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Date	November 2016	Marks available	2	Reference code	16N.2.SL.TZ0.6
Level	Standard level	Paper	Paper 2	Time zone	0 - no time zone
Command term	Define and State	Question number	6	Adapted from	N/A

Question

(i) Define gravitational field strength.

(ii) State the SI unit for gravitational field strength.

[2]

A planet orbits the Sun in a circular orbit with orbital period T and orbital radius R. The mass of the Sun is M.

(i) Show that $T = \sqrt {\frac{{4{\pi ^2}{R^3}}}{{GM}}}$ .

(ii) The Earth’s orbit around the Sun is almost circular with radius 1.5×10¹¹ m. Estimate the mass of the Sun.

[4]

Markscheme

(i) «gravitational» force per unit mass on a «small or test» mass

(ii) N kg^–1

Award mark if N kg^-1 is seen, treating any further work as neutral.
Do not accept bald m s^–2

i
clear evidence that v in ${v^2} = \frac{{4{\pi ^2}{R^2}}}{{{T^2}}}$ is equated to orbital speed $\sqrt {\frac{{GM}}{R}}$
OR
clear evidence that centripetal force is equated to gravitational force
OR
clear evidence that a in $a = \frac{{{v^2}}}{R}$ etc is equated to g in $g = \frac{{GM}}{{{R^2}}}$ with consistent use of symbols
Minimum is a statement that $\sqrt {\frac{{GM}}{R}}$ is the orbital speed which is then used in $v = \frac{{2\pi R}}{T}$
Minimum is F_c = F_g ignore any signs.
Minimum is g = a.

substitutes and re-arranges to obtain result
Allow any legitimate method not identified here.
Do not allow spurious methods involving equations of shm etc

$\ll T = \sqrt {\frac{{4{\pi ^2}R}}{{\left( {\frac{{GM}}{{{R^2}}}} \right)}}} = \sqrt {\frac{{4{\pi ^2}{R^3}}}{{GM}}} \gg$

ii
«T = 365 × 24 × 60 × 60 = 3.15 × 10⁷s»

$M = \, \ll \frac{{4{\pi ^2}{R^3}}}{{G{T^2}}} = \gg \,\, = \frac{{4 \times {{3.14}^2} \times {{\left( {1.5 \times {{10}^{11}}} \right)}^3}}}{{6.67 \times {{10}^{ - 11}} \times {{\left( {3.15 \times {{10}^7}} \right)}^2}}}$
2×10³⁰«kg»

Allow use of 3.16 x 10⁷ s for year length (quoted elsewhere in paper).
Condone error in power of ten in MP1.
Award [1 max] if incorrect time used (24 h is sometimes seen, leading to 2.66 x 10³⁵ kg).
Units are not required, but if not given assume kg and mark POT accordingly if power wrong.
Award [2] for a bald correct answer.
No sf penalty here.

Examiners report

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Syllabus sections

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

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