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A-LevelPhysicsGravitational fields9702/43 · May/June 2019 · Q18 Marks

Two point masses are isolated in space and are separated by a distance x. State an expression relating the gravitational force F between the two masses to the magnitudes M and m of the masses. State the name of any other symbol used.

  1. (a)State an expression relating the gravitational force F between the two masses to the magnitudes M and m of the masses. State the name of any other symbol used.[1]
  2. (b)A spacecraft is to be put into a circular orbit about a spherical planet. The planet may be considered to be isolated in space. The mass of the planet, assumed to be concentrated at its centre, is 7.5 x 1023 kg. The radius of the planet is 3.4 × 106 m.
  3. (b)(i)The spacecraft is to orbit the planet at a height of 2.4 × 105m above the surface of the planet. At this altitude, there is no atmosphere. Show that the speed of the spacecraft in its orbit is 3.7 × 103ms-1.[2]
  4. (b)(ii)One possible path of the spacecraft as it approaches the planet is shown in Fig. 1.1. [Figure 1.1] The spacecraft enters the orbit at point A with speed 3.7 × 103ms-1. At point B, a distance of 5.00 × 107m from the centre of the planet, the spacecraft has a speed of 4.1 × 103ms¯¹. The mass of the spacecraft is 650 kg. For the spacecraft moving from point B to point A, show that the change in gravitational potential energy of the spacecraft is 8.3 x 10º J.[3]
  5. (c)By considering changes in gravitational potential energy and in kinetic energy of the spacecraft, determine whether the total energy of the spacecraft increases or decreases in moving from point B to point A. A numerical answer is not required.[2]

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Mark scheme

Marking points from the official mark scheme (May/June 2019). The mark or mark type is shown in brackets.

  1. [B1] (F =) GMm / x², where G is the (universal) gravitational constant
  2. [C1] GMm / x² = mv²/x or v² = GM/x
  3. [A1] v2 = (6.67 x 10-11 × 7.5 × 1023) / (3.4 × 106 + 240 × 103) so v = 3.7 × 103 m s-1
  4. [C1] potential energy = (-)GMm / x
  5. [M1] EA = (-)(6.67 × 10–11 × 7.5 × 1023 × 650) / (3.64 × 10º) or Ев = (-)(6.67 × 10–11 × 7.5 × 1023 × 650) / (5.00 × 107)
  6. [A1] correct substitution and subtraction EB – EA shown, leading to ∆Ep = 8.3 × 10º J
  7. [(C1) $ = (−)GM/x and potential energy = mp]
  8. [(M1) Δφ = (6.67 × 10–11 × 7.5 × 1023) × [(1 / (3.64 × 10º)) – (1 / (5.00 × 107))] (= 1.27 × 107 J kg-1)]
  9. [(A1) ΔΕΡ = 1.27 × 107 × 650 = 8.3 × 10º J]
  10. [B1] kinetic energy or potential energy decreases
  11. [B1] kinetic energy and potential energy decrease so total energy decreases

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About this question

Topic
Gravitational fields, A-Level Physics (syllabus 9702). Structured question, 8 marks.
Source
A-Level Physics, May/June 2019 Paper 4 Variant 3, question 1.

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