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

g_sun = 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

_SBT_s⁴(R_s/l)²

where _SB is the Stefan-Boltzmann constant (= 5.67 x 10^-8 W/(m²K)⁴), T_s is the sun's surface temperature, and R_s is the solar radius. The corresponding pressure of sunlight at the distance l is

(_SBT_s⁴/c)(R_s/l)².

Now, let a be the acceleration due to light pressure at distance l, and g_s 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_SBT_s⁴/mc)(R_s/l)².

If g_so is the solar surface gravity, then

g_s = g_so(R_s/l)²,

and a/g_s = A_SBT_s⁴/mcg_so.

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)
T_s = 5800 °K
m = 5.24 x 10^-16 kg
c = 3 x 10⁸ m/sec
g_so = 2.74 x 10² m/sec²

With these values, we find a/g_s = 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/g_s 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.

 

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⁴ This value is correct for photons that are absorbed. For photons that are reflected, this value must be multiplied by 2.

⁵ See footnote 4.