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Question 21M.2.HL.TZ1.c.i

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Date May 2021 Marks available [Maximum mark: 3] Reference code 21M.2.HL.TZ1.c.i
Level HL Paper 2 Time zone TZ1
Command term Show that Question number c.i Adapted from N/A
c.i.
[Maximum mark: 3]
21M.2.HL.TZ1.c.i

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                                      =8.0×1024 kg
Mass of star                                          =3.2×1030 kg
Distance from the star to the planet R  =4.4×1010 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.

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

[3]

Markscheme

Vplanet = «−»(6.67×10-11)(8×1024)9.1×106=«−» 5.9 × 107 «J kg−1» ✓

Vstar = «−»(6.67×10-11)(3.2×1030)4.4×1010=«−» 4.9 × 109 «J kg−1» ✓

Vplanet + Vstar = «−» 4.9 «09» × 109 «J kg−1» ✓


Must see substitutions and not just equations.

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