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Date	November 2018	Marks available	2	Reference code	18N.2.HL.TZ0.8
Level	Higher level	Paper	Paper 2	Time zone	0 - no time zone
Command term	Outline	Question number	8	Adapted from	N/A

Question

There is a proposal to place a satellite in orbit around planet Mars.

The satellite is to have an orbital time T equal to the length of a day on Mars. It can be shown that

T² = kR³

where R is the orbital radius of the satellite and k is a constant.

The ratio $\frac{distance of Mars from the Sun}{distance of Earth from the Sun}$ $\frac{{{\text{distance of Mars from the Sun}}}}{{{\text{distance of Earth from the Sun}}}}$ = 1.5.

Outline what is meant by gravitational field strength at a point.

[2]

a.i.

Newton’s law of gravitation applies to point masses. Suggest why the law can be applied to a satellite orbiting Mars.

[2]

a.ii.

Mars has a mass of 6.4 × 10²³ kg. Show that, for Mars, k is about 9 × 10^–13s²m^–3.

[3]

b.i.

The time taken for Mars to revolve on its axis is 8.9 × 10⁴ s. Calculate, in m s^–1, the orbital speed of the satellite.

[2]

b.ii.

Show that the intensity of solar radiation at the orbit of Mars is about 600 W m^–2.

[2]

c.i.

Determine, in K, the mean surface temperature of Mars. Assume that Mars acts as a black body.

[2]

c.ii.

The atmosphere of Mars is composed mainly of carbon dioxide and has a pressure less than 1 % of that on the Earth. Outline why the mean temperature of Earth is strongly affected by gases in its atmosphere but that of Mars is not.

[3]

c.iii.

Markscheme

force per unit mass ✔

acting on a small/test/point mass «placed at the point in the field» ✔

a.i.

Mars is spherical/a sphere «and of uniform density so behaves as a point mass» ✔

satellite has a much smaller mass/diameter/size than Mars «so approximates to a point mass» ✔

a.ii.

« $\frac{m v^{2}}{r} = \frac{G M m}{r^{2}}$ $\frac{{m{v^2}}}{r} = \frac{{GMm}}{{{r^2}}}$ hence» $v = \sqrt{\frac{G M}{R}}$ $v = \sqrt {\frac{{GM}}{R}}$ . Also $v = \frac{2 π R}{T}$ $v = \frac{{2\pi R}}{T}$

$m ω^{2} r = \frac{G M m}{r^{2}}$ $m{\omega ^2}r = \frac{{GMm}}{{{r^2}}}$ hence $ω^{2} = \frac{G M}{R^{3}}$ ${\omega ^2} = \frac{{GM}}{{{R^3}}}$ ✔

uses either of the above to get $T^{2} = \frac{4 π^{2}}{G M} R^{3}$ ${T^2} = \frac{{4{\pi ^2}}}{{GM}}{R^3}$

uses $k = \frac{4 π^{2}}{G M}$ $k = \frac{{4{\pi ^2}}}{{GM}}$ ✔

k = 9.2 × 10⁻¹³ / 9.3 × 10⁻¹³

Unit not required

b.i.

$R^{3} = \frac{T^{2}}{k} = \frac{{(8.9 \times 10^{4})}^{2}}{9.25 \times 10^{- 13}}$ ${R^3} = \frac{{{T^2}}}{k} = \frac{{{{\left( {8.9 \times {{10}^4}} \right)}^2}}}{{9.25 \times {{10}^{ - 13}}}}$ R = 2.04 × 10⁷ «m» ✔

v = « $ω r = \frac{2 π \times 2.04 \times 10^{7}}{89000} =$ $\omega r = \frac{{2\pi \times 2.04 \times {{10}^7}}}{{89000}} =$ » 1.4 × 10³ «m s^–1»

v = « $\sqrt{\frac{G M}{R}} = \sqrt{\frac{6.67 \times 10^{- 11} \times 6.4 \times 10^{23}}{2.04 \times 10^{7}}} =$ $\sqrt {\frac{{GM}}{R}} = \sqrt {\frac{{6.67 \times {{10}^{ - 11}} \times 6.4 \times {{10}^{23}}}}{{2.04 \times {{10}^7}}}} =$ » 1.4 × 10³ «m s^–1» ✔

b.ii.

use of $I \propto \frac{1}{r^{2}}$ $I \propto \frac{1}{{{r^2}}}$ «1.36 × 10³ × $\frac{1}{{1.5}^{2}}$ $\frac{1}{{{{1.5}^2}}}$ » ✔

604 «W m^–2» ✔

c.i.

use of $\frac{600}{4}$ $\frac{{600}}{4}$ for mean intensity ✔

temperature/K = « $\sqrt[4]{\frac{600}{4 \times 5.67 \times 10^{- 8}}} =$ $\sqrt[4]{{\frac{{600}}{{4 \times 5.67 \times {{10}^{ - 8}}}}}} =$ » 230 ✔

c.ii.

reference to greenhouse gas/effect ✔

recognize the link between molecular density/concentration and pressure ✔

low pressure means too few molecules to produce a significant heating effect

low pressure means too little radiation re-radiated back to Mars ✔

The greenhouse effect can be described, it doesn’t have to be named

c.iii.

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

c.iii.

Syllabus sections

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

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