. The radius of the Earth is 6.4 × 106 m.
(a)
Explain why the V values are negative.
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(3)
(b)
Use data from the graph to show that the mass of the Earth is approximately 6 × 1024 kg.
(3)
Page 16 of 52
(c)
(i)
Calculate the work done in raising a satellite of mass 2100 kg from the surface of the
Earth to a height of 850 km above the surface of the Earth.
work done ......................................................... J
(3)
(ii)
Calculate the change in the kinetic energy of the satellite when it moves from its
850 km orbit to one at a height of 700 km above the Earth’s surface. Make it clear
whether the change in kinetic energy is an increase or decrease.
kinetic energy change ......................................................... J
(4)
(iii)
Without performing any further calculations explain how the change in kinetic energy
relates to the change of the potential energy when the satellite’s orbit alters as in part
(c)(ii).
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(2)
(Total 15 marks)
Page 17 of 52
18
A satellite is in orbit at a height h above the surface of a planet of mass M and radius R.
What is the velocity of the satellite?
A
B
C
D
(Total 1 mark)
19
A satellite of mass m travels in a circular orbit of radius r around a planet of mass M. Which one
of the following expressions gives the angular speed of the satellite?
A
B
C
D
(Total 1 mark)
Page 18 of 52
20
(a)
Explain why the mass of an object is constant but its weight may change.
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(3)
(b)
The table gives the gravitational potentials, V, at three different distances, r, from the centre
of the Earth.
(i)
distance from centre of Earth
r / km
gravitational potential
V / 107 J kg–1
7500
–5.36
12500
–3.22
22500
–1.79
Explain why the gravitational potential at a point in a gravitational field is negative.
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(2)
Page 19 of 52
(ii)
Show that the data in the table are consistent with V
r –1.
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(3)
(iii)
A satellite of mass 450 kg is moved from an orbit of radius 7500 km around the Earth
to an orbit of radius 12 500 km.
Use data from the table to show that the potential energy of the satellite increases,
by about 10 GJ.
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(2)
Page 20 of 52
(c)
The kinetic energy of a 450 kg satellite orbiting the Earth with a radius of 7500 km is 12 GJ.
(i)
Calculate the kinetic energy of the 450 kg satellite when it is in an orbit of radius
12 500 km.
mass of the Earth = 6.0 × 1024 kg
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kinetic energy ............................................ GJ
(4)
(ii)
Calculate the change in kinetic energy of the satellite when it moves into the higher
orbit.
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change in kinetic energy ............................................ GJ
(1)
(iii)
Calculate the total energy that has to be supplied to move the 450 kg satellite from
an orbit of radius 7500 km to an orbit of radius 12 500 km.
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total energy ............................................ GJ
(1)
(Total 16 marks)
21
(a)
(i)
State what is meant by the term escape velocity.
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(1)
Page 21 of 52
(ii)
Show that the escape velocity, v, at the Earth’s surface is given by v =
where M is the mass of the Earth
and R is the radius of the Earth.
(2)
(iii)
The escape velocity at the Moon’s surface is 2.37 × 10 3 m s–1 and the radius of the
Moon is 1.74 × 106 m.
Determine the mean density of the Moon.
mean density ......................................... kg m–3
(2)
(b)
State two reasons why rockets launched from the Earth’s surface do not need to achieve
escape velocity to reach their orbit.
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(2)
(Total 7 marks)
22
(a)
Explain why astronauts in an orbiting space vehicle experience the sensation of
weightlessness.
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(2)
Page 22 of 52
(b)
A space vehicle has a mass of 16 800 kg and is in orbit 900 km above the surface of the
Earth.
mass of the Earth = 5.97 × 1024 kg
radius of the Earth = 6.38 × 106 m
(i)
Show that the orbital speed of the vehicle is approximately 7400 m s–1.
(4)
Page 23 of 52
(ii)
The space vehicle moves from the orbit 900 km above the Earth’s surface to an orbit
400 km above the Earth’s surface where the orbital speed is 7700 m s –1.
