Exam-style Questions: Circular Motion

 

  1. This question is about the planet Jupiter and one of the moons that orbits it, called lo.

    Copyright S-cool

     

    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,Copyright S-cool, where m is the mass of lo. This force is the gravitational attraction between lo and Jupiter.

    i) Show thatCopyright S-coolwhere 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)

  2. 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

 

Answer outline and marking scheme for question:

 

  1. a) Time = 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)

  2. 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)