Star Observation Calculation Homework

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Homework Statement


The intensity of the Sun's radiation is about 1380 Wm-2 at Earth's distance, 1.5x1011m. Earth absorbs this radiation as a black body, and radiates its own energy back into space.

a) How much Energy per second falls on the Earth's surface? (Diameter of Earth = 12800km)
b) What temperature would Earth's surface need to be to radiate this much energy back into space (i.e equilibrium)
c) Why is Earth's actual average surface temperature a bit higher than this
d) Ceres, the largest asteroid, orbits around 4.1 x1011m from the sun:
(i) What would be the Sun's intensity on Ceres
(ii) Find the average temperature of its surface.

Homework Equations


P=AoT4 where o = 5.6697x10-8Js-1m-2 K-4
I = P / 4 x pi x D2 where D is the distance from the star

The Attempt at a Solution


a) P = 1380 x 4pi x (1.5x1011)2
don't think this is right...
b) T = fourth root of P / 4pi x r2 x o
c) global warming?? greenhouse effect??
d) (i) Just use the ratio of Earth's distance : Ceres' distance, plug in the intensity
(ii) Don't even know where to start, considering we don't have the radius...
 
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Hi,
I agree with you in the first question since energy/sec is just watt, so assuming the sun shine on the whole Earth Pon earth = P*A
Also does agree with 2, it's a straighforward application of stephan's law,
Because It doesn't radiate all the energy that it absorbs due to something, maybe global warming and a small contribution go to heat the Earth's surface
And repeat the same scénario to answer the last two, good luck !
 
Last edited:
Noctisdark said:
Hi,
I agree with you in the first question since energy/sec is just watt, so assuming the sun shine on the whole Earth Pon earth = P*A
Also does agree with 2, it's a straighforward application of stephan's law,
Because It doesn't radiate all the energy that it absorbs and a small contribution go to heat the Earth's surface
And repeat the same scénario to answer the last two, good luck !
Hey there, My attempt at question 1 and 2 must be incorrect as I come out with a Power of 3.9x1026, and therefore a surface temperature of over 1million Kelvin which just isn't right? HELP!
 
No idea, we have no radius of ceres, maybe google it ?
Just to correct one thing 1380 is the sun intensity at earth, so I = 1380, calculate P (You've done that), assume that R being the radius of ceres and for it to radiate back that energy into space then P = 4πR2*σ*T4, and solve for T
 

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