Question

Find the radiation pressure of solar radiation on the surface of earth. Solar constant is $1.4kW{m}^{-2}$

• A. $4.7\times {10}^{-5}Pa$
• B. $4.7\times {10}^{-6}Pa$
• C. $2.37\times {10}^{-6}Pa$
• D. $9.4\times {10}^{-6}Pa$

Answer 1

Answer: (B)

$4.7\times {10}^{-6}Pa$

METHOD-1:

${Pressure}_{absorbed}=\frac{E_f}{c}$

$E_f=$ energy flux, $C=$ Speed of light

${Pressure}_{absorbed}=\frac{1.4\times 1000}{3\times {10}^8}=4.66\times {10}^{-6}Pa$

METHOD-2: (checking the unit,if formula is not remembered)

We know that:

$Power=force\times velocity$-----(1)

$Pressure=\frac{force}{area}$------(2)

$Force=\frac{power}{velocity}=\frac{1.4\times 1000}{3\times {10}^8}=4.66\times {10}^{-6}newton$

Put in equation (2)

${Pressure}_{absorbed}=\frac{4.66\times {10}^{-6}N}{m^2}=4.66\times {10}^{-6}Pa$