Energy from cosmic microwave background

[chris1969]

Hello everyone

I've a question about the cosmic microwave background which my nephew asked me a couple of days ago. How much energy hits the Earth each year from the CMB? I'm assuming here that it is measurable in this way, but apologies if not.

Thanks in advance for your help.

Chris

 

[anorlunda]

On Wikipedia, I found this number for the intensity of the CMB. $4.005 x 10^{-14} \frac{J}{m^3}$

Now, can you look up the Earth's surface area and do the rest of the math?

 

[MarkJW]

On Wikipedia, I found this number for the intensity of the CMB. $4.005 x 10^{-14} \frac{J}{m^3}$

Now, can you look up the Earth's surface area and do the rest of the math?

Unfortunately you indicated a volumetric density. I think the number is reasonable though as https://dspace.carthage.edu/bitstream/handle/123456789/37/Adam%20Ferg.pdf?sequence=1 states ~10^-14 W/m^2.

 

[Vanadium 50]

Unfortunately you indicated a volumetric density.

Easy enough to fix - multiply by the speed of light.

 

[Orodruin]

Easy enough to fix - multiply by the speed of light.

Not that easy, the direction is random. There is another factor of order one.

 

[chris1969]

Thanks ever so much everyone. Can I check my logic/arithmetic please?

1) Energy density of CMB is 4.005 * 10^-14 (Joules per cube metre)
2) Radius of Earth is about 6,371 km
3) Surface area of Earth is about 1.28 * 10^14 metres squared
4) Speed of light is 300,000 m/s

Multiplying (1) * (3) * (4) would give about 1.53 * 10^6 Joules per second from the CMB.

Am I on the right lines here, or have I made a basic error somewhere.

Thanks once again. It's been over 20 years since I did high school physics, so I more than a little rusty.

Chris

 

[jwinter]

I would go with Stefan-Boltzmann and use the surface area of the Earth as the surface area of the black body that is shining on us - in which case you get a power of:

Area_of_Earth x Sigma x T^4 = (5.10*10^14) * (5.67*10^-8) * (2.726^4) = 1.6 Gigawatts

In case your cousin wonders whether the sea will eventually boil from all that energy hitting the earth, you might care to point out that the Earth (being at say 300°K) is shining back into space of order 100 million times what it is receiving from the CMB (being at 2.7°K). ((300/2.7)^4 from Stefan-Boltzmann law)