Question

Radiation from the Sun reaching Earth (just outside the atmosphere) has an intensity of 1.39 kW/m2. (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. Assume that the speed of light and Earth radius are 2.998 × 108 m/s and 6.37 thousand km respectively.

Answer 1

Answer:

(a) 5.91 × 10⁸ N; (b) 3.542 × 10²² N

Explanation:

Data:

  I = 1.39 kW/m² = 1.39 × 10³ N·m⁻³s⁻¹

 c = 2.998 × 10⁸ m/s  

 r = 6370 km = 6.37 × 10⁶ m

G  =  6.674 ×10⁻¹¹ N⋅m²kg⁻²

M = 1.989 × 10³⁰ kg  = mass of Sun

m = 5.972 × 10²⁴ kg = mass of Earth

d = 1.496 × 10¹¹ m     = distance from Earth to Sun

Calculations:  

(a) Force exerted by the radiation pressure

(i) Radiation pressure

All the incident radiation is absorbed, so

$\begin{array}{rcl}p & = & \dfrac{I}{c}\\\\& = & \dfrac{\text{1.39 \times 10 ^{3} N \cdot m ^{-3} s ^{-1} }}{2.998 \times 10^{8}\text{ m \cdot s}^{-1}}\\\\& = & 4.636 \times 10^{-6}\text{ N \cdot m}^{-2}\\\end{array}$

(ii) Area of Earth's disc

$\begin{array}{rcl}A & = & \pi r^{2}\\& = & \pi \times (6.37 \times 10^{6} \text{ m})^{2}\\& = & 1.275 \times 10^{14} \text{ m}^{2}\\\end{array}$

(iii) Radiation force

$\begin{array}{rcl}p & = & \dfrac{F}{A}\\\\F & = & pA\\ & = & 4.636 \times 10^{-6}\text{ N \cdot m}^{-2} \times 1.275 \times 10^{14} \text{ m}^{2}\\ & = & 5.91 \times 10^{8}\text{ N}\\\end{array}$

(b) Gravitational Force

$\begin{array}{rcl}F & = & \dfrac{GMm}{r^{2}}\\\\& = & \dfrac{\text{6.674 \times 10 ^{-11} N \cdot m ^{2} kg ^{-2}\times 1.989 \times 10 ^{30} kg \times 5.972 \times 10 ^{24} kg}}{(\text{1.496 \times 10 ^{11} m})^{2}}\\\\& = & 3.542 \times 10^{22} \text{ N}\\\end{array}$

Answer 2

Answer:

$F=5.8\times 10^{8}\ N$

$F=35.57\times 10^{21}\ N$

Explanation:

Given that

Intensity I

$I= 1.39\ KW/m^2$

$Speed\ of \ light = 2.99\times 10^8\ m/s$

Radius of earth,R = 6370 Km

We know that surface area of earth, A

$A=\pi R^2$

$A=\pi (6370\times 10^3)^2\ m^2$

$A=1.27\times 10^{14}\ m^2$

As we know that pressure due to intensity given as

$P=\dfrac{I}{V}$

V =Velocity of light

$V=3\times 10^8\ m/s$

$P=\dfrac{1.39\times 1000}{=3\times 10^8}$

$P=4.6\times 10^{-6}\ Pa$

We know that force F

F = P .A

$F=4.6\times 10^{-6}\times 1.27\times 10^{14}\ N$

$F=5.8\times 10^{8}\ N$

b)Gravitational force F

$F=G\dfrac{m.M}{r^2}$

$M = mass\ of\ sun = 2\times 10^{30} kg\\m = mass\ of\ earth = 6\times 10^{24}kg$

$r =1.5\times 10^{11}\ m$

$G =6.67\times 10^{-11}Nm^2/kg^2$

So F

$F=6.67\times 10^{-11}\times \dfrac{2\times 10^{30}\times 6\times 10^{24}kg}{1.5\times 10^{11}}$

$F=35.57\times 10^{21}\ N$