IBDP Physics 6.2 – Newton’s law of gravitation: IB Style Question Bank HL Paper 2

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.

a.i.

State what is meant by gravitational field strength.[1]

a.ii.

Show that V = –g(R + h).[2]

a.iii.

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]

b.

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]

c.

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

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]

d.

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]

 
Answer/Explanation

Markscheme

a.i.

the «gravitational» force per unit mass exerted on a point/small/test mass[1 mark]

a.ii.

at height h potential is V = –\(\frac{{GM}}{{(R + h)}}\)

field is g = \(\frac{{GM}}{{{{(R + h)}^2}}}\)

«dividing gives answer»

Do not allow an answer that starts with g = –\(\frac{{\Delta V}}{{\Delta r}}\) and then cancels the deltas and substitutes R + h[2 marks]

a.iii.

correct shape and sign

non-zero negative vertical intercept

[2 marks]

b.

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]

c.

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

«\(\frac{1}{2}\)mv2 + mV = 0 ⇒» v = \(\sqrt { – 2V} \)

v = «\(\sqrt {2 \times 6.0 \times {{10}^7}} \) =» 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]

d.

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 = –\(\frac{{GM}}{{(R + h)}}\)»

force on planet «\(\frac{{GMm}}{{{{(R + h)}^2}}}\)» = 1.4 × 1013 «N»

MP2 must be explicit[2 marks]

Question

The gravitational potential due to the Sun at its surface is –1.9 x 1011 J kg–1. The following data are available.

Mass of Earth= 6.0 x 1024 kg
Distance from Earth to Sun= 1.5 x 1011 m
Radius of Sun= 7.0 x 108 m

a.

Outline why the gravitational potential is negative.[2]

b.i.

The gravitational potential due to the Sun at a distance r from its centre is VS. Show that

rVS = constant.[1]

b.ii.

Calculate the gravitational potential energy of the Earth in its orbit around the Sun. Give your answer to an appropriate number of significant figures.[2]

b.iii.

Calculate the total energy of the Earth in its orbit.[2]

b.iv.

An asteroid strikes the Earth and causes the orbital speed of the Earth to suddenly decrease. Suggest the ways in which the orbit of the Earth will change.[2]

c.

Outline, in terms of the force acting on it, why the Earth remains in a circular orbit around the Sun.[2]

 
Answer/Explanation

Markscheme

a.

potential is defined to be zero at infinity

so a positive amount of work needs to be supplied for a mass to reach infinity

b.i.

VS = \( – \frac{{GM}}{r}\) so r x VS «= –GM» = constant because G and M are constants

b.ii.

GM = 1.33 x 1020 «J m kg–1»

GPE at Earth orbit «= –\(\frac{{1.33 \times {{10}^{20}} \times 6.0 \times {{10}^{24}}}}{{1.5 \times {{10}^{11}}}}\)» = «–» 5.3 x 1033 «J»

Award [1 max] unless answer is to 2 sf.

Ignore addition of Sun radius to radius of Earth orbit.

b.iii.

ALTERNATIVE 1
work leading to statement that kinetic energy \(\frac{{GMm}}{{2r}}\), AND kinetic energy evaluated to be «+» 2.7 x 1033 «J»

energy «= PE + KE = answer to (b)(ii) + 2.7 x 1033» = «–» 2.7 x 1033 «J»

ALTERNATIVE 2
statement that kinetic energy is \( = – \frac{1}{2}\) gravitational potential energy in orbit

so energy «\( = \frac{{{\text{answer to (b)(ii)}}}}{2}\)» = «–» 2.7 x 1033 «J»

Various approaches possible.

b.iv.

«KE will initially decrease so» total energy decreases
OR
«KE will initially decrease so» total energy becomes more negative

Earth moves closer to Sun

new orbit with greater speed «but lower total energy»

changes ellipticity of orbit

c.

centripetal force is required

and is provided by gravitational force between Earth and Sun

Award [1 max] for statement that there is a “centripetal force of gravity” without further qualification.

Question

The diagram shows a planet near two stars of equal mass M.

Each star has mass M=2.0×1030kg. Their centres are separated by a distance of 6.8×1011m. The planet is at a distance of 6.0×1011m from each star.

a.

On the diagram above, draw two arrows to show the gravitational field strength at the position of the planet due to each of the stars.[2]

b.

Calculate the magnitude and state the direction of the resultant gravitational field strength at the position of the planet.[3]

 
Answer/Explanation

Markscheme

a.

two arrows each along the line connecting the planet to its star AND directed towards each star

arrow lines straight and of equal length

Do not allow kinked, fuzzy curved lines.

b.

\(g = \ll \frac{{GM}}{{{r^2}}} = \frac{{6.67 \times {{10}^{ – 11}} \times 2.0 \times {{10}^{30}}}}{{{{\left( {6.0 \times {{10}^{11}}} \right)}^2}}} \gg \) OR 3.7×10-4Nkg-1

\({g_{{\rm{net}}}} = \ll 2g\cos \theta = 2 \times 3.7 \times {10^{ – 4}} \times \frac{{\sqrt {{{6.0}^2} – {{3.4}^2}} }}{{6.0}} = \gg 6.1 \times {10^{ – 4}}{\rm{Nk}}{{\rm{g}}^{ – 1}}\)

directed vertically down «page» OR towards midpoint between two stars OR south

Allow rounding errors.