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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.

Question

HLPaper 2

Explain why the power incident on the planet is

$\frac{P}{4\pi d^2} \times \frac{A}{4}$

[2]

Verified

Solution

$\frac{P}{4\pi d^2}$ is the power received by the planet/at a distance d «from star» 1 mark

$\frac{A}{4}$ is the projected area/cross sectional area of the planet 1 mark

The albedo of the planet is $α_p$ . The equilibrium surface temperature of the planet is T. Derive the expression

$T = \sqrt[4]{\frac{P(1-\alpha_p)}{16\pi d^2 e \sigma}}$

where e is the emissivity of the planet.

[2]

Verified

Solution

use of $e\sigma AT^4$ OR $\frac{P}{4\pi d^2} \times \frac{A}{4} \times (1-\alpha_p)$ 1 mark

with correct manipulation to show the result 1 mark

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 $α_M = 0.13$ . The solar constant is $1.36 × 10^3 \, W \, m^{-2}$ . Calculate the surface temperature of the Moon.