Question

The intensity of sunlight reaching the earth is $1360\mathrm{W}/{\mathrm{m}}^2$. Assuming all the sunlight is absorbed, what is the radiation pressure force on the earth$?$ Give your answer (a) in newtons and (b) as a fraction of the sun's gravitational force on the earth.

Short answer

a. Pressure force in newton is$45.3\times {10}^{-6}\mathrm{N}/{\mathrm{m}}^2$

b. Gravitational force on earth is $3.5\times {10}^{22}\mathrm{N}$.

Step-by-step solution

Calculation for pressure force (part a)

a.

Given info:

Intensity is $13600\frac{\mathrm{W}}{{\mathrm{m}}^2}$.

Radiation pressure $\mathrm{Pressure}\to {\mathrm{P}}_{\mathrm{rad}}$is the force enclosed on an item per unit,

On an object that absorbs all light, the radiation pressure is:

${\mathrm{P}}_{\mathrm{rad}}=\frac{\mathrm{F}}{\mathrm{A}}$

$=\frac{\frac{\mathrm{P}}{\mathrm{A}}}{\mathrm{c}}$

$=\frac{\mathrm{I}}{\mathrm{c}}$

${\mathrm{P}}_{\text{rad}}=\frac{13600{\displaystyle \frac{\mathrm{W}}{{\mathrm{m}}^2}}}{3\times {10}^8{\displaystyle \frac{\mathrm{m}}{\mathrm{s}}}}$

${\mathrm{P}}_{\text{rad}}=45.3\times {10}^{-6}\mathrm{N}/{\mathrm{m}}^2$

Calculation for pressure force (part b)

b.

Gravitation force is,

${\mathrm{F}}_{\mathrm{g}}=\frac{\mathrm{G}\times {\mathrm{M}}_{\mathrm{e}}\times {\mathrm{M}}_{\mathrm{s}}}{{\mathrm{d}}^2}$

${\mathrm{F}}_{\mathrm{g}}=\frac{6.67\times {10}^{-11}\times 5.98\times {10}^{24}\times 1.98\times {10}^{30}}{{\left(150\times {10}^9\right)}^2{\mathrm{m}}^2}$

${\mathrm{F}}_{\mathrm{g}}=3.5\times {10}^{22}\mathrm{N}$