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Formula for solar radiation pressure

  1. Sep 4, 2005 #1


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    Formula for solar radiation pressure

    I'm looking for a formula that gives the amount of force in Newtons that is produced from the pressure of solar radiation.

    I've googled it and found:
    [tex] F_R = C_R \frac{I}{c}S[/tex]
    where I is the radiation intensity, c is the speed of light, and S is the cross-sectional (sun-facing) area of the object being pushed, and Cr is the solar radiation coefficient which equals 1.0. (Does this value have units? I'm assuming it doesn't. And why clutter up the formula by multiplying it by 1?)

    I don't quite know what I (intensity) is. Googling it, I find that its units are the Candela (CD). But that it is also a measure of energy which is expressed in Watts. (aka Joules / second, or Nm/s)


    [tex]N = \frac{Nm/s} {m/s} m^2 [/tex]

    [tex]N = N m^2[/tex] which makes no sense. Unless the solar radiation coefficient has units of /m^2. Maybe that's why it's thrown in?

    Furthermore, Intensity should change with distance. So I'm guessing that for Intensity I would use:

    Luminosity of the Sun (3.9e26 watts) * (area of a sphere of 1 solar radii / area of a sphere of radius = distance), which would give me Intensity in Watts.

    Ultimately, I'm trying to figure out how much solar radiation pushes against small objects at any given instant, causing their orbits to spiral outward.

    Any thoughts? And any idea which is stronger: pressure from solar radiation or pressure from solar wind? Am I correct in assuming that pressure from solar radiation is basically constant, while pressure from solar wind varies with solar activity? Which is more responsible for creating the tails on comets?
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  3. Sep 4, 2005 #2


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    In MKS (SI) force should be in Newtons, while pressure is in Pa, which is N/m2.

    Radiation intensity might be in photons/unit area or energy/unit area. I would expect the former.

    from Wikipedia

    I checked an old textbook, which gives

    Force - 1/c ( L r2/d2 ), where L is solar luminosity, and L/(4[itex]\pi[/itex]d2 is the total radiant energy from the sun, and r is the radius of the object (round particle).

    So L/(4c[itex]\pi[/itex]d2 would be pressure.

    Remember, photon momentum p = E/c.

    The solar wind (http://en.wikipedia.org/wiki/Solar_Wind) has the stronger effect on a comets tail - but one should do a calculation to show the pressure of solar wind.
    Last edited: Sep 4, 2005
  4. Sep 4, 2005 #3


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    Thanks for your helpful response, Astronuc.

    Would I be correct to assume that pressure = force if my cross-section were exactly 1 square meter?
  5. Sep 4, 2005 #4


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    That's strange, I would expect it to be:

    [tex]force = \frac{flux*area}{c}[/tex]

    or just the energy absorbed per unit time divided by the speed of light (dp/dt). This would be with

    [tex]flux= \frac{L}{4\pi d^2}[/tex]


    [tex]area = \pi r^2[/tex]

    This would instead lead to

    [tex]force = \frac{L}{c}(\frac{r^2}{4d^2})[/tex]

    That is, with an extra factor of four in the denominator. The equation for pressure would be the same, however.

    They would have different units, but would be numerically the same in mks units.
  6. Sep 5, 2005 #5


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    Intensity is POWER/Area ; the "C_R" in your formula
    is probably supposed to be the reflection coefficient
    (1 for fully absorbed, 2 for fully reflected ...).

    At a distance "a" from the Sun, in astronomical units,
    the radiation pressure is C_R (4.57E-6 N/m^2)/a^2
    as Astronuc's quote implies, for fully absorbed light.

    This pressure does NOT double if the surface reflects!
    A sphere reflects directly backward (180 degrees)
    only light that hits the *center* of the sphere;
    by impact parameter .7r , the deflection is 90deg.
    The outer half of the cross-sectional Area "S"
    is less effective for momentum than if it absorbed!
    So "C_R" is shape-dependent and non-trivial.

    Wikipedia claims (Solar wind) that the Sun emits
    800 kg/s in the solar-wind protons & electrons;
    at a speed of 450 km/s , at 1 a.u., that's
    (mv/t)/(4 pi r^2) = 1.27E-15 N/m^2 = wimpy.
    (Is 800 kg/s way too small? 1 m^2 at Earth gets)
    ( 1.2 Million protons, 0.2 Million He++, 1.4 M e- )
    ( per second, on avg. We only get +1 mol/day? )
  7. Sep 7, 2005 #6


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    Space Tiger - thanks for the correction. I believe I = L / (4[itex]\pi[/itex]d2), where d is distance from sun.

    Tony - Sorry for the crappy post. I was in a rush then and didn't get back in time to edit it.

    lightgrav makes a good point - what I posted is the incident energy and assumes it is absorbed. Photons could be reflected or scattered (depending on surface characteristics, e.g. composition, and particle shape) and so the momentum transfer could be different.

    I'd have to check the solar wind numbers. Here is what Hyperphysics has about Solar Wind
    Last edited: Sep 7, 2005
  8. Sep 8, 2005 #7


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    Yes, solar pressure is nearly constant if you're talking about objects orbiting the Earth. The variation in the object's distance from the Sun is insignicant when compared to the distance between the Sun and Earth. But solar pressure wouldn't cause the objects to spiral outward - it would just make the orbits more elliptical (in about 200 years, some of the GPS satellites will reach the geostationary belt at apogee).

    If you're talking about objects that orbit the Sun, the pressure in a nearly circular orbit could be considered nearly constant. That wouldn't apply to comets.

    The solar radiation coefficient is an indication of how reflective the object is. A transparent object has a reflectivity of 0, a totally absorbant object has a reflectivity of 1, and a perfect mirror has a reflectivity of 2. A certain amount of momentum is transferred with each photon absorbed. A certain amount of momentum is transferred with each photon emitted. If the object is transparent, every photon absorbed on one side is emitted out the opposite side giving a net of zero. If perfectly reflective, every photon absorbed is emitted back out the same side, doubling the amount of momentum transferred.
  9. Sep 8, 2005 #8


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    Thanks, ST, Astronuc LG and Bob. Playing around with these formulas in a simulation, I can see that it's a very weak force. But if I create 1 kg objects with the radius of Earth, they get blown away quickly :rofl:

    I could probably just google this, but how is aldebo and reflectivity related? Or are they the same?
  10. Sep 8, 2005 #9


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    Albedo is the measure of the reflectivity of an object (celestial body) and is expressed as the ratio of the amount of light reflected by an object to that of the amount of light incident upon it.

    A value of 1 represents a perfectly reflecting (white) surface, whilst a value of zero represents a perfectly absorbing (black) surface.

    Source - www.r-clarke.org.uk/glossary.htm

    See also - http://www.nineplanets.org/help.html
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