Question

Radiation from the Sun reaching Earth (just outside the atmosphere) has an intensity of $\mathbf{1}\mathbf{.}\mathbf{4}\mathbf{}{\mathbf{\text{kW/m}}}^{\mathbf{\text{2}}}$. (a) Assuming that Earth (and its atmosphere) behaves like a flat disk perpendicular to the Sun’s rays and that all the incident energy is absorbed, calculate the force on Earth due to radiation pressure. (b) For comparison, calculate the force due to the Sun’s gravitational attraction.

Short answer

1. Force on earth due to the radiation pressure is $6.0\times {10}^8\text{N}$.
2. Force due to the Sun’s gravitational attraction is $3.6\times {10}^{22}\text{N}$.

Step-by-step solution

Listing the given quantities

The intensity of radiation $\mathrm{I}=1.4{\text{kW/m}}^{\text{2}}=1.4\times {10}^3{\text{W/m}}^{\text{2}}$.

Understanding the concepts of radiation pressure and gravitational force

We first find the relation between the radiation pressure and force. Then substituting the value of radiation pressure and the area of the earth’s surface, we can calculate the force on the earth due to radiation pressure. Using the gravitational force of attraction, we can find the force between the sun and the earth.

(a) Calculations of the force on earth due to the radiation pressure

If the radiation is totally absorbed by the earth’s surface, the force is,

$\mathrm{F}=\frac{\mathrm{IA}}{\mathrm{c}}$

$\mathrm{I}$is the intensity of radiation, and $\mathrm{A}$is the area of the earth’s surface perpendicular to the path of the radiation.

The radiation pressure ${\mathrm{p}}_{\mathrm{r}}$is the force per unit area.

$\mathrm{F}={\mathrm{p}}_{\mathrm{r}}\mathrm{A}$ (1)

The radiation pressure ${\mathrm{p}}_{\mathrm{r}}$is the force per unit area.

${\mathrm{p}}_{\mathrm{r}}=\frac{\mathrm{I}}{\mathrm{c}}$

Substituting this value of radiation pressure in (1), we get,

$$\begin{gathered} F=\frac{IA}{C} \\ =\frac{I\pi {R_e}^2}{c} \end{gathered}$$

Substitute the values in the expression, and we get,

$$\begin{gathered} \mathrm{F}=\frac{\left(1.4\times {10}^3\frac{\text{W}}{{\text{m}}^{\text{2}}}\right)\mathrm{\pi }{\left(6.37\times {10}^6\text{m}\right)}^2}{3\times {10}^8\text{m/s}} \\ =6.0\times {10}^8\text{N} \end{gathered}$$

Thus, the force on earth due to the radiation pressure is $6.0\times {10}^8\text{N}$.

(b) Calculations of the force due to the Sun’s gravitational attraction

The gravitational force of attraction between the sun and the earth is,

$F_{grav}=\frac{G{M}_{sun}{M}_{earth}}{d_{es}^2}$

Here,

M_sun= 2x10³⁰ kg

M_earth= 5.98x10²⁴kg

The distance between the earth and the sun is l.

The gravitational constant is,$G=6.67\times {10}^{-11}\frac{\text{N}·{\text{m}}^{\text{2}}}{{\text{kg}}^{\text{2}}}$.

Substitute the values in the expression, and we get,

$$\begin{gathered} F_{grav}=\frac{6.67\times {10}^{-11}N{m}^2/K{g}^2}{{\left(1.5\times {10}^{11}m\right)}^2} \\ =3.6\times {10}^{22}N \end{gathered}$$

Thus, the force due to the Sun’s gravitational attraction is $3.6\times {10}^{22}\text{N}$.

This is greater than the force exerted by the radiation pressure.