Exam-style Questions: Circular Motion
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This question is about the planet Jupiter and one of the moons that orbits it, called lo.

lo orbits Jupiter at a speed, v of 1.7 x 104 m s-1 at an orbital radius, r, of 4.2 x 108m.
a) Show that lo takes approximately 43 hours to orbit Jupiter once.
(2 Marks)
b) lo is held in its orbit by a centripetal force,
, where m is the mass of lo. This force is the gravitational attraction between lo and Jupiter.i) Show that
where M is the mass of Jupiter.ii) Show that the mass of Jupiter is about 2.0 x 1027 kg.
(4 Marks)
c) Show that the gravitational potential at the top of Jupiter's atmosphere, 7.1 x 107 m from the centre of the planet, is about -2 x 109 Jkg-1.
Assume that Jupiter is a sphere.
(2 Marks)
d) In July 1994, comet Shoemaker-Levy 9 crashed into Jupiter causing dramatic heating of the planet's atmosphere. During the approach to the planet, the comet broke up. One piece that struck the planet had a mass of about 4 x 1012 kg.
This fragment crossed the orbit of lo heading directly towards Jupiter with a velocity of 10 km s-1.
i) Show that kinetic energy of the fragment at that moment is 2 x 1020 J.
(1 Mark)
ii) Explain why the fragment will enter the atmosphere of Jupiter with a velocity greater than 10km s-1.
(2 Marks)
(Marks available: 11)
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a) A cyclist goes round a circular track, of radius 10 m, at a constant speed of 8.0 m s-1.
What is the acceleration of the cyclist and what is its direction?
(3 Marks)
b) What is the resultant force on the cycle and the rider if together they have a mass of 90 kg?
(2 Marks)
(Marks available: 5)
Answer outline and marking scheme for question:
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a) Time = 2ð r/1.7 x 104
= 155230s
= 43.1 hours
(2 Marks)
b) i) -GMm/r2
= -mv2/r
So: Gm/r2 = v2/r
Therefore M = v2r/G
ii) M = (1.7 x 104 )2 x 4.2 x 108/ 6.67 x 10-11
= 1.82 x 1027 kg
(4 Marks)
c) Vg = -GM/r = -6.67 x 10-11 x 1.9 x 1027/7.1 x 107
= - 1.79 x 109 K kg-1
(2 Marks)
d) i) ½ mv2 = ½ x 4 x 1012 x 100002 = 2 x 1020 J
(1 Mark)
ii) The fragment will have gained kinetic energy (1 Mark) as it lost gravitational potential energy during the approach to the planet (1 Mark).
(Or force argument: Attracted by gravity (1 Mark) causes it to accelerate (1 Mark)
(2 Marks)
(Marks available: 11)
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a)
a = v2 = 82 = 6.4 m s-2 r 10 Direction: towards the centre of the track.
(3 Marks)
b) F = ma (1 Mark) = 90 x 6.4 = 576 N (1 Mark)
(2 Marks)
(Marks available: 5)