User interface language: English | Español

Date May 2018 Marks available 1 Reference code 18M.2.HL.TZ2.6
Level Higher level Paper Paper 2 Time zone 2
Command term State Question number 6 Adapted from N/A

Question

A planet has radius R. At a distance h above the surface of the planet the gravitational field strength is g and the gravitational potential is V.

State what is meant by gravitational field strength.

[1]
a.i.

Show that V = –g(R + h).

[2]
a.ii.

Draw a graph, on the axes, to show the variation of the gravitational potential V of the planet with height h above the surface of the planet.

[2]
a.iii.

A planet has a radius of 3.1 × 106 m. At a point P a distance 2.4 × 107 m above the surface of the planet the gravitational field strength is 2.2 N kg–1. Calculate the gravitational potential at point P, include an appropriate unit for your answer.

[1]
b.

The diagram shows the path of an asteroid as it moves past the planet.

                                                                    M18/4/PHYSI/HP2/ENG/TZ2/06.c

When the asteroid was far away from the planet it had negligible speed. Estimate the speed of the asteroid at point P as defined in (b).

[3]
c.

The mass of the asteroid is 6.2 × 1012 kg. Calculate the gravitational force experienced by the planet when the asteroid is at point P.

[2]
d.

Markscheme

the «gravitational» force per unit mass exerted on a point/small/test mass

[1 mark]

a.i.

at height h potential is V = – G M ( R + h )

field is g G M ( R + h ) 2

«dividing gives answer»

 

Do not allow an answer that starts with g = – Δ V Δ r and then cancels the deltas and substitutes R + h

[2 marks]

a.ii.

correct shape and sign

non-zero negative vertical intercept

 

M18/4/PHYSI/HP2/ENG/TZ2/06.a.iii/M

[2 marks]

a.iii.

V«–2.2 × (3.1 × 106 + 2.4 × 107) =» «» 6.0 × 107 J kg–1

 

Unit is essential

Allow eg MJ kg1 if power of 10 is correct

Allow other correct SI units eg m2s2, N m kg1

[1 mark]

b.

total energy at P = 0 / KE gained = GPE lost

« 1 2 mv2mV = 0 ⇒» v 2 V

v« 2 × 6.0 × 10 7 =» 1.1 × 104 «ms–1»

 

 

Award [3] for a bald correct answer

Ignore negative sign errors in the workings

Allow ECF from 6(b)

[3 marks]

c.

ALTERNATIVE 1

force on asteroid is «6.2 × 1012 × 2.2 =» 1.4 × 1013 «N»

«by Newton’s third law» this is also the force on the planet

ALTERNATIVE 2

mass of planet = 2.4 x 1025 «kg» «from V = – G M ( R + h ) »

force on planet « G M m ( R + h ) 2 » = 1.4 × 1013 «N»

 

MP2 must be explicit

[2 marks]

d.

Examiners report

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

Syllabus sections

Core » Topic 6: Circular motion and gravitation » 6.2 – Newton’s law of gravitation
Show 36 related questions
Core » Topic 6: Circular motion and gravitation
Core

View options