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Force of solar radiation on earth

  • Thread starter indie452
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  • #1
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Homework Statement



ok, well...im not sure if this is posted in the right section but ive just come across a past exam question that is supposed to be in classical mech and relativity section of the paper...

If 1.3kW/m2 of solar radiation falls on the Earth at midday at the equator what is the total force from this solar radiation on the Earth? (you may assume that all the radiation is absorbed)
And so what is the rate of loss of mass from the sun due to solar radiation?

Earth radius [r] = 6400km
Sun-Earth distance [R] = 1.5x1011
[Assume circular orbit]
Earth mass [m] = 6x1024kg
Sun mass [M] = 2x1030kg


The Attempt at a Solution



okay so i had no idea how to start at first by using classical mech methods, so i decided to try and use methods from my quantum course:

P = [stephans const]*T4
so T = 4th root[1300 / 5.67x10-8] = 389.1258K

[wavelength]*T = 2.8x10-3
so [wavelength] = 7.1956x10-6m

and so E = hc/[wavelength] = 2.764x10-20J

and Fd = change in E -> energy all gets absorbed so it is converted into other forms of E
so E after is zero

and so F = E/d = 1.84x10-31N


but i have no idea if this is right, or even if this method is valid...any help would be greatly appreciated...

Also i didnt know how to get started on the 2nd part of the qu

btw...after typing this up in this post i dont think this is right at all cause i think some of my units are a bit messed up...
 
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Answers and Replies

  • #2
124
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force of solar radiation on the earth

1. Homework Statement

ok, well...im not sure if this is posted in the right section but ive just come across a past exam question that is supposed to be in classical mech and relativity section of the paper...

If 1.3kW/m2 of solar radiation falls on the Earth at midday at the equator what is the total force from this solar radiation on the Earth? (you may assume that all the radiation is absorbed)
And so what is the rate of loss of mass from the sun due to solar radiation?

Earth radius [r] = 6400km
Sun-Earth distance [R] = 1.5x1011
[Assume circular orbit]
Earth mass [m] = 6x1024kg
Sun mass [M] = 2x1030kg


3. The Attempt at a Solution

okay so i had no idea how to start at first by using classical mech methods, so i decided to try and use methods from my quantum course:

P = [stephans const]*T4
so T = 4th root[1300 / 5.67x10-8] = 389.1258K

[wavelength]*T = 2.8x10-3
so [wavelength] = 7.1956x10-6m

and so E = hc/[wavelength] = 2.764x10-20J

and Fd = change in E -> energy all gets absorbed so it is converted into other forms of E
so E after is zero

and so F = E/d = 1.84x10-31N


but i have no idea if this is right, or even if this method is valid...any help would be greatly appreciated...

Also i didnt know how to get started on the 2nd part of the qu
 
  • #3
diazona
Homework Helper
2,175
6


Do you understand the context of the equations you're using? For instance, when you computed T from P: T represents the temperature of what? Then when you computed the wavelength, it represents the wavelength of what? And then E is the energy of what?

Also, think about this: how did you get the value of P in the first place? Because you're only given power per unit area, in watts per meter squared, not power. You must keep track of units in calculations like this.
 
  • #4
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0


yeah i see what you mean there were a few of us all sat there trying to do this qu and none of us could manage it...i think we just started grasping at straws which is never good i know...its just that our classical mech and relativity lecturer wrote this qu on a past paper but we havent done anything like it yet, so i am unsure whether it will come up this year but i wanted to be able to understand it just incase cause you never know when it can be helpful.

do you have any suggestions on how to start it properly?
 
  • #5
diazona
Homework Helper
2,175
6


Well, here's my first thought: basically you have a whole bunch of photons colliding with the Earth every second. Figure out what the energy and momentum of that bunch of photons are, then just use the same procedure you normally do for totally inelastic collisions.
 

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