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Date May 2021 Marks available 1 Reference code 21M.2.HL.TZ1.2
Level Higher level Paper Paper 2 Time zone 1
Command term Calculate 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                                      =8.0×1024kg
Mass of star                                          =3.2×1030kg
Distance from the star to the planet R  =4.4×1010m.

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]
a.

Calculate the value of the centripetal force.

[1]
b.

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]
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

a.

F=(6.67×10-11)(8×1024)(3.2×1030)(4.4×1010)2=8.8×1023 «N» 

b.

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

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

Vplanet + Vstar = «−» 4.9 «09» × 10«J kg−1» 


Must see substitutions and not just equations.

c.i.

use of vesc = 2V 

v = 9.91 × 104 «m s−1» 

 

 

c.ii.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.i.
[N/A]
c.ii.

Syllabus sections

Core » Topic 6: Circular motion and gravitation » 6.1 – Circular motion
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Core » Topic 6: Circular motion and gravitation
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