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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 Suggest 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  distance of Mars from the Sun 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 × 10–13 s2 m–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.

« m v 2 r = G M m r 2 hence»  v = G M R . Also  v = 2 π R T

OR

m ω 2 r = G M m r 2 hence ω 2 = G M R 3

 

uses either of the above to get  T 2 = 4 π 2 G M R 3

OR

uses  k = 4 π 2 G M  ✔

 

k = 9.2 × 10−13 / 9.3 × 10−13

 

 

Unit not required

 

b.i.

R 3 = T 2 k = ( 8.9 × 10 4 ) 2 9.25 × 10 13   R = 2.04 × 107 «m» ✔

 

v = « ω r = 2 π × 2.04 × 10 7 89000 = » 1.4 × 103 «m s–1»

OR

v = « G M R = 6.67 × 10 11 × 6.4 × 10 23 2.04 × 10 7 = » 1.4 × 103 «m s–1» ✔

b.ii.

use of  I 1 r 2  «1.36 × 103 × 1 1.5 2 » ✔

604 «W m–2» ✔

c.i.

use of 600 4 for mean intensity ✔

temperature/K = « 600 4 × 5.67 × 10 8 4 = » 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

OR

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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Core » Topic 6: Circular motion and gravitation
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