Chapter 30, Problem 41P

(a)

To determine

The expression for the force on the Earth due to the pressure of the radiation of the Earth by the Sun and compare this force of the Sun on Earth.

Answer to Problem 41P

The force on the Earth is $5.83\times {10}^8\text{N}$ , the gravitational force on the sun is $3.529\times {10}^{22}\text{N}$ and ratio for the radiation pressure on the earth to the gravitational force is $1.65\times {10}^{-14}$ .

Explanation of Solution

Given:

The intensity $I$ of the sunlight is $1.37\text{kW}/{\text{m}}^{\text{2}}$ .

Formula Used:

The expression for force acting on the Earth can be expressed by,

  $F_{\text{Earth}}=P_{\text{Earth}}\left(\pi R_{\text{Earth}}^2\right)$

The expression for the pressure on the Earth is given by,

  $P_{\text{Earth}}=\frac{I}{c}$

The expression for the gravitational force of the sun on the Sun on the Earth is given by,

  $F_{\text{g,Earth}}=\frac{GM{m}_{\text{Earth}}}{r^2}$

The expression for the ratio of the pressure on the Earth to the gravitational force on the Earth by the sun is given by,

  $\frac{F_{\text{Earth}}}{F_{g,\text{Earth}}}$

Calculation:

The expression to determine the force on the Earth is calculated as,

  $\begin{array}{c}{F}_{\text{Earth}}=P_{\text{Earth}}\left(\pi R_{\text{Earth}}^2\right)\\ =\frac{I\pi R_{\text{Earth}}^2}{c}\end{array}$

The force on the Earth is calculated as,

  $\begin{array}{c}{F}_{\text{Earth}}=\frac{I\pi R_{\text{Earth}}^2}{c}\\ =\frac{\left(1.37\text{kW}/{\text{m}}^{\text{2}}\left(\frac{{10}^3\text{W}/{\text{m}}^{\text{2}}}{\text{1}\text{kW}/{\text{m}}^{\text{2}}}\right)\right)\pi {\left(6.37\times {10}^6\text{m}\right)}^2}{2.998\times {10}^8\text{m}/\text{s}}\\ =\frac{\left(1.37\times {10}^3\text{W}/{\text{m}}^{\text{2}}\right)\pi {\left(6.37\times {10}^6\text{m}\right)}^2}{2.998\times {10}^8\text{m}/\text{s}}\\ =5.83\times {10}^8\text{N}\end{array}$

The gravitational force on the Sun is calculated as,

  $\begin{array}{c}{F}_{\text{g,Earth}}=\frac{GM{m}_{\text{Earth}}}{r^2}\\ =\frac{\left(6.67\times {10}^{11}\text{N}\cdot \text{m}/\text{kg}\right)\left(1.99\times {10}^{30}\text{kg}\right)\left(5.98\times {10}^{24}\text{kg}\right)}{{\left(1.50\times {10}^{11}\text{m}\right)}^2}\\ =3.529\times {10}^{22}\text{N}\end{array}$

The ratio for the radiation pressure on the earth to the gravitational force on the Earth by the is calculated as,

  $\begin{array}{c}\frac{F_{\text{Earth}}}{F_{\text{g,Earth}}}=\frac{5.83\times {10}^8\text{N}}{3.529\times {10}^{22}\text{N}}\\ =1.65\times {10}^{-14}\end{array}$

Conclusion:

Therefore, the force on the Earth is $5.83\times {10}^8\text{N}$ , the gravitational force on the sun is $3.529\times {10}^{22}\text{N}$ and ratio for the radiation pressure on the earth to the gravitational force is $1.65\times {10}^{-14}$ .

(b)

To determine

The force on the Mars due to the pressure of the radiation by Sun and compare the force of the Sun on the Earth.

Answer to Problem 41P

The force on the Mars is $7.18\times {10}^7\text{N}$ , the gravitational force on the sun is $1.68\times {10}^{21}\text{N}$ and ratio for the radiation pressure on the Mars to the gravitational force is $4.27\times {10}^{-14}$ .

Explanation of Solution

Given:

The distance of the Mars from the Sun is $2.28\times {10}^8\text{km}$ .

