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Sunlight Exerts PressureProblem: 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 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 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
(
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 =
(A
If gso is the solar surface gravity, then gs
= gso(Rs/l)2,
and a/gs
= A 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
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 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.
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