Calculate the total change that occurs in the energy of the space vehicle.
Assume that the vehicle remains outside the atmosphere after the change of orbit.
Use the value of 7400 m s–1 for the speed in the initial orbit.
change in energy ................................................... J
(4)
(Total 10 marks)
Page 24 of 52
23
Read the following passage and answer the questions that follow
Satellites used for telecommunications are usually in geostationary orbits. Using
suitable dishes to transmit the signals, communication over most of the Earth’s
surface is possible at all times using only 3 satellites.
Satellites used for meteorological observations and observations of the Earth’s
surface are usually in low Earth orbits. Polar orbits, in which the satellite passes
over the North and South Poles of the Earth, are often used.
5
One such satellite orbits at a height of about 12 000 km above the Earth’s surface
circling the Earth at an angular speed of 2.5 × 10–4 rad s–1. The microwave signals
from the satellite are transmitted using a dish and can only be received within a
limited area, as shown in the image below.
10
The signal of wavelength λ is transmitted in a cone of angular width θ, in radian,
given by
where d is the diameter of the dish.
The satellite transmits a signal at a frequency of 1100 MHz using a 1.7 m diameter
dish. As this satellite orbits the Earth, the area over which a signal can be
received moves. There is a maximum time for which a signal can be picked up by
a receiving station on Earth.
(a)
15
Describe two essential features of the orbit needed for the satellite to appear
geostationary.
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(2)
(b)
Calculate the time taken, in s, for the satellite mentioned in line 7 in the passage to
complete one orbit around the Earth.
time taken = _____________s
(1)
Page 25 of 52
(c)
Show that at a distance of 12 000 km from the satellite the beam has a width of 1900 km.
(3)
(d)
The satellite is in a polar orbit and passes directly over a stationary receiver at the South
Pole.
Show that the receiver can remain in contact with the satellite for no more than about 20
minutes each orbit.
radius of the Earth = 6400 km
maximum time = ___________________________ minute
(3)
(e)
The same satellite is moved into a higher orbit.
Discuss, with reasons, how this affects the signal strength and contact time for the receiver
at the South Pole.
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(4)
(Total 13 marks)
Page 26 of 52
24
(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)
(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)
Page 27 of 52
(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)
25
Two satellites P and Q, of equal mass, orbit the Earth at radii R and 2R respectively. Which one
of the following statements is correct?
A
P has less kinetic energy and more potential energy than Q.
B
P has less kinetic energy and less potential energy than Q.
C
P has more kinetic energy and less potential energy than Q.
D
P has more kinetic energy and more potential energy than Q.
(Total 1 mark)
Page 28 of 52
Mark schemes
1
2
3
4
D
[1]
D
[1]
A
[1]
(a)
(i)
relationship between them is Ep = mV (allow ΔEp = mΔV) [or V
is energy per unit mass (or per kg)] (1)
1
(ii)
value of Ep is doubled (1)
value of V is unchanged (1)
2
(b)
(i)
use of V =
gives rA =
(1)
= 3.3(2) × 107 (m) (1)
2
(ii)
since V
(1)
(which is ≈ 1.1 × 104 km)
1
(iii)
centripetal acceleration gB =
(1)
[allow use of 1.1 × 107 m from (b)(ii)]
= 3.2 (m s–2) (1)
[alternatively, since gB = (–)
(1)
= 3.2 (m s–2) (1)]
2
Page 29 of 52
(iv)
