Chapter 34, Problem 42P

(a)

To determine

The intensity of solar radiation incident on Mars.

Answer to Problem 42P

The intensity of solar radiation incident on Mars is $0.59\text{W}/{\text{m}}^{\text{2}}$.

Explanation of Solution

Given info: The intensity of solar radiation incident on the Earth is $1.370\text{W}/{\text{m}}^{\text{2}}$

Write the formula to calculate the power of the sun radiation on the Earth is,

$P_{\text{E}}=I_{\text{E}}\left(4\pi r_{\text{E}}{}^2\right)$

Here,

$P_{\text{E}}$ is the power of the sun radiation on the Earth.

$I_{\text{E}}$ is the intensity of solar radiation incident on the Earth.

$r_{\text{E}}$ is the distance of the Earth from the Sun.

Write the formula to calculate the power of the sun radiation on the Mars is,

$P_{\text{M}}=I_{\text{M}}\left(4\pi r_{\text{M}}{}^2\right)$

Here,

$P_{\text{M}}$ is the power of the sun radiation on the Mars.

$I_{\text{M}}$ is the intensity of solar radiation incident on the Mars.

$r_{\text{M}}$ is the distance of the Mars from the Sun.

The power of the sun radiation is equal in all the planets.

$P_{\text{M}}=P_{\text{E}}$

Substitute $I_{\text{E}}\left(4\pi r_{\text{E}}{}^2\right)$ for $P_{\text{E}}$ and $I_{\text{M}}\left(4\pi r_{\text{M}}{}^2\right)$ for $P_{\text{M}}$ in the above equation.

$I_{\text{M}}\left(4\pi r_{\text{M}}{}^2\right)=I_{\text{E}}\left(4\pi r_{\text{E}}{}^2\right)$

Rearrange the above expression for $I_{\text{M}}$.

$I_{\text{M}}=\frac{I_{\text{E}}{r}_{\text{E}}{}^2}{r_{\text{M}}{}^2}$

Substitute $1.370\text{W}/{\text{m}}^{\text{2}}$ for $I_{\text{E}}$, $1.496\times {10}^{11}\text{m}$ for $r_{\text{E}}$ and $2.28\times {10}^{11}\text{m}$ for $r_{\text{M}}$ in the above equation to find the value of $I_{\text{M}}$.

$\begin{array}{c}{I}_{\text{M}}=\frac{\left(1.370\text{W}/{\text{m}}^{\text{2}}\right){\left(1.496\times {10}^{11}\text{m}\right)}^2}{{\left(2.28\times {10}^{11}\text{m}\right)}^2}\\ =0.59\text{W}/{\text{m}}^{\text{2}}\end{array}$

Conclusion:

Therefore, the intensity of solar radiation incident on Mars is $0.59\text{W}/{\text{m}}^{\text{2}}$.

(b)

To determine

The power of the sun radiation incident on the Mars.

Answer to Problem 42P

The power of the sun radiation incident on the Mars is $3.853\times {10}^{23}\text{W}$.

Explanation of Solution

Given info: The intensity of solar radiation incident on the Earth is $1.370\text{W}/{\text{m}}^{\text{2}}$

Write the formula to calculate the power of the sun radiation on the Mars is,

$P_{\text{M}}=I_{\text{M}}\left(4\pi r_{\text{M}}{}^2\right)$

Here,

$P_{\text{M}}$ is the power of the sun radiation on the Mars.

$I_{\text{M}}$ is the intensity of solar radiation incident on the Mars.

$r_{\text{M}}$ is the distance of the Mars from the Sun.

Substitute $2.28\times {10}^{11}\text{m}$ for $r_{\text{M}}$ and $0.59\text{W}/{\text{m}}^{\text{2}}$ for $I_{\text{M}}$ in the above equation to find the value of $P_{\text{M}}$.

$\begin{array}{c}{P}_{\text{M}}=\left(0.59\text{W}/{\text{m}}^{\text{2}}\right)\left(4\pi \right){\left(2.28\times {10}^{11}\text{m}\right)}^2\\ =3.853\times {10}^{23}\text{W}\end{array}$

Conclusion:

Therefore, the power of the sun radiation incident on the Mars is $3.853\times {10}^{23}\text{W}$.

