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 distance of Mars from the Sundistance of Earth from the Sundistance of Mars from the Sundistance of Earth from the Sun = 1.5.
Outline what is meant by gravitational field strength at a point.
Newton’s law of gravitation applies to point masses. Suggest why the law can be applied to a satellite orbiting Mars.
Mars has a mass of 6.4 × 1023 kg. Show that, for Mars, k is about 9 × 10–13 s2 m–3.
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.
Show that the intensity of solar radiation at the orbit of Mars is about 600 W m–2.
Determine, in K, the mean surface temperature of Mars. Assume that Mars acts as a black body.
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.
Markscheme
force per unit mass ✔
acting on a small/test/point mass «placed at the point in the field» ✔
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» ✔
«mv2r=GMmr2mv2r=GMmr2 hence» v=√GMRv=√GMR. Also v=2πRTv=2πRT
OR
mω2r=GMmr2mω2r=GMmr2 hence ω2=GMR3ω2=GMR3 ✔
uses either of the above to get T2=4π2GMR3T2=4π2GMR3
OR
uses k=4π2GMk=4π2GM ✔
k = 9.2 × 10−13 / 9.3 × 10−13
Unit not required
R3=T2k=(8.9×104)29.25×10−13R3=T2k=(8.9×104)29.25×10−13 R = 2.04 × 107 «m» ✔
v = «ωr=2π×2.04×10789000=ωr=2π×2.04×10789000=» 1.4 × 103 «m s–1»
OR
v = «√GMR=√6.67×10−11×6.4×10232.04×107=√GMR=√6.67×10−11×6.4×10232.04×107=» 1.4 × 103 «m s–1» ✔
use of I∝1r2I∝1r2 «1.36 × 103 × 11.5211.52» ✔
604 «W m–2» ✔
use of 60046004 for mean intensity ✔
temperature/K = «4√6004×5.67×10−8=4√6004×5.67×10−8=» 230 ✔
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