Math & Science Home  Proficiency Tests  Mathematical Thinking in Physics  Aeronauts 2000 

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/m^{2}) leads rather directly to an intuitive expression for pressure exerted on objects by sunlight. W/m^{2}
= j/(m^{2} sec) = (nt m)/(m^{2} sec) = (nt/m^{2})(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 pressure^{3}. At 1 AU, = 1.36 x 10^{3} W/m^{2}. The corresponding pressure must be p =
4.53 x 10^{6} nt/m^{2}
(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/cm^{3}, 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} m^{2}, 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/sec^{2}.
The acceleration due to the sun's gravity at 1 AU is g_{sun}
= 5.92 x 10^{3} m/sec^{2}
so that the pressure of sunlight overwhelms the solar gravity by a factor^{4} 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 _{SB}T_{s}^{4}(R_{s}/l)^{2}
where _{SB} is the StefanBoltzmann constant (= 5.67 x 10^{8} W/(m^{2}K)^{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 (_{SB}T_{s}^{4}/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_{SB}T_{s}^{4}/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_{SB}T_{s}^{4}/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} m^{2
}_{SB}
= 5.67 x 10^{8} W/(m^{2}K^{4})
T_{s} = 5800 °K m = 5.24 x 10^{16} kg c = 3 x 10^{8} m/sec g_{so} = 2.74 x 10^{2} m/sec^{2} 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 10^{3} km and m = 5.98 x 10^{24} kg, and the ratio^{5} 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.


Please send suggestions/corrections to: Web Related: David.Mazza@grc.nasa.gov Technology Related: Joseph.C.Kolecki@grc.nasa.gov Responsible NASA Official: Theresa.M.Scott (Acting) 