(c)

To determine

The radiation force that acts on Mars.

Answer to Problem 42P

The radiation force that acts on Mars is $2.807\times {10}^5\text{N}$.

Explanation of Solution

Given info: The intensity of solar radiation incident on the Earth is $1.370\text{W}/{\text{m}}^{\text{2}}$.

Write the formula to calculate the radiation force that acts on Mars is,

$F_{\text{M}}=\frac{4\pi I_{\text{M}}{R}_{\text{M}}{}^2}{c}$

Here,

$F_{\text{M}}$ is radiation force that acts on Mars.

$I_{\text{M}}$ is intensity of solar radiation incident on Mars.

$R_{\text{M}}$ is the radius of the Mars.

$c$ is the speed of light.

Substitute $3.37\times {10}^6\text{m}$ for $R_{\text{M}}$, $3\times {10}^8\text{m}/\text{s}$ for $c$ and $0.59\text{W}/{\text{m}}^{\text{2}}$ for $I_{\text{M}}$ in the above equation to find the value of $F_{\text{M}}$.

$\begin{array}{c}{F}_{\text{M}}=\frac{4\pi \left[0.59\text{W}/{\text{m}}^{\text{2}}\times \left(\frac{1\text{N}/\text{m}\cdot \text{s}}{1\text{W}/{\text{m}}^{\text{2}}}\right)\right]{\left(3.37\times {10}^6\text{m}\right)}^2}{3\times {10}^8\text{m}/\text{s}}\\ =2.807\times {10}^5\text{N}\end{array}$

Conclusion:

Therefore, the radiation force that acts on Mars is $2.807\times {10}^5\text{N}$.

(d)

To determine

The comparison of the gravitational attraction exerted by the Sun on Mars with the radiation force that acts on Mars.

Answer to Problem 42P

The gravitational force exerted on the Mars is $5.84\times {10}^{15}$ times the radiation force that acts on Mars.

Explanation of Solution

Given info: The intensity of solar radiation incident on the Earth is $1.370\text{W}/{\text{m}}^{\text{2}}$

Write the formula to calculate the gravitational force exerted on the Mars is,

$F_{\text{gM}}=\frac{GM{m}_{\text{M}}}{r_{\text{M}}{}^2}$

Here,

$F_{\text{gM}}$ is the gravitational force exerted on the Mars.

$M$ is the mass of the Sun.

$m_{\text{M}}$ is the mass of the Mars.

$G$ is the gravitational constant.

Substitute $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}/{\text{kg}}^{\text{2}}$ for $G$, $1.99\times {10}^{30}\text{kg}$ for $M$, $6.42\times {10}^{23}\text{kg}$ for $m_{\text{M}}$ and $2.28\times {10}^{11}\text{m}$ for $r_{\text{M}}$ in the above equation to find the value of $F_{\text{g}}$.

$\begin{array}{c}{F}_{\text{gM}}=\frac{\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}/{\text{kg}}^{\text{2}}\right)\left(1.99\times {10}^{30}\text{kg}\right)\left(6.42\times {10}^{23}\text{kg}\right)}{{\left(2.28\times {10}^{11}\text{m}\right)}^2}\\ =16.39\times {10}^{20}\text{N}\end{array}$

Thus the gravitational force exerted on the Mars is $16.39\times {10}^{20}\text{N}$.

The ratio of gravitational force exerted on the Mars to the radiation force that acts on Mars is,

${\left(\text{Ratio}\right)}_1=\frac{F_{\text{gM}}}{F_{\text{M}}}$

Substitute $16.39\times {10}^{20}\text{N}$ for $F_{\text{gM}}$ and $2.807\times {10}^5\text{N}$ for $F_{\text{M}}$ in the above equation to find the value of ${\left(\text{Ratio}\right)}_1$

$\begin{array}{c}{\left(\text{Ratio}\right)}_1=\frac{16.39\times {10}^{20}\text{N}}{2.807\times {10}^5\text{N}}\\ =5.84\times {10}^{15}\end{array}$

Thus the gravitational force exerted on the Mars is $5.84\times {10}^{15}$ times the radiation force that acts on Mars.

