Chapter 7.2, Problem 18SSC

(a)

To determine

The ratio of the gravitational fields due to Earth and the Sun at the center of the Moon.

Answer to Problem 18SSC

  $\frac{{\text{g}}_S}{{\text{g}}_E}=2.26$

Explanation of Solution

Given:

The distance between the Moon and the Earth’s center is, $r_E=3.9\times {10}^5\text{ km}=3.9\times {10}^8\text{ m}$

The distance between the Earth and the Sun’s center is, ${\text{r}}_S=14.96\times {10}^7\text{ km = }14.96\times {10}^{10}\text{ m}$

Mass of the Earth is, ${\text{m}}_E=5.97\times {10}^{24}\text{ kg}$

Mass of the Sun is, ${\text{m}}_S=1.99\times {10}^{30}\text{ kg}$

During a full moon, the Sun, Earth, and the Moon are in line with each other.

  

Figure 1

Formula used:

Gravitational field due to Earth is, ${\text{g}}_E=\frac{G{m}_E}{r_E^2}$$\left(1\right)$

Where, $G$ is gravitational constant, $G=6.67\times {10}^{-11}{\text{ Nm}}^2/{\text{kg}}^2$ $m_E$ is the mass of the Earth $r_E$ is the distance between the Moon and the Earth’s center

Gravitational field due to the Sun is, ${\text{g}}_S=\frac{G{m}_S}{r_S^2}$$\left(2\right)$

Where, $G$ is gravitational constant, $G=6.67\times {10}^{-11}{\text{ Nm}}^2/{\text{kg}}^2$ $m_S$ is mass of the Sun $r_S$ is the distance between the Earth and the Sun’s center

The ratio of the gravitational fields due to Earth and the Sun at the center of the Moon is

  $\frac{{\text{g}}_S}{{\text{g}}_E}=\frac{\left(\frac{G{m}_S}{r_S^2}\right)}{\left(\frac{G{m}_E}{r_E^2}\right)}=\frac{m_S{r}_E^2}{m_E{r}_S^2}$$\left(3\right)$

Calculation:

Substituting the numerical values in equation $\left(3\right)$ ,

  $\frac{{\text{g}}_S}{{\text{g}}_E}=\frac{\left(1.99\times {10}^{30}\text{ kg}\right){\left(3.9\times {10}^8\text{ m}\right)}^2}{\left(5.97\times {10}^{24}\text{ kg}\right){\left(14.96\times {10}^{10}\text{ m}\right)}^2}$

  $=\frac{\left(1.99\times {10}^{30}\text{ kg}\right)\left(15.21\times {10}^{16}{\text{ m}}^2\right)}{\left(5.97\times {10}^{24}\text{ kg}\right)\left(2.23\times {10}^{22}{\text{ m}}^2\right)}$

  $=\frac{29.81\times {10}^{46}\text{ kg}\cdot {\text{km}}^2}{13.31\times {10}^{46}\text{ kg}\text{.}\cdot {\text{m}}^2}=2.26$

Conclusion:

The ratio of the gravitational fields due to Earth and the Sun at the center of the Moon is $2.26$ .

(b)

To determine

The net gravitational field due to the Sun and Earth at the center of the Moon.

Answer to Problem 18SSC

  ${\text{g}}_{net}\text{= 8}\text{.5}\times {10}^{-3}\text{ N/kg}$

Explanation of Solution

Given:

The distance between the Moon and the Earth’s center is, $r_E=3.9\times {10}^5\text{ km}=3.9\times {10}^8\text{ m}$

The distance between the Earth and the Sun’s center is, ${\text{r}}_S=14.96\times {10}^7\text{ km = }14.96\times {10}^{10}\text{ m}$

Mass of the Earth is, ${\text{m}}_E=5.97\times {10}^{24}\text{ kg}$

Mass of the Sun is, ${\text{m}}_S=1.99\times {10}^{30}\text{ kg}$

During a full moon, the Sun, Earth, and the Moon are in line with each other.

Hence the distance between theMoon and the Sun’s center is, ${\text{r}}_{SM}=\left(14.96\times {10}^{10}\text{+}3.9\times {10}^8\right)\text{m = }\left(14.96\times {10}^{10}\text{+}0.039\times {10}^{10}\right)\text{ m = 14}\text{.99}\times {10}^{10}\text{ m}$

Formula used:

Gravitational field due to Earth at the center of the Moon is,

  ${\text{g}}_E=\frac{G{m}_E}{r_E^2}$$\left(4\right)$

Where, $G$ is gravitational constant, $G=6.67\times {10}^{-11}{\text{ Nm}}^2/{\text{kg}}^2$ $m_E$ is mass of the Earth $r_E$ is the distance between the Moon and the Earth’s center

Gravitational field due to the Sun at the center of the Moon is,

  ${\text{g}}_S=\frac{G{m}_S}{r_{SM}^2}$$\left(5\right)$

Where, $G$ is gravitational constant, $G=6.67\times {10}^{-11}{\text{ Nm}}^2/{\text{kg}}^2$ $m_S$ is mass of the Sun $r_{SM}$ is the distance between the Moon and the Sun’s center

Calculation:

Gravitational field due to Earth at the center of the Moon can be calculated by substituting the numerical values in equation $\left(4\right)$ ,

  ${\text{g}}_E=\frac{\left(6.67\times {10}^{-11}{\text{ Nm}}^2/{\text{kg}}^2\right)\left(5.97\times {10}^{24}\text{ kg}\right)}{{\left(3.9\times {10}^8\text{ m}\right)}^2}$

  $=\frac{39.81\times {10}^{13}{\text{Nm}}^2/\text{kg}}{15.21\times {10}^{16}{\text{ m}}^2}$

  $=2.6\times {10}^{-3}\text{ N/kg}$

Gravitational field due to the Sun at the center of the Moon can be calculated by substituting the numerical values in equation $\left(5\right)$ ,

  ${\text{g}}_S=\frac{\left(6.67\times {10}^{-11}{\text{ Nm}}^2/{\text{kg}}^2\right)\left(1.99\times {10}^{30}\text{ kg}\right)}{{\left(\text{14}\text{.99}\times {10}^{10}\text{ m}\right)}^2}$

  $=\frac{13.27\times {10}^{19}{\text{ Nm}}^2/\text{kg}}{2.24\times {10}^{22}{\text{ m}}^2}$

  $=5.9\times {10}^{-3}\text{ N/kg}$

The net gravitational field due to the Sun and Earth at the center of the Moon is,

  ${\text{g}}_{net}=g_E+g_S$

  $=2.6\times {10}^{-3}\text{ N/kg}+5.9\times {10}^{-3}\text{ N/kg}$

  $\text{= 8}\text{.5}\times {10}^{-3}\text{ N/kg}$

Conclusion:

The net gravitational field due to the Sun and Earth at the center of the Moon is $\text{ 8}\text{.5}\times {10}^{-3}\text{ N/kg}$ .