use of ΔEp = mΔV gives ΔEp = 330 × (–12.0 – (–36.0)) × 106 (1)
(which is 7.9 × 109 J or ≈ 8 GJ)
1
(c)
g is not constant over the distance involved
(or g decreases as height increases
or work done per metre decreases as height increases
or field is radial and/or not uniform) (1)
1
[10]
5
6
7
C
[1]
D
[1]
(a)
direction changing, velocity vector
B1
1
(b)
Newton’s law equation
M1
centripetal force equation
M1
cancel mass of Triton
A1
3
Page 30 of 52
(c)
ω = 2πf or ω = 2 π/T
M1
ω2r3 = constant or ω2 =
M1
or statement of Kepler III for B3
= 5.2(2)
M1
4
[8]
8
(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)
G and M are constants, hence T2
(1)
R3 (1)
3
Page 31 of 52
(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)
period = 24 hours or equals period of Earth’s rotation (1)
remains in fixed position relative to surface of Earth (1)
equatorial orbit (1)
same angular speed as Earth or equatorial surface (1)
max 2
(b)
= mω2r (1)
(i)
T=
(1)
(1)
(gives r = 42.3 × 103 km)
Page 32 of 52
(ii)
ΔV = GM
(1)
= 6.67 × 10–11 × 6 × 1024 ×
= 5.31 × 107 (J kg–1) (1)
ΔEp = mΔV (= 750 × 5.31 × 107) = 3.98 × 1010 J (1)
(allow C.E. for value of ΔV)
[alternatives:
calculation of
or calculation of
(6.25 × 107) or
(4.69 × 1010) or
(9.46 × 106) (1)
(7.10× 109) (1)
calculation of both potential energy values (1)
subtraction of values or use of mΔV with correct answer (1)]
6
[8]
10
11
12
13
A
[1]
D
[1]
C
[1]
A
[1]
Page 33 of 52
14
(a)
ω
=
[or
]
= 1.1 × 10–3 (1.08 × 10–3) (1) [= 6.2 (6.19) × 10–2]
rad s–1 [accept s–1] (1)
[degree s–1]
3
(b)
(i)
or
(1)
gives r3 =
(1)
r = 6.99 × 106 (m) (1)
3
(ii)
F (= mω2r) = 1.1 × 104 × (1.08 × 10–3)2 × 6.99 × 106 (1)
= 9.0 × 104 (8.97 × 104) (N) (1)
[or
(1)
= 9.0 × 104 (8.98 × 104) (N) (1)]
2
[8]
15
(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 34 of 52
(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]
Page 35 of 52
16
17
B
[1]
(a)
work done per unit mass in bringing object from infinity to point
B1
potential at infinity zero by definition
B1
work has been done by the field so potential at all points closer than
infinity negative
B1
3
(b)
use of point on graph allow within ± small square
C1
substitution into V = −
C1
range from 590 – 6.90 × 1024 (kg)
A1
3
(c)
(i)
∆Ep =
C1
addition of radius of Earth to give 7.25 × 106 (m)
C1
1.54 × 1010 (J)
A1
3
Page 36 of 52
(ii)
equates
and
C1
to give ∆EK =
C1
1.25 × 109 J
A1
positive or increase
B1
4
(iii)
(lower altitude so) gpe decreases ke increases
C1
loss of gpe is twice gain in ke
A1
2
[15]
18
19
A
[1]
D
[1]
Page 37 of 52
20
(a)
mass depends only on the amount of matter present owtte
B1
weight is force between body and Earth/depends on g/mg/
gravitational field strength or answers in terms of Newton’s
gravitational law
B1
g (etc) varies at different points on and above the Earth or is
different on different planets etc
B1
3
(b)
(i)
reference is ‘infinity’ where potential is 0
B1
energy has to be put in/work has to be done to move
mass to infinity or a bodies energy/PE decreases as
a body moves from infinity towards the Earth
B1
2
(ii)
need to show Vr to be constant, clear from algebra
or final statement
B1
two sets of data used correctly
B1
all three sets of data used correctly (4.02, 4.025, 4.028)
B1
3
(iii)
energy change per kg = (5.36 – 3.22) × 107 (J)
B1
total change = 963 (960) × 107 J
B1
2
Page 38 of 52
(c)
(i)
GMm/r2 = mv2/r or v = (GM/r)
C1
v2 = 3.2 × 107m2s–2 or v = 5670 ms–1
C1
use of KE = ½ mv2 using their v
C1
7.2 GJ
A1
4
(ii)
KE changes by 4.8 GJ (allow ecf, 12 – their ci)
B1
1
(iii)
total energy (supplied) = (4.8) GJ (cnao)
(allow 5.2 GJ using 10 GJ for change in Ep)
(allow variations due to rounding off if physics
