Chapter 7, Problem 52PQ

(a)

To determine

The rate of change of gravitational force $F\left(r\right)$ with respect to the distance of separation $r$ between the masses.

Answer to Problem 52PQ

The rate of change of gravitational force $F\left(r\right)$ with respect to the distance of separation $r$ between the masses is $\underset{\_}{\frac{dF}{dr}=-2\frac{GMm}{r^3}}$.

Explanation of Solution

Write the expression for the gravitational force between two masses.

  $F\left(r\right)=\frac{GMm}{r^2}$ (I)

Here, $F\left(r\right)$ is the gravitational force, $G$ is the gravitational constant, $M$ is the mass of one object, $m$ is the mass of the other object, and $r$ is the distance of separation between the masses.

Differentiate equation (I) with respect to $r$ to find rate of change of gravitational force $F\left(r\right)$ with respect to the distance of separation $r$ between the masses.

  $\begin{array}{c}\frac{dF}{dr}=\frac{d}{dr}\left(\frac{GMm}{r^2}\right)\\ =\left(GMm\right)\frac{d}{dr}\left(\frac{1}{r^2}\right)\\ =GMm\left(\frac{-2}{r^3}\right)\\ =-2\frac{GMm}{r^3}\end{array}$ (II)

Conclusion:

Therefore, the rate of change of gravitational force $F\left(r\right)$ with respect to the distance of separation $r$ between the masses is $\underset{\_}{\frac{dF}{dr}=-2\frac{GMm}{r^3}}$.

(b)

To determine

The value of $dF\left(r\right)/dr$ between Sun and Earth, and that between Earth and Moon.

Answer to Problem 52PQ

For the Sun-Earth system $\underset{\_}{\frac{dF}{dr}=-4.7\times {10}^{11}\text{N/m}}$, and for Earth-Moon system $\underset{\_}{\frac{dF}{dr}=-10.8\times {10}^{11}\text{N/m}}$.

Explanation of Solution

The distance between the Sun and Earth is $1.5\times {10}^{11}\text{m}$, the mass of Sun is $2.0\times {10}^{30}\text{kg}$, the mass of Earth is $6.0\times {10}^{24}\text{kg}$, the distance between Earth and Moon is $3.8\times {10}^8\text{m}$, and the mass of moon is $7.4\times {10}^{22}\text{kg}$.

Equation (II) gives the expression for rate of change of gravitational force with respect to the distance of separation.

  $\frac{dF}{dr}=-2\frac{GMm}{r^3}$

Conclusion:

Substitute $6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}{\text{/kg}}^{\text{2}}$ for $G$, $2.0\times {10}^{30}\text{kg}$ for $M$, $6.0\times {10}^{24}\text{kg}$ for $m$, and $1.5\times {10}^{11}\text{m}$ for $r$ in equation (II) to find $dF\left(r\right)/dr$ for Sun-Earth system.

  $\begin{array}{c}{\left(\frac{dF}{dr}\right)}_{\text{Sun-Earth}}=-2\frac{\left(6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)\left(2.0\times {10}^{30}\text{kg}\right)\left(6.0\times {10}^{24}\text{kg}\right)}{{\left(1.5\times {10}^{11}\text{m}\right)}^3}\\ =-4.7\times {10}^{11}\text{N/m}\end{array}$

Substitute $6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}{\text{/kg}}^{\text{2}}$ for $G$, $6.0\times {10}^{24}\text{kg}$ for $M$, $7.4\times {10}^{22}\text{kg}$ for $m$, and $3.8\times {10}^8\text{m}$ for $r$ in equation (II) to find $dF\left(r\right)/dr$ for Earth-Moon system.

  $\begin{array}{c}{\left(\frac{dF}{dr}\right)}_{\text{Earth-Moon}}=-2\frac{\left(6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)\left(6.0\times {10}^{24}\text{kg}\right)\left(7.4\times {10}^{22}\text{kg}\right)}{{\left(3.8\times {10}^8\text{m}\right)}^3}\\ =-10.8\times {10}^{11}\text{N/m}\end{array}$

Therefore, for the Sun-Earth system $\underset{\_}{\frac{dF}{dr}=-4.7\times {10}^{11}\text{N/m}}$, and for Earth-Moon system $\underset{\_}{\frac{dF}{dr}=-10.8\times {10}^{11}\text{N/m}}$.

(c)

To determine

When the Earth-Moon distance remains the same, but the Earth is moved closer to the Sun, whether there is any point where $dF\left(r\right)/dr$ for the two forces has same value.

Answer to Problem 52PQ

When the Earth-Moon distance remains the same, but the Earth is moved closer to the Sun, there is a point $\underset{\_}{r=1.1\times {10}^{11}\text{m}}$, where $dF\left(r\right)/dr$ for the two forces has same value.

Explanation of Solution

Given that the value of $dF\left(r\right)/dr$ is $-10.8\times {10}^{11}\text{N/m}$

The position corresponding to the condition when the Earth is moved closer to the Sun, but $dF\left(r\right)/dr$ for the two forces has same value can be computed by solving equation (II) for $r$.

  $r=\sqrt[3]{-2\frac{GMm}{\left(\frac{dF}{dr}\right)}}$ (III)

Conclusion:

Substitute $6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}{\text{/kg}}^{\text{2}}$ for $G$, $2.0\times {10}^{30}\text{kg}$ for $M$, $6.0\times {10}^{24}\text{kg}$ for $m$, $1.5\times {10}^{11}\text{m}$ for $r$, and $-10.8\times {10}^{11}\text{N/m}$ for $dF\left(r\right)/dr$ in equation (III) to find $r$.

  $\begin{array}{c}r=\sqrt[3]{-2\frac{\left(6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)\left(2.0\times {10}^{30}\text{kg}\right)\left(6.0\times {10}^{24}\text{kg}\right)}{\left(-10.8\times {10}^{11}\text{N/m}\right)}}\\ =1.1\times {10}^{11}\text{m}\end{array}$

Therefore, when the Earth-Moon distance remains the same, but the Earth is moved closer to the Sun, there is a point $\underset{\_}{r=1.1\times {10}^{11}\text{m}}$, where $dF\left(r\right)/dr$ for the two forces has same value.