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.

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

(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

and its acceleration away from the sun, due to the pressure of sunlight, is

The acceleration due to the sun's gravity at 1 AU is

so that the pressure of sunlight overwhelms the solar gravity by a factor⁴ of

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

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

The perturbation of the Earth due to the pressure of sunlight is too small to detect by any ordinary means.

⁵ See footnote 4.