is correct in previous parts)
B1
1
[16]
21
(a)
(i)
(Minimum) Speed (given at the Earth’s surface) that will allow an object to leave /
escape the (Earth’s) gravitational field (with no further energy input)
Not gravity
Condone gravitational pull / attraction
B1
1
Page 39 of 52
(ii)
½ mv2 =
B1
Evidence of correct manipulation
At least one other step before answer
B1
2
(iii)
Substitutes data and obtains M = 7.33 × 1022(kg)
or
Volume = (1.33 × 3.14 × (1.74 × 106)3 or 2.2 × 1019
or ρ =
C1
3300 (kg m-3 )
A1
2
(b)
(Not given all their KE at Earth’s surface) energy continually
added in flight / continuous thrust provided / can use fuel
(continuously)
B1
Less energy needed to achieve orbit than to escape from
Earth’s gravitational field / it is not leaving the gravitational
field
B1
2
[7]
22
(a)
Idea that both astronaut and vehicle are travelling at same (orbital) speed or have the same
(centripetal) acceleration / are in freefall
Not falling at the same speed
B1
No (normal) reaction (between astronaut and vehicle)
B1
2
Page 40 of 52
(b)
(i)
Equates centripetal force with gravitational force using
appropriate formulae
E.g.
=
or mrω2
B1
Correct substitution seen e.g. v2 =
B1
(Radius of) 7.28 × 106 seen or 6.38 × 106 + 0.9 × 106
B1
7396 (m s−1) to at least 4 sf
Or v2 = 5.47 × 107 seen
B1
4
Page 41 of 52
(ii)
ΔPE = 6.67 × 10−11 × 5.97 × 1024 × 1.68 × 104 (1 / (7.28 × 106)
− 1 / (6.78 × 106) )
C1
−6.8 × 1010 J
C1
ΔKE =0.5 × 1.68 × 104 ×(77002−74002) = 3.81 × 1010J
C1
ΔKE − ΔPE = (−) 2.99 × 1010 (J)
A1
OR
Total energy in original orbit shown to be (−)GMm / 2r
or mv2 / 2 − GMm / r
C1
Initial energy
= − 6.67 × 10−11 × 5.97 × 1024 × 1.68 × 104 / (2 × 7.28 × 106) =
4.59 × 1011
C1
Final energy
= − 6.67 × 10−11 × 5.97 × 1024 × 1.68 × 104 / (2 × 6.78 × 106) =
4.93 × 1011
3.4 × 1010(J)
Condone power of 10 error for C marks
A1
4
[10]
23
(a)
Equatorial orbit ✓
Moving west to east ✓
Period 24 hours✓
ANY TWO
2
(b)
T
= 2.5 × 104 s ✓
1
Page 42 of 52
(c)
λ
=0.27 (3)m )✓
θ
= 0.16(1) rad = 92 ° ✓
(linear) width = Dθ = 12000 km 0.16(1) rad ) = 1.9(3) × 103 km ✓
3
(d)
Angle subtended by beam at Earth’s centre
= beam width / Earth’s radius = 1.9(3) × 10 3 / 6400 ) ✓
0.30 rad (or 17°) ✓
Time taken = α / ω = 0.30 / 2.5(4) × 10-4 = 1.18 × 103 s
= 20 mins ✓
Alternative:
Speed of point on surface directly below satellite = ωR
= 2.5(4) × 10-4 × 6400 × 103 )
= 1.63 × 103 m s-1✓
Time taken = width / speed
= 1.93 × 106 m / 1.63 × 103 m s-1 ✓
= 1.18 × 103 s
(accept 1.2 × 103 s or 20 mins) ✓
or
Satellite has to move through angle of 1900 / 6400 radian = 0.29
rad✓
Fraction of one orbit = 0.30 / 2 × 3.14✓
Time = 0.048 × 2.5 × 104 = 1.19 × 103 s✓
Time=
× 2.5 × 104 = 1.18 × 103 s
or
Circumference of Earth = 2π × 6370 ✓
= 40023 km
Width of beam at surface = 1920 km ✓
Time =
×2.48×104
= 1180 s = 19.6 min ✓
3
Page 43 of 52
(e)
Signal would be weaker ✓ (as distance it travels is greater)
Energy spread over wider area/intensity decreases with increase of distance ✓
Signal received for longer (each orbit) ✓
Beam width increases with satellite height/satellite moves at lower angular speed ✓)
4
[13]
24
(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
(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
Page 44 of 52
(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
(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]
25
C
[1]
Page 45 of 52
Examiner reports
1
2
3
4
This question was about satellites. The former required correct algebraic expressions for the
centripetal acceleration and speed of a satellite in circular orbit around a planet. Just over
four-fifths of the responses were correct.
