A-LevelPhysicsGravitational fields9702/42 · Oct/Nov 2020 · Q19 Marks

The Earth may be considered to be a uniform sphere of radius 6.4 × 10⁶m with its mass of 6.0 x 10²⁴ kg concentrated at its centre. A satellite of mass 2.4 x 10³ kg is launched from the Equator. It is placed in an equatorial orbit at a height of 5.6 × 10⁶ m above the Earth's surface.

(a)Define gravitational potential at a point.[2]
(b)(i)Calculate the change ∆Ep in gravitational potential energy of the satellite for its movement from the surface of the Earth to its position in the equatorial orbit.[3]
(b)(ii)Determine the speed of the satellite when in orbit.[3]
(c)Before the satellite in (b) is launched, its speed at the Equator due to the Earth's rotation is 470ms⁻¹. Suggest why the energy required to launch the satellite depends on whether the satellite, in its orbit, is travelling from west to east or from east to west.[1]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2020). The mark or mark type is shown in brackets.

[B1] work done per unit mass
[B1] (work done) moving mass from infinity (to the point)
[C1] gravitational potential energy = (−)GMm / r
[C1] ΔEP = 6.67 × 10⁻¹¹ × 6.0 × 10²⁴ × 2.4 × 10³ × [(6.4 × 10⁶)⁻¹ – (1.2 × 10⁷)⁻¹] OR 6.67 × 10⁻¹¹ × 6.0 × 10²⁴ × [(6.4 × 10⁶)⁻¹ – (1.2 × 10⁷)⁻¹] OR ΔEP = mΔφ
[A1] ΔEP = 7.0 × 10¹⁰ J
[C1] GMm/r² = mv² / r
[C1] v² = GM/r = (6.67 x 10⁻¹¹ × 6.0 × 10²⁴) / (1.2 × 10⁷)
[A1] v = 5800 m s⁻¹
[B1] any one point from: smaller gain in energy required if orbit is west to east / smaller change in velocity if orbit is west to east / smaller gain in energy if orbit is in same direction as Earth's rotation / smaller change in velocity if orbit is in same direction as Earth's rotation / satellite already moving west to east at launch / Earth's rotation is from west to east

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