DP Physics (first assessment 2025)

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Question 21N.2.SL.TZ0.6

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Date November 2021 Marks available [Maximum mark: 9] Reference code 21N.2.SL.TZ0.6
Level SL Paper 2 Time zone TZ0
Command term Estimate, Explain, Show that Question number 6 Adapted from N/A
6.
[Maximum mark: 9]
21N.2.SL.TZ0.6

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

(a.i)

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

[1]

Markscheme

incident intensity 13609.32 OR 15.7≈16 «W m−2» ✓

 

Allow the use of 1400 for the solar constant.

(a.ii)

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]

Markscheme

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.iii)

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

[1]

Markscheme

σT 4 = 3.07

OR

T = 86 «K» ✓

(b.i)

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

Show that T2=4π2R 3GM where M is the mass of Saturn.

[2]

Markscheme

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

correct rearrangement to reach the expression given ✓

 

Allow use of GMR=2πRT for MP1.

(b.ii)

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

[2]

Markscheme

T=15.9×24×3600 «s» ✓

M=4π21.2×10936.67×10-11×15.9×24×36002=5.4×1026 «kg» ✓

 

Award [2] marks for a bald correct answer.

Allow ECF from MP1.

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