This question where the purpose was to calculate the Earth’s orbital speed, combined circular
motion with gravitation. 62% of the students were successful, whilst incorrect answers were
spread fairly evenly between the three incorrect responses.
This question with a facility of 41%, was also demanding. Here several factors - kinetic energy,
weight, time period and speed - had to be considered for two satellites in different circular orbits.
The three incorrect answers had a fairly even distribution of responses.
Many very good answers were seen in part (a) (i), expressed either fully in words or simply by
quoting Ep = mV. The corresponding equation for an incremental change, ΔEp = mΔV, was also
acceptable but mixed variations on this such as Ep = mΔV (which showed a lack of
understanding) were not. The consequences of doubling m were generally well understood in
part (a) (ii), where most candidates scored highly, but some inevitably thought that Ep would be
unchanged whilst V would double.
Candidates who were not fully conversant with the metric prefixes used with units had great
difficulty in part (b), where it was necessary to know that 1 MJ =106 J, 1 GJ =109 J, and (even) 1
km = 103 m. Direct substitution into V = (–) GM/r (having correctly converted the value of V to J
kg–1) usually gave a successful answer for the radius of orbit A in part (b) (i). A similar approach
1/r
was often adopted in part (b) (ii) to find the radius of orbit B, although the realisation that V
facilitated a quicker solution. Some candidates noticed that VB = 3 VA and guessed that rB = rA/3,
but this was not allowed when there was no physical reasoning to support the calculation.
Part (b) (iii) caused much difficulty, because candidates did not always appreciate that the
centripetal acceleration of a satellite in stable orbit is equal to the local value of g, which is equal
to GM/r2. This value turns out to equal to V/r, which provided an alternative route to the answer.
Many incredible values were seen, some of them greatly exceeding 9.81 m s–2.
Part (c) was generally well understood, with some very good and detailed answers from the
candidates. Alternative answers were accepted: either that g is not constant over such large
distances, or that the field of the Earth is radial rather than uniform.
Page 46 of 52
7
8
(a)
Most candidates were able to correctly answer this part.
(b)
Although the majority of candidates were able to quote either the Newton’s law of
gravitation or centripetal force equation only the better candidates equated these and
showed that the mass of Triton cancelled.
(c)
Only the best candidates were able to show this in a convincing way. A limited number of
candidates gained a little credit for stumbling through one or two appropriate relationships.
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
Two appropriate features of a geo-synchronous orbit were usually given by the candidates in
part (a), but the marks for them were often the last that could be awarded in this question. The
required radius in part (b)(i) came readily to the candidates who correctly equated the
gravitational force on the satellite with mω2r, applied T = 2π/ω, and completed the calculation by
substituting T = 24 hours and the values given in the question. Other candidates commonly
presented a tangled mass of unrelated algebra in part (b)(i), from which the examiners could
rescue nothing worthy of credit.
