1. Structured Questions
16. A satellite of mass m is launched into a circular orbit about the Earth of mass ME, separated by a
distance r, as shown in Fig. 16.1. Assuming the Earth to be at rest, the total energy of the Earth-
satellite system is the sum of the kinetic energy and the gravitational potential energy of the satellite.
satellite
v
m
r
Earth, ME
Fig. 16.1
(a) State what is meant by the gravitational potential energy of the satellite. [1]
(b) Applying Newton‟s 2nd Law of Motion, show that the kinetic energy of the system can be
GME m
represented by the expression . [2]
2r
GME m
(c) Hence, show that the total energy of the system can be represented by the expression  .
2r
[2]
(d) Explain why the total energy of the Earth-satellite system is negative. [2]
17. (a) Define the following terms:
(i) gravitational field strength [1]
(ii) gravitational potential [1]
(b) The Earth may be considered as an isolated sphere with a radius of 6.4 x 103 km and a mass of
6.0 x 1024 kg which is concentrated at its centre. An object, projected vertically from the surface
of the Earth, reaches a maximum height of 1.3 x 104 km.
Neglecting the effects of air resistance, determine
(i) the change in the gravitational potential of the object. [3]
(ii) the initial projection speed of the object from the Earth‟s surface. Ignore the rotation of the
Earth. [2]
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2. 18. (a) (i) State the relationship between gravitational field strength g and gravitational potential .
[1]
GM
(ii) The equation for the gravitational potential in the field of a point mass is    .
r
Explain the negative sign in the equation. [2]
(b) (i) Derive an expression for the gravitational field strength g at the surface of the Earth.
Express your answer in terms of the universal gravitational constant G, the mass of the
Earth ME and the radius of the Earth RE. State two assumptions you make. [4]
(ii) A common equation for the gravitational potential energy of an object above the surface of
the Earth is U  mgh . Suggest a reason why this equation should not be applied when h
is more than several kilometres. [1]
(c) A space vehicle of mass 500 kg moves in a circular orbit at a height of 1000 km above the
surface of the Earth. Given that the mass of Earth is 6.0 × 1024 kg and radius of Earth is 6400
km, determine
(i) gravitational potential energy U of the space vehicle [2]
(ii) linear speed v of the space vehicle [3]
(iii) period T of the orbit of the space vehicle [2]
(iv) The engine of the space vehicle provides energy for it to reach an infinite distance from
the Earth. Determine the minimum energy delivered by the engine. [3]
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3. 19. (a) Fig. 19.1 shows how the gravitational potential VG in the Earth‟s gravitational field varies with
distance r from the Earth‟s centre for regions close to the orbit of a geostationary satellite. The
satellite is launched with the help of a rocket
Fig. 19.1
(i) the work required to lift the rocket and satellite of total mass 20 000 kg from r = 4.0  107
m to 4.4  107 m. [2]
(ii) the velocity of escape from a satellite orbit at r = 4.4  107 m. [2]
(b) The lowest Earth-orbiting satellites have an orbital period of about 90 minutes.
(i) State with a reason whether the lowest Earth-orbiting satellite is a geostationary satellite.
[2]
(ii) Given the mass of Earth to be 6.0  1024 kg, show that the radius at which these lowest
satellites orbit the Earth is about 6.7  106 m. [2]
(iii) Disused satellites contribute to “space junk” left orbiting the Earth. By comparing the
motion of a 1000 kg lowest-Earth orbiting satellite with the energy that explosive TNT can
yield, suggest why “space junk” presents a significant risk to future space missions.
(One tonne of explosive TNT yields 4.1  109 J.) [2]
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4. 20. This question is about a space shuttle used to take tourists to the 'edge of space' 100 km above the
surface of the Earth.
Fig. 20.1 shows the space shuttle carrying three people taken to a height of about 10 km above the
Earth's surface on the back of a carrier craft. At this height where the air pressure is only 25% of that
at the surface of the Earth, the space shuttle is released from the carrier craft
Fig. 20.1
Fig. 20.2 shows the space shuttle starting to ignite its fuel and rising to a height of 100 km above the
Earth's surface.
Fig. 20.2
(a) Suggest and explain an advantage of releasing the space shuttle at the height at which air
pressure is significantly reduced. [2]
(b) Fig. 20.3 shows the variation with the height h near the Earth‟s surface of the gravitational
potential Vg.
h / km
Vg / MJ kg-1
Fig. 20.3
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5. (i) State the feature of the graph that shows that the gravitational field strength is
approximately uniform over the range of the flight. [1]
(ii) Hence show that the value of gravitational field strength, g, over this height range is
9.7 N kg-1. [2]
(c) Using values from the graph, or otherwise, show that the energy required by the space shuttle
of mass 3800 kg to rise from 10 km to 100 km above the Earth is about 3.3 x 109 J. [2]
(d) At one instant, the engines of the space shuttle exert a vertical (radial) thrust of 74 kN. Calculate
the acceleration on the shuttle at this instant. You may neglect the effect of resistive forces in
your calculation. [2]
(e) (i) the engines turned off, the space shuttle travels vertically (radially) away from the Earth in
very thin atmosphere. Explain why the velocity of the space shuttle decreases even
though the air resistance is negligible. [2]
(ii) During this period when the space shuttle travels through a region of low atmospheric
pressure without any thrust acting on it, the people in the cabin of the space shuttle float
and experience „weightlessness'. However when the space shuttle descends nearer the
Earth‟s surface, the atmospheric pressure increases and the experience of
weightlessness by the people in the cabin disappears. Explain the changing weight
experienced by the people in the cabin. [4]
(f) If at the height of 100 km above the Earth‟s surface the space shuttle is made to orbit round the
Earth instead, calculate the angular speed of the space shuttle, given that the radius of the
Earth is 6.4 x 106 m. [3]
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