Conclusion:

Therefore, the gravitational force exerted on the Mars is $5.84\times {10}^{15}$ times the radiation force that acts on Mars.

(e)

To determine

The comparison of the ratio of the gravitational force exerted by the Sun on Earth to the radiation force that acts on Earth with the ratio found in part (d).

Answer to Problem 42P

The ratio for the Earth is greater than the ratio of for the Mars.

Explanation of Solution

Given info: The intensity of solar radiation incident on the Earth is $1.370\text{W}/{\text{m}}^{\text{2}}$

Write the formula to calculate the radiation force that acts on Earth is,

$F_{\text{E}}=\frac{4\pi I_{\text{E}}{R}_{\text{E}}{}^2}{c}$

Here,

$F_{\text{E}}$ is radiation force that acts on Earth.

$I_{\text{E}}$ is intensity of solar radiation incident on Earth.

$R_{\text{E}}$ is the radius of the Earth.

$c$ is the speed of light.

Substitute $6.37\times {10}^6\text{m}$ for $R_{\text{E}}$, $3\times {10}^8\text{m}/\text{s}$ for $c$ and $1.370\text{W}/{\text{m}}^{\text{2}}$ for $I_{\text{E}}$ in the above equation to find the value of $F_{\text{E}}$.

$\begin{array}{c}{F}_{\text{E}}=\frac{4\pi \left[1.370\text{W}/{\text{m}}^{\text{2}}\times \left(\frac{1\text{N}/\text{m}\cdot \text{s}}{1\text{W}/{\text{m}}^{\text{2}}}\right)\right]{\left(6.37\times {10}^6\text{m}\right)}^2}{3\times {10}^8\text{m}/\text{s}}\\ =2.33\times {10}^6\text{N}\end{array}$

Thus the radiation force that acts on Earth is $2.33\times {10}^6\text{N}$.

Write the formula to calculate the gravitational force exerted on the Earth is,

$F_{\text{gE}}=\frac{GM{m}_{\text{E}}}{r_{\text{E}}{}^2}$

Here,

$F_{\text{gE}}$ is the gravitational force exerted on the Earth.

$M$ is the mass of the Earth.

$m_{\text{E}}$ is the mass of the Earth.

$G$ is the gravitational constant.

Substitute $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}/{\text{kg}}^{\text{2}}$ for $G$, $1.99\times {10}^{30}\text{kg}$ for $M$, $5.98\times {10}^{24}\text{kg}$ for $m_{\text{E}}$ and $1.496\times {10}^{11}\text{m}$ for $r_{\text{E}}$ in the above equation to find the value of $F_{\text{g}}$.

$\begin{array}{c}{F}_{\text{gE}}=\frac{\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}/{\text{kg}}^{\text{2}}\right)\left(1.99\times {10}^{30}\text{kg}\right)\left(5.98\times {10}^{24}\text{kg}\right)}{{\left(1.496\times {10}^{11}\text{m}\right)}^2}\\ =35.46\times {10}^{21}\text{N}\end{array}$

Thus the gravitational force exerted on the Earth is $35.46\times {10}^{21}\text{N}$.

The ratio of gravitational force exerted on the Earth to the radiation force that acts on earth is,

${\left(\text{Ratio}\right)}_2=\frac{F_{\text{gE}}}{F_{\text{E}}}$

Substitute $35.46\times {10}^{21}\text{N}$ for $F_{\text{gE}}$ and $2.33\times {10}^6\text{N}$ for $F_{\text{E}}$ in the above equation to find the value of ${\left(\text{Ratio}\right)}_2$

$\begin{array}{c}{\left(\text{Ratio}\right)}_2=\frac{35.46\times {10}^{21}\text{N}}{2.33\times {10}^6\text{N}}\\ =15.22\times {10}^{15}\end{array}$

Thus the gravitational force exerted on the Earth is $15.22\times {10}^{15}$ times the radiation force that acts on Earth.

Thus, the ratio for the Earth is greater than the ratio of for the Mars.

Conclusion:

Therefore, the ratio for the Earth is greater than the ratio of for the Mars.