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Date	May 2021	Marks available	3	Reference code	21M.2.HL.TZ1.2
Level	Higher level	Paper	Paper 2	Time zone	1
Command term	Show that	Question number	2	Adapted from	N/A

Question

A planet is in a circular orbit around a star. The speed of the planet is constant. The following data are given:

Mass of planet    $= 8.0 \times 10^{24}$ kg
Mass of star    $= 3.2 \times 10^{30}$ kg
Distance from the star to the planet R   $= 4.4 \times 10^{10}$ m.

A spacecraft is to be launched from the surface of the planet to escape from the star system. The radius of the planet is 9.1 × 10³ km.

Explain why a centripetal force is needed for the planet to be in a circular orbit.

[2]

Calculate the value of the centripetal force.

[1]

Show that the gravitational potential due to the planet and the star at the surface of the planet is about −5 × 10⁹J kg⁻¹.

[3]

c.i.

Estimate the escape speed of the spacecraft from the planet–star system.

[2]

c.ii.

Markscheme

«circular motion» involves a changing velocity ✓

«Tangential velocity» is «always» perpendicular to centripetal force/acceleration ✓

there must be a force/acceleration towards centre/star ✓

without a centripetal force the planet will move in a straight line ✓

$F = \frac{(6.67 \times 10^{- 11}) (8 \times 10^{24}) (3.2 \times 10^{30})}{(4.4 \times 10^{10})^{2}} = 8.8 \times 10^{23}$ «N» ✓

V_planet = «−» $\frac{(6.67 \times 10^{- 11}) (8 \times 10^{24})}{9.1 \times 10^{6}} =$ «−» 5.9 × 10⁷«J kg⁻¹» ✓

V_star = «−» $\frac{(6.67 \times 10^{- 11}) (3.2 \times 10^{30})}{4.4 \times 10^{10}} =$ «−» 4.9 × 10⁹«J kg⁻¹» ✓

V_planet + V_star= «−» 4.9 «09» × 10⁹«J kg⁻¹» ✓

Must see substitutions and not just equations.

c.i.

use of v_esc = $\sqrt{2 V}$ ✓

v = 9.91 × 10⁴ «m s⁻¹» ✓

c.ii.

Examiners report

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c.i.

[N/A]

c.ii.

Syllabus sections

Additional higher level (AHL) » Topic 10: Fields » 10.1 – Describing fields

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