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3 years ago

Show that the equilibrium temperature of the surface of the moon is 273 K assuming it has an albedo of 0.08.

Physics

1 answer:

const2013 [10]3 years ago

4 0

Answer:

The equilibrium temperature of the surface of the moon can be found by the formula as follows:

$T=(\frac{K_s(1-alebdo)}{4 \sigma})^{\frac{1}{4}}$

Where $K_s = 1366 W/m^2$ is the solar constant and $\sigma = 5.67 \times 10^ {-8}W/m^2K^{-4}$ is the Stefan's Boltzmann constant.

$T=(\frac{1366(1-0.08)}{4\times 5.67 \times 10^ {-8}})^{\frac{1}{4}} = (55.41\times 10^8)^{\frac{1}{4}} = 272.7 K=273 K$

Thus, the equilibrium temperature of the surface of the moon is 273 K.

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Initial vertical component of velocity:

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Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is $20.6\; {\rm m\cdot s^{-1}}$ (upwards,) the vertical velocity right before landing would be $(-20.6\; {\rm m\cdot s^{-1}})$ (downwards.) The change in vertical velocity is:

$\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}$ .

Since there is no drag on this projectile, the vertical acceleration of this projectile would be $g$ . In other words, $a = g = -9.81\; {\rm m\cdot s^{-2}}$ .

Hence, the time it takes to achieve a (vertical) velocity change of $\Delta v_{y}$ would be:

$\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}$ .