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Sunlight Exerts Pressure
Problem: Estimate the pressure exerted by sunlight on objects at 1 AU from the sun. Discuss the result.
Solution: A dimensional analysis of the solar constant (in units of W/m2) leads rather directly to an intuitive expression for pressure exerted on objects by sunlight.
W/m2 = j/(m2 sec) = (nt m)/(m2 sec) = (nt/m2)(m/sec)
The last term on the right appears to be made up of a pressure and a velocity which, in this case, must be taken to be the velocity of light. Thus, if = solar constant, then /c must be the corresponding pressure3. At 1 AU, = 1.36 x 103 W/m2. The corresponding pressure must be
p = 4.53 x 10-6 nt/m2
(about 6.56 x 10-10 psi).
Now, with this value in hand, let us consider the forces on a 1 diameter particle with a density of 1 gm/cm3, such as might be found in the tail of a comet. The projected area, A, of the 1 particle is 7.85 x 10-13 m2, and its mass, m, is 5.24 x 10-13 gm = 5.24 x 10-16 kg. The force exerted by sunlight on this particle is
f = pA = 3.56 x 10-18 nt
and its acceleration away from the sun, due to the pressure of sunlight, is
a = 6.79 x 10-3 m/sec2.
The acceleration due to the sun's gravity at 1 AU is
gsun = 5.92 x 10-3 m/sec2
so that the pressure of sunlight overwhelms the solar gravity by a factor4 of
6.79/5.92 = 1.15.
We see here why a comet's tail points away from the sun!
In fact, the ratio of the two accelerations is independent of the distance from the sun for any given particle. The solar constant at a distance l from the sun is
where SB is the Stefan-Boltzmann constant (= 5.67 x 10-8 W/(m2K)4), Ts is the sun's surface temperature, and Rs is the solar radius. The corresponding pressure of sunlight at the distance l is
Now, let a be the acceleration due to light pressure at distance l, and gs be the solar gravitation at the same distance. For a particle of cross sectional area A and mass m, the acceleration, a, is
a = (ASBTs4/mc)(Rs/l)2.
If gso is the solar surface gravity, then
gs = gso(Rs/l)2,
and a/gs = ASBTs4/mcgso.
This last expression is the ratio of the accelerations sought. N.B., The ratio is independent of distance.
As a check on our previous calculation, we use the values:
A = 7.85 x 10-13 m2
SB = 5.67 x 10-8 W/(m2K4)
Ts = 5800 °K
m = 5.24 x 10-16 kg
c = 3 x 108 m/sec
gso = 2.74 x 102 m/sec2
With these values, we find a/gs = 1.17, in agreement with our previous result.
Having done the calculation for a 1 diameter particle, let us now repeat it for the Earth. For Earth, r = 6.37 x 103 km and m = 5.98 x 1024 kg, and the ratio5 a/gs comes out to
1.63 x 10-14
The perturbation of the Earth due to the pressure of sunlight is too small to detect by any ordinary means.
4 This value is correct for photons that are absorbed. For photons that are reflected, this value must be multiplied by 2.
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