Power Radiation Ratio of Venus and Earth

[rnjscksdyd]

Homework Statement

Venus and Earth may be regarded as behaving as black bodies. The mean temperature at the surface of Venus is about 600K and at the surface of Earth is about 300K. Which of the following is the best estimate for the ratio
\frac{power.radiated.per.unit.area.on.Earth}{power.radiated.per.unit.area.on.Venus}

(A) 1/2
(B) 1/4
(C) 1/8
(D) 1/16

I had to put dots in between words.. It won't let me space them

Homework Equations

σAT^4

The Attempt at a Solution

Venus: σAT^4 = σA(600)^4
Earth: σAT^4 = σA(300)^4

The actual answer is (D)
and I got this by
\frac{σA(300)^4}{σA(600)^4}=1/16

If the above working is right, I want to know where 'A' or 'Area' disappeared to. I thought Area of Earth and Venus are different so they can't cancel out? Or is power radiated always referring to the unit surface area, or 1m^2 ?

p.s. sorry for asking too many questions, my exam is next week :(

 

[tms]

Homework Statement

Venus and Earth may be regarded as behaving as black bodies. The mean temperature at the surface of Venus is about 600K and at the surface of Earth is about 300K. Which of the following is the best estimate for the ratio
\frac{power.radiated.per.unit.area.on.Earth}{power.radiated.per.unit.area.on.Venus}

(A) 1/2
(B) 1/4
(C) 1/8
(D) 1/16

I had to put dots in between words.. It won't let me space them

Here is how to do it:
\frac{\text{power radiated per unit area on Earth}}{\text{power radiated per unit area on Venus}}
\frac{\text{power radiated per unit area on Earth}}{\text{power radiated per unit area on Venus}}

Although it looks kind of ugly here.

Homework Equations

σAT^4

If the above working is right, I want to know where 'A' or 'Area' disappeared to. I thought Area of Earth and Venus are different so they can't cancel out? Or is power radiated always referring to the unit surface area, or 1m^2 ?

Assuming there are no losses, as is presumably the case in this problem, you can calculate or measure the flux through any closed surface, so you can pick the same surface for both Venus and Earth. Note that the question asks not for the total power, but the power per unit area, which is just \sigma T^4 in both cases.

 

[rnjscksdyd]

Here is how to do it:
\frac{\text{power radiated per unit area on Earth}}{\text{power radiated per unit area on Venus}}
\frac{\text{power radiated per unit area on Earth}}{\text{power radiated per unit area on Venus}}

Although it looks kind of ugly here.




Assuming there are no losses, as is presumably the case in this problem, you can calculate or measure the flux through any closed surface, so you can pick the same surface for both Venus and Earth. Note that the question asks not for the total power, but the power per unit area, which is just \sigma T^4 in both cases.

I get it now! :D thank you so much ;D