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Date	November 2021	Marks available	2	Reference code	21N.2.SL.TZ0.6
Level	Standard level	Paper	Paper 2	Time zone	0 - no time zone
Command term	Estimate	Question number	6	Adapted from	N/A

Question

Titan is a moon of Saturn. The Titan-Sun distance is 9.3 times greater than the Earth-Sun distance.

Show that the intensity of the solar radiation at the location of Titan is 16 W m⁻²

[1]

a.i.

Titan has an atmosphere of nitrogen. The albedo of the atmosphere is 0.22. The surface of Titan may be assumed to be a black body. Explain why the average intensity of solar radiation absorbed by the whole surface of Titan is 3.1 W m⁻²

[3]

a.ii.

Show that the equilibrium surface temperature of Titan is about 90 K.

[1]

a.iii.

The orbital radius of Titan around Saturn is $R$ $R$ and the period of revolution is $T$ $T$ .

Show that $T^{2} = \frac{4 π^{2} R^{3}}{G M}$ $T^{2} = \frac{4 π^{2} R^{3}}{G M}$ where $M$ $M$ is the mass of Saturn.

[2]

b.i.

The orbital radius of Titan around Saturn is 1.2 × 10⁹m and the orbital period is 15.9 days. Estimate the mass of Saturn.

[2]

b.ii.

Markscheme

incident intensity $\frac{1360}{9 . 3^{2}}$ $\frac{1360}{9 . 3^{2}}$ OR $15.7 \approx 16$ $15.7 \approx 16$ «W m⁻²» ✓

Allow the use of 1400 for the solar constant.

a.i.

exposed surface is ¼ of the total surface ✓

absorbed intensity = (1−0.22) × incident intensity ✓

0.78 × 0.25 × 15.7 OR 3.07 «W m⁻²» ✓

Allow 3.06 from rounding and 3.12 if they use 16 W m⁻².

a.ii.

σT ⁴ = 3.07

T = 86 «K» ✓

a.iii.

correct equating of gravitational force / acceleration to centripetal force / acceleration ✓

correct rearrangement to reach the expression given ✓

Allow use of $\sqrt{\frac{G M}{R}} = \frac{2 π R}{T}$ $\sqrt{\frac{G M}{R}} = \frac{2 π R}{T}$ for MP1.

b.i.

$T = 15.9 \times 24 \times 3600$ $T = 15.9 \times 24 \times 3600$ «s» ✓

$M = \frac{4 π^{2} {(1.2 \times 10^{9})}^{3}}{6.67 \times 10^{- 11} \times {(15.9 \times 24 \times 3600)}^{2}} = 5.4 \times 10^{26}$ $M = \frac{4 π^{2} {(1.2 \times 10^{9})}^{3}}{6.67 \times 10^{- 11} \times {(15.9 \times 24 \times 3600)}^{2}} = 5.4 \times 10^{26}$ «kg» ✓