Question

A particle in the solar system is under the combined influence of the Sun's gravitational attraction and the radiation force due to the Sun's rays. Assume that the particle is a sphere of density 830 kg/m³ and that all the incident light is absorbed. If its radius is less than some critical radius R, the particle will be blown out of the solar system. Calculate this critical radius. Note that the total radiation power of the Sun is 3.90 × 10²⁶ W and its mass is 1.99 × 10³⁰ kg. The universal gravitational constant is given by 6.673 × 10^-11 N·m²/kg².

Answer 1

given

the particle is a sphere of density 830 kg/m³ =

power of the Sun is 3.90 X 10²⁶ W = P

mass is 1.99 X 10³⁰ kg = M

The universal gravitational constant is given by 6.673 X 10^-11 N·m²/kg² = G

the mass of the particleis m = density X volume

m = V

and the volume is V = 4/3 r³

then mass m = X 4/3 r³

the force is F_g = G M m / R²

F_g = G M ( X 4/3 r³ ) / R²

the radiation pressure is P_r = P / 4 R² C

and F = P X A

F = P X A / 4 R² C

F = P r² / 4 R² C

F = P r² / 4 R² C

now equating F_g = F

G M ( X 4/3 r³ ) / R² = P r² / 4 R² C

by simplyfying we get

r = 3 P / ( 16 G M C )

r = 3 X 3.90 X 10²⁶ / ( 16 X 3.14 X 6.673 X 10^-11 X 1.99 X 10³⁰ X 830 X 3 X 10⁸ )

r = 7.04 X 10^-7 m

the critical radius is r = 7.04 X 10^-7 m