In part (b)(ii) an incredible proportion of the candidates assumed that it was possible to calculate
the increase in the potential energy by the use of mg Δh, in spite of the fact that the satellite had
be raised vertically through almost 36,000 km. These attempts gained no marks. Other efforts
started promisingly by the use of V = –GM / r, but made the crucial error of using (4.23 × 107 –
6.4 × 106) as r in the denominator. Some credit was available to candidates who made progress
with a partial solution that proceeded along the correct lines, such as evaluating the gravitational
potential at a point in the orbit of the satellite. Confusion between the mass of the Earth and the
mass of the satellite was common when doing this.
Page 47 of 52
10
11
The geo-synchronous satellite in this question did not seriously trouble many of the candidates,
since the facility was 80%. Wrong responses were almost evenly split between the remaining
three distractors, with none attracting more than 8% of the candidates.
This question was about gravitational forces. Application of the inverse square law was
completed successfully by 70% of the candidates in the former question. Candidates had to
appreciate that the condition described would be met when the centripetal force acting on
material is just equal to its weight, so ω2R = GM/R2. Only 48% of them were successful, but the
question discriminated very well.
12
13
14
This question tested students’ understanding of the effect of the descent towards a planet on the
speed and orbit time. 74% of them knew that the speed would increase and the orbit time would
decrease, because v ∝ r−1/2 whilst T ∝ r3/2.
This question required candidates to select an incorrect statement about what would happen to a
comet as it approached the Sun. Distractor C was chosen by 31% of the candidates; this
suggests they thought that the comet would make a line-of-centres approach instead of looping
around the Sun.
This question as a whole was very rewarding for the candidates who were sufficiently familiar
with the principles of gravitation to understand the mathematical conditions for a satellite in stable
orbit, as required in part (b) (i). These candidates made good progress with all parts of the
question, whereas many other candidates were only able to score well on parts (a) and (b) (ii). In
part (a), the correct conversion of the orbital time of the Hubble satellite into seconds followed by
correct use of ω = 2π/T, with a correct unit for angular speed, brought full marks for the majority
of the candidates. Confusion of angular speed ω with linear speed v continues to be a problem,
and giving the unit of ω as m s-1 inevitably caused the loss of one mark.
Page 48 of 52
Part (b) (i) required candidates to appreciate that the radius of the orbit of a satellite can be found
from the orbit equation GMm/r2 = mω2r. The angular speed ω had been determined in part (a),
whilst the values for G and the Earth’s mass M could be taken from the Data and Formulae
Booklet. Because the question had indicated that the Hubble telescope is in orbit close to the
Earth, some candidates assumed that the radius of its orbit would be that of the Earth, 6.37 × 106
m.
Another common unsuccessful response was to attempt to determine the answer using the orbit
relationship T2/r3 = constant, incorrectly treating the surface of the Earth as a satellite orbit and
using T = 24 hours and r = 6.37 × 106 m.
Candidates who used F = mω2r, or F = GMm/r2, had very little difficulty in part (b) (ii), where both
marks were still accessible to those who had worked out wrong values for ω and/or r in the
earlier parts of the question. Attempts at this part using F = mv2/r were often incorrect because of
inability to correctly work out the linear speed, v.
15
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.
16
In this question, equating the centripetal force on a satellite with the gravitational force on it
should lead easily to a correct algebraic expression for the speed. Two thirds of candidates were
successfully able to do this.
Page 49 of 52
17
Most students realised that, in part (a), the potential at infinity is zero but few could elucidate why
the values are negative. Even fewer students mentioned that potential is work done per unit
mass in bringing a small test mass from infinity to the various points. Many students suggested
that ‘gravitational potential is a negative force’ – there appears to be much confusion over why
gravitational potential is negative.
Those students who understood what to do in part (b) usually gained full marks – and this was
the clear majority. Those choosing points at the extremes of the graph often were outside the
accepted tolerance for the mass as a result of making an imprecise estimate of the coordinates
of their chosen point.
Most students made good attempt at part (c) (i). The most common error was to forget to add the
radius of the Earth to height of the satellite’s orbit.
Part (c) (ii) was not well understood, few students were able to calculate the change in kinetic
energy either by calculating the velocities or relating the centripetal force to the gravitational
attraction to obtain Ek=
Again, many forgot to add the radius of the Earth to the satellite’s
altitude. Many students did not make it clear whether the change was an increase or a decrease.
