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Homework Help: Photon hitting a solar sail

  1. Sep 21, 2006 #1
    The problem is that I'm doing what seems to be perfectly reasonable algebra but I'm getting a result that makes no sense.

    Let's say a photon travels from the left and hits a stationary solar sail which is perpendicular to the light. There is perfect reflection and a change in momenta for both objects. For momentum we have:

    p_iphoton + p_iSS = p_fphoton + p_fSS =>
    h/c*nu1 + m*(0)^2 = h/c*nu2 + mv
    h/c*(nu1-nu2) = mv


    KE_iphoton + KE_iSS = KE_fphoton + KE_fSS
    h*nu1 + 0 = h*nu2 + mv^2
    h*(nu1-nu2) = mv^2 = c*mv => v = c

    (KE_iphoton is the photon's initial kinetic energy, p_iphoton is its initial momentum, KE_fphoton is the final KE, etc.; h = planck's constant, c = speed of light, m = mass of solar sail, v = final velocity of solar sail, nu1 = initial frequency of photon, nu2 = final frequency of photon.)

    Obviously, the final speed of the solar sail isn't the speed of light. What did I do wrong?
  2. jcsd
  3. Sep 22, 2006 #2


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    You did not take direction into account with your momentum conservation calculation.
  4. Sep 22, 2006 #3

    Andrew Mason

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    As Andrevh pointed out, the change in momentum is the sum of, not difference of, the magnitudes of the momentum before and after due to the 180 degree change in direction.

    Also, you are using [itex]h/c\nu[/itex] for momentum of the photon which is not correct. Momentum of the photon is [itex]E/c = h\nu/c = h/\lambda[/itex].

    I don't follow your math here. You have to divide the change in momentum of the photon by the sail's mass to find the speed change in the sail. The m in your equation is the sail's mass not the mass of the photon.

    [tex]\Delta P_{ph}/m_s = v_s = P_s/m_s[/tex]

    [tex]KE_s = \frac{1}{2}m_sv_s^2 = P_s^2/2m_s[/tex]

  5. Sep 23, 2006 #4
    Yes, I see it now. Thank you and Andrevh. I'm guessing you'd have a difference of frequencies in the energy equation, but a sum of frequencies in the momentum equation, so you can't substitute a change in photon energy with a change in photon momentum, at least not like I did.

    That's what I had, except that rather than clarify things with parentheses like I should have, I relied on order of operations. I used [itex]h/c*\nu = (h/c)*\nu = (h*\nu)/c.[/itex]

    Let me clarify. In the last line of equations, I had
    [tex]h*(\nu_1-\nu_2) = mv^2 = c*mv => v = c[/tex]
    I believe the first two terms are, in fact, equal to each other, right? Now, based on my erroneous assumption that
    [tex]\Delta P_{ph} = (h/c) * (\nu_1 - \nu_2)[/tex]
    I got
    [tex]\Delta P_{ph} = c * \Delta E_{ph}[/itex]
    which implies that mv^2 = cmv.
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