The following data is an extract from a planetary fact sheet published by NASA.

Distance from sun, \(r\) \((\quantity{10^6}{km})\) Orbital period, \(P\) \((\mathrm{days})\) Orbital speed, \(v\) \((\mathrm{km\,s^{-1}})\) Mean surface temperature, \(T\) \((\mathrm{^\circ C})\)
Mercury \(57.9\) \(88\) \(47.4\) \(167\)
Venus \(108.2\) \(224.7\) \(35\) \(464\)
Earth \(149.6\) \(365.2\) \(29.8\) \(15\)
Mars \(227.9\) \(687\) \(24.1\) \(-65\)
Jupiter \(778.6\) \(4331\) \(13.1\) \(-110\)
Saturn \(1433.5\) \(10\,747\) \(9.7\) \(-140\)
Uranus \(2872.5\) \(30\,589\) \(6.8\) \(-195\)
Neptune \(4495.1\) \(59\,800\) \(5.4\) \(-200\)
Pluto \(5906.4\) \(90\,560\) \(4.7\) \(-225\)

Physics suggests that the orbital period, \(P\), and distance from the sun, \(r\), are related by a formula of the form \[P=A \,r^k\] where \(A\) and \(k\) are constants. By plotting a suitable graph, estimate the values of \(A\) and \(k\).

  • A straight-forward graph of \(P\) against \(r\) will probably show a curve from which it is hard to estimate values.
  • Can you transform the data to give a straight line graph?
  • You will probably find it helpful to use either graph paper or graphing software to obtain accurate values.

Use your results to predict the orbital period of Ceres, a dwarf planet in the asteroid belt with \(r=414.0\).

The surface temperature is generally lower for planets further away from the sun. Use the data to find an exponential law (\(T=Ae^{-kr}\)) or a power law (\(T=Ar^k\)) to model this.

  • You might need to use the absolute temperature, measured in Kelvin. A difference of \(\quantity{1}{K}\) is the same as a difference of \(\quantity{1}{^\circ C}\) but \(\quantity{0}{^\circ C}\approx\quantity{273}{K}\).

Use your results to predict the surface temperature of Ceres.