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Photon momentum

  1. Aug 3, 2006 #1
    Hey everyone. I'm kind of new to the whole physics world, but I've got a question I've been wondering about.

    So the Sun attracts the Earth because of gravity (or bends space time that the Earth is traveling through), and my question is this: Since photons have momentum, and momentum must be conserved when the photons strike the Earth, how much collectively are the photons pushing the Earth away from the Sun? Obviously it must be very little, but does it cause even a noticable error when calculating the orbit of the earth? That is, if the sun did not emit light, would it appear as though it's gravitational pull has gotten stronger than when it -was- emitting light? Maybe a dumb question, but something I was pondering...thank you!
  2. jcsd
  3. Aug 4, 2006 #2


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    In principle you're right. Don't even need to talk about photons for that: classically, already, electromagnetic waves carry momentum as well as energy (btw, the photon momentum comes from that correspondence).
    The transferred momentum on an object due to absorption or reflection of light is called radiation pressure.

    However, as you suggest, it must be very small as compared to the gravitational pull. Didn't do the calculation, though.
  4. Aug 4, 2006 #3
    Agreed, someone should do the calculation just for fun
  5. Aug 4, 2006 #4
    Haha....yeah, sooooooomeone should do it...hm....so basically, the photons are pushing us away, but we just leave it out of mind because it's a pretty small amount?

    After a long enough time, would we move slightly away, or isn't it even that much of an effect?
  6. Aug 4, 2006 #5


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    Ok, I did this very quickly, so check if I didn't make any error...

    Look at http://scienceworld.wolfram.com/physics/RadiationPressure.html

    The radiation pressure (force per unit of surface) equals the energy flux divided by the lightspeed.

    If we take it that the earth receives about 1000 W per square meter of radiation from the sun, then that converts to 1000 W / (3.10^8 m/s) = 3.3 10^(-6) Newton per square meter, or Pascal (~ 33 10^(-9) mbar, which is a relatively good lab vacuum).

    If we take the earth radius to be 6500 km, then the cross section of the earth is 1.3 10^14 m^2, so this results ini a total force on the earth of:

    4.4 10^8 N

    (which is about the weight of a 40,000 ton mass, say, a big ship)

    The earth mass is about 6 10^24 kg,
    so this results in an acceleration of about 10^(-16) m/s^2
  7. Aug 4, 2006 #6

    This is interesting to me. Can you calculate the radiation pressure on the earth from all other directions. I'm intrested in knowing if the pressures at all other points on the earth, not directly in line with the sun, are greater or lesser than the radiation pressure from the sun.

    Seems like even though all those stars are so far away, the sheer amount of radiation pressure would be much greater, considering every object in line with the earth (and this is a very high percent of the observable universe) exerts some pressure.

    Edit: I don't actually mean every point just say the same surface area exactally opposite from the sun.
    Last edited: Aug 4, 2006
  8. Aug 4, 2006 #7


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    Given that the radiation pressure is the energy flux divided by the lightspeed, and that by far the main source of energy flux comes from the sun, and not from distant stars, by far the main contribution must come from the sun, no ?

    The pressure is proportional to the energy flux...
  9. Aug 4, 2006 #8

    I really have no idea, thats why I asked :P I thank you for the reply though.
  10. Aug 4, 2006 #9
    Thanks vanesch! :smile: I'm gonna have to read up a little bit on radiation pressure now. Thanks again for the calculations too
  11. Aug 5, 2006 #10


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    The radiation pressure is related to the amount of radiation received, so the comparison here is valid. In fact, you could get a rough idea of the difference between the radiation pressure from the Sun and that from the stars by comparing the brightness of the light we receive from the Sun to that of the stars.

    It would also be good for us to keep in mind that the stars are distributed almost evevnly throughout the universe (even though we only see them from the Earth's dark side), so they provide about the same pressure from all directions.
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