Chapter 6, Problem 57P

To determine

The speed of the rock.

Answer to Problem 57P

The speed of the rock is $\underset{\_}{3.3\times {10}^4\text{m}/\text{s}}$.

Explanation of Solution

The total energy of the system is conserved. That is the initial and final energies are equal.

    $\frac{1}{2}{m}_{\text{rock}}{v}_i^2-\frac{G{M}_{\text{Earth}}{m}_{\text{rock}}}{d_{\text{Earth}}}-\left(-\frac{G{M}_{\text{moon}}{m}_{\text{rock}}}{d_{\text{moon}}}\right)=\frac{1}{2}{m}_{\text{rock}}{v}_f^2-\frac{G{M}_{\text{Earth}}{m}_{\text{rock}}}{h}-\left(-\frac{G{M}_{\text{moon}}{m}_{\text{rock}}}{R-h}\right)$        (I)

Here, $m_{\text{rock}}$ is the mass of the rock, $G$ is the gravitational constant, $v_i$ is the initial speed of the rock. $v_f$ is the final speed of the rock, $M_{\text{Earth}}$ is the mass of Earth, $M_{\text{moon}}$ is the mass of the moon, $R$ is the distance between Earth and moon, $h$ is the height of the rock above the Earth, $d_{\text{Earth}}$ is the distance to the Earth, and $d_{\text{moon}}$ is the distance to the moon.

Add the term $\frac{G{M}_{\text{Earth}}{m}_{\text{rock}}}{h}$ in both sides of the equation (I), substrate $\frac{G{M}_{\text{moon}}{m}_{\text{rock}}}{R-h}$ from the both sides, and substitute, $0$ for $v_i$.

    $\begin{array}{c}\frac{1}{2}{m}_{\text{rock}}{v}_f^2=\frac{1}{2}{m}_{\text{rock}}\times 0-\frac{G{M}_{\text{Earth}}{m}_{\text{rock}}}{d_{\text{Earth}}}+\frac{G{M}_{\text{moon}}{m}_{\text{rock}}}{d_{\text{moon}}}-\frac{G{M}_{\text{moon}}{m}_{\text{rock}}}{R-h}+\frac{G{M}_{\text{Earth}}{m}_{\text{rock}}}{h}\\ v_f^2=2G\left(\frac{M_{\text{moon}}}{d_{\text{moon}}}-\frac{M_{\text{Earth}}}{d_{\text{Earth}}}-\frac{M_{\text{moon}}}{R-h}+\frac{M_{\text{Earth}}}{h}\right)\end{array}$        (II)

Given that, the gravitational forces on the rock from Earth and Moon are equal in magnitude and opposite direction.

    $\begin{array}{c}-\frac{G{M}_{\text{Earth}}{m}_{\text{rock}}}{d_{\text{Earth}}^2}=-\frac{G{M}_{\text{Moon}}{m}_{\text{rock}}}{d_{\text{Moon}}^2}\\ \frac{d_{\text{Earth}}^2}{M_{\text{Earth}}}=\frac{d_{\text{Moon}}^2}{M_{\text{Moon}}}\end{array}$        (III)

Multiply both sides if equation (III) by $\frac{M_{\text{Moon}}}{d_{\text{Moon}}^2}$.

    $\begin{array}{c}\frac{d_{\text{Earth}}^2}{M_{\text{Earth}}}\frac{M_{\text{Moon}}}{d_{\text{Moon}}^2}=\frac{d_{\text{Moon}}^2}{M_{\text{Moon}}}\frac{M_{\text{Moon}}}{d_{\text{Moon}}^2}\\ d_{\text{Earth}}=d_{\text{Moon}}\sqrt{\frac{M_{\text{Earth}}}{M_{\text{Moon}}}}\end{array}$        (IV)

The distance between the Earth and moon is equal to the distance from the rock to the Earth plus the distance between the rock and the Moon.

    $R=d_{\text{Earth}}+d_{\text{Moon}}$        (V)

Use equation (IV) in (V).

    $\begin{array}{c}R=d_{\text{Moon}}\sqrt{\frac{M_{\text{Earth}}}{M_{\text{Moon}}}}+d_{\text{Moon}}\\ =d_{\text{Moon}}\left(1+\sqrt{\frac{M_{\text{Earth}}}{M_{\text{Moon}}}}\right)\end{array}$        (VI)

Similarly, $R$ can be written as,

    $\begin{array}{c}R=d_{\text{Earth}}\sqrt{\frac{M_{\text{Earth}}}{M_{\text{Moon}}}}+d_{\text{Earth}}\\ =d_{\text{Earth}}\left(1+\sqrt{\frac{M_{\text{Earth}}}{M_{\text{Moon}}}}\right)\end{array}$        (VII)

Solve the equation (VII).

    $d_{\text{Earth}}=\frac{R}{1+\sqrt{\frac{M_{\text{Earth}}}{M_{\text{Moon}}}}}$        (VIII)

Similarly $d_{\text{moon}}$ can be written as,

    $d_{\text{Moon}}=\frac{R}{1+\sqrt{\frac{M_{\text{Earth}}}{M_{\text{Moon}}}}}$        (IX)

Conclusion:

Substitute, $5.974\times {10}^{24}\text{kg}$ for $M_{\text{Earth}}$, $7.349\times {10}^{22}\text{kg}$ for $M_{\text{moon}}$, and $3.845\times {10}^8\text{m}$ for $R$ in the equation (VIII).

    $\begin{array}{c}{d}_{\text{Moon}}=\frac{3.845\times {10}^8\text{m}}{1+\sqrt{\frac{5.974\times {10}^{24}\text{kg}}{7.349\times {10}^{22}\text{kg}}}}\\ =0.3845\times {10}^8\text{m}\end{array}$

Substitute, $5.974\times {10}^{24}\text{kg}$ for $M_{\text{Earth}}$, $7.349\times {10}^{22}\text{kg}$ for $M_{\text{moon}}$, $6.674\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2$ for $G$, $0.3845\times {10}^8\text{m}$ for $d_{\text{Moon,}}{d}_{\text{Earth}}$, $7.2\times {10}^5\text{m}$ for $h$, and $3.845\times {10}^8\text{m}$ for $R$ in the equation (II).

    $\begin{array}{c}{v}_f^2=2\left(6.674\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\\ \times \left(\frac{7.349\times {10}^{22}\text{kg}}{0.3845\times {10}^8\text{m}}-\frac{5.974\times {10}^{24}\text{kg}}{0.3845\times {10}^8\text{m}}-\frac{7.349\times {10}^{22}\text{kg}}{3.845\times {10}^8\text{m}-7.2\times {10}^5\text{m}}+\frac{5.974\times {10}^{24}\text{kg}}{7.2\times {10}^5\text{m}}\right)\\ =3.3\times {10}^4\text{m}/\text{s}\end{array}$

Therefore, the speed of the rock is $\underset{\_}{3.3\times {10}^4\text{m}/\text{s}}$.