Most realised that there was loss of potential energy and increase in kinetic energy in part (c) (iii).
Of those students correctly obtaining the factor of two in the kinetic energy equation, few went on
to say that the decrease in potential energy was twice the increase in kinetic energy.
18
19
20
This question three quarters of the students were successful when dealing with the algebra
giving the velocity motion of a satellite in stable orbit of radius (R + h). This question had
appeared in a 2002 examination, when the students found it marginally harder and it was slightly
less discriminating.
This question, with a facility of 71%, required the angular speed of a satellite in circular orbit to
be found and appeared to cause little difficulty.
In part (a), most candidates were able to make some reference to weight being mass multiplied
by gravitational field strength although this was often expressed simply as W = mg or, too loosely,
as mass × gravity. Many did not go on to explain why gravitational field strength was not
constant.
Relatively few stated that mass was dependent only on the matter that was contained in the
object.
A few pointed out that, in fact, there could be relativistic increase in mass and these were
rewarded.
Page 50 of 52
A majority of the candidates identified zero potential at infinity in part (b)(i), but explanations of
why this led to negative values closer to the Earth were often unconvincing. Candidates needed
to say more than ‘the field is attractive’.
Many candidates had difficulties with the straightforward exercise in part (b)(ii). Rather than
simply analysing the data to show that Vr at each position produces a constant, many used the
equation V = GM/r. Although candidates were not penalised for a correct approach using this
method as long as they were thorough, the additional arithmetic often led to errors. In this type of
question it is important that candidates give an appropriate reason why the analysis
demonstrates consistency of the data with the law that is proposed. This was often not the case
and responses were frequently a jumble of calculations from which the examiner was,
presumably, required to draw their own conclusion.
Many candidates overcomplicated part (b)(iii) owing to inadequate understanding of the
information that was provided in the table. Some candidates made no use of the table data at all
and used GMm(1/r1 – 1/r2). Although this was against the spirit of the question, candidates were
allowed one of the available marks provided that this was completed successfully but many ran
into problems with the arithmetic. Some used change in PE = mgh.
Part (c)(i) was done very poorly. Although knowing that they had to use ½ mv2, many did not
know that the speed of the satellite in an orbit can be found using GMm/r2 = mv2/r.
Because of their inability to produce an answer to part (c)(i) there were a high proportion of the
candidates who omitted parts (c)(ii) and (c)(iii). Most of those who found an answer to (c)(i), even
if incorrect, were able to score here by realising that the answer was the difference between 12
GJ and their answer to (c)(i).
There were few correct answers to part (c)(iii) because most candidates paid no attention to the
signs of the changes, the KE change being a decrease and the 10 GJ PE change an increase.
21
(a)
(b)
(i)
Well done.
(ii)
Candidates scored 2 or zero. The latter invariably used centripetal force =
gravitational force.
(iii)
Many promising calculation were ruined by failure to cube the radius when finding the
volume.
Most candidates did not realise that escape velocity was not needed because the rocket
was not escaping!
Page 51 of 52
22
(a)
Most candidates mentioned the lack of reaction force but some answers were spoilt by
claiming that there was no resultant force or even no gravity.
(b)
(i)
Each step needed to be clearly shown, starting with the statement that gravitational
force = centripetal force.
There were several cases of the use of 900 km for the radius.
(ii)
24
Common errors were treating potential energy as positive, the use of the wrong radii
and the use of mgh.
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.
25
This question provided poorer discrimination between candidates’ abilities than any other
question in this test. Candidates ought to know that satellites speed up as they move into lower
orbits, and therefore gain kinetic energy if their mass is unchanged. It should also be clear that
satellites lose gravitational potential energy as they move closer to Earth. Therefore it is
surprising that only 55% of the candidates gave the correct answer. The fairly even spread of
responses amongst the other distractors suggests that many candidates were guessing.
Page 52 of 52