Date | May 2021 | Marks available | 2 | Reference code | 21M.2.SL.TZ1.4 |
Level | Standard level | Paper | Paper 2 | Time zone | 1 |
Command term | Calculate | Question number | 4 | Adapted from | N/A |
Question
A planet orbits at a distance d from a star. The power emitted by the star is P. The total surface area of the planet is A.
Explain why the power incident on the planet is
P4πd2×A4.P4πd2×A4.
The albedo of the planet is αpαp. The equilibrium surface temperature of the planet is T. Derive the expression
T=4√P(1-αp)16πd2eσT=4√P(1−αp)16πd2eσ
where e is the emissivity of the planet.
On average, the Moon is the same distance from the Sun as the Earth. The Moon can be assumed to have an emissivity e = 1 and an albedo αMαM = 0.13. The solar constant is 1.36 × 103 W m−2. Calculate the surface temperature of the Moon.
Markscheme
P4πd2P4πd2 is the power received by the planet/at a distance d «from star» ✓
A4A4 is the projected area/cross sectional area of the planet ✓
use of eσAT4 OR P4πd2×A4×(1-αp) ✓
with correct manipulation to show the result ✓
4√1.36×103×0.874×5.67×10-8 ✓
T = 268.75 «K» ≅ 270 «K» ✓