Question

A small spinning asteroid is in a circular orbit around a star, much like the earth's motion around our sun. The asteroid has a surface area of 7.70 m2. The total power it absorbs from the star is 3800 W.
Assuming the surface is an ideal absorber and radiator, calculate the equilibrium temperature of the asteroid (in K)

Answer 1

Answer:

Temperature will be 305 K  

Explanation:

We have given The asteroid has a surface area $A=7.70m^2$

Power absorbed P = 3800 watt

Boltzmann constant $\sigma =5.67\times 10^{-8}Wm/K^4$

According to Boltzmann rule power radiated is given by

$P=\sigma AT^4$

$3800=5.67\times 10^{-8}\times 7.70\times T^4$

$T^4=87.0381\times 10^8$

$T=305K$

So temperature will be 305 K