Formula used:

The expression to determine the value of the $I_{\text{Mars}}$ is given by,

  $I_{\text{Mars}}=I{\left(\frac{r_{\text{Earth}}}{r_{\text{Mars}}}\right)}^2$

The expression for force acting on the Mars can be expressed by,

  $F_{\text{Mars}}=P_{\text{Mars}}\left(\pi R_{\text{Mars}}^2\right)$

The expression for the pressure on the Mars is given by,

  $P_{\text{Mars}}=\frac{I_{\text{Mars}}}{c}$

The expression for the gravitational force of the sun on the Mars is given by,

  $F_{\text{g,Mars}}=\frac{GM{m}_{\text{Mars}}}{r^2}$

The expression for the ratio of the pressure on the Mars to the gravitational force on the Mars by the sun is given by,

  $\frac{F_{\text{Mars}}}{F_{\text{g,Mars}}}$

Calculation:

The value of the $I_{\text{Mars}}$ is calculated as,

  $\begin{array}{c}{I}_{\text{Mars}}=I{\left(\frac{r_{\text{Earth}}}{r_{\text{Mars}}}\right)}^2\\ =1.37\times {10}^3\text{W}/{\text{m}}^{\text{2}}{\left(\frac{1.50\times {10}^{11}\text{m}}{2.28\times {10}^{11}\text{m}}\right)}^2\\ =592.97\text{W}/{\text{m}}^{\text{2}}\end{array}$

The expression to determine the force on the Mars is calculated as,

  $\begin{array}{c}{F}_{\text{Mars}}=P_{\text{Mars}}\left(\pi R_{\text{Mars}}^2\right)\\ =\frac{I_{\text{Mars}}\pi R_{\text{Mars}}^2}{c}\end{array}$

The force on the Mars is calculated as,

  $\begin{array}{c}{F}_{\text{Mars}}=\frac{I_{\text{Mars}}\pi R_{\text{Mars}}^2}{c}\\ =\frac{\left(592.97\text{W}/{\text{m}}^{\text{2}}\right)\pi {\left(3.4\times {10}^6\text{m}\right)}^2}{2.998\times {10}^8\text{m}/\text{s}}\\ =7.18\times {10}^7\text{N}\end{array}$

The gravitational force on the Sun is calculated as,

  $\begin{array}{c}{F}_{\text{g,Mars}}=\frac{GM{m}_{\text{Mars}}}{r^2}\\ =\frac{\left(6.67\times {10}^{11}\text{N}\cdot \text{m}/\text{kg}\right)\left(1.99\times {10}^{30}\text{kg}\right)\left(5.98\times {10}^{24}\text{kg}\right)}{{\left(2.28\times {10}^8\text{km}\right)}^2}\\ =1.68\times {10}^{21}\text{N}\end{array}$

The ratio for the radiation pressure on the Mars to the gravitational force on the Mars by the is calculated as,

  $\begin{array}{c}\frac{F_{\text{Mars}}}{F_{\text{g,Mars}}}=\frac{7.18\times {10}^7\text{N}}{1.68\times {10}^{21}\text{N}}\\ =4.27\times {10}^{-14}\end{array}$

Conclusion:

Therefore, the force on the Mars is $7.18\times {10}^7\text{N}$ , the gravitational force on the sun is $1.68\times {10}^{21}\text{N}$ and ratio for the radiation pressure on the Mars to the gravitational force is $4.27\times {10}^{-14}$ .

(c)

To determine

The planet that has the larger ratio of the radiation pressure to the gravitational attraction.

Answer to Problem 41P

The planet Mars has the larger ratio of radiation force to the gravitational force as the mass of Mars is smaller than the mass of the Earth.

Explanation of Solution

Calculation:

The ratio for the radiation pressure on the earth to the gravitational force on the Earth is given by,

  $\frac{F_{\text{Earth}}}{F_{\text{g,Earth}}}=1.65\times {10}^{-14}$

The ratio for the radiation pressure on the Mars to the gravitational force on the Mars is given by,

  $\frac{F_{\text{Mars}}}{F_{\text{g,Mars}}}=4.27\times {10}^{-14}$

The planet Mars has the larger ratio of radiation force to the gravitational force as the mass of Mars is smaller than the mass of the Earth.

Conclusion:

Therefore, the planet Mars has the larger ratio of radiation force to the gravitational force as the mass of Mars is smaller than the mass of the Earth.