Question

Radiation from the Sun reaching Earth (just outside the atmosphere) has an intensity of $1.4{\text{kW/m}}^2$. If $F_R$ is the force on Earth due to radiation pressure and $F_g$ is the force due to the Suns gravitational attraction, then select the correct option (Assuming that Earth (and its atmosphere) behaves like a flat disk perpendicular to the Suns rays and that all the incident energy is absorbed)
$[\text{Take}{\text{R}}_{earth}=6400\text{km},M_{earth}=5.98\times {10}^{24}\text{kg},M_{sun}=2\times {10}^{30}\text{kg},\text{Distance between the centres of Sun and Earth}=1.5\times {10}^8\text{km}]$

• A. $F_R=F_g$
• B. $F_R<<F_g$
• C. $F_R=6\times {10}^8\text{N}$
• D. $F_R=3.6\times {10}^{22}\text{N}$

Answer 1

Answer: (B), (C)

$F_R=6\times {10}^8\text{N}$

The radiation pressure produces a force equal to,

$F_R=p_R\left(\pi R_{earth}^2\right)=\frac{I}{c}\times \left(\pi R_{earth}^2\right)$

$\therefore F_R=\frac{(1.4\times {10}^3)\times [\pi \times (6400\times {10}^3{)}^2]}{3\times {10}^8}=6\times {10}^8\text{N}$

The gravitational pull of the Sun on the Earth is,

$F_g=\frac{G{M}_{earth}{M}_{Sun}}{d_{se}^2}$

$\therefore F_g=\frac{(6.67\times {10}^{-11})\times (5.98\times {10}^{25})\times (2\times {10}^{30})}{(1.5\times {10}^{11}{)}^2}=3.6\times {10}^{22}\text{N}$

Hence, $\left(B\right)$ and $\left(C\right)$ are the correct answers.