Chapter 11, Problem 103P

(a)

To determine

ToShow: That the potential energy shared by an element of the rod of mass $dm$ and the point particle of mass $m_0$ located at $x_0\ge \frac{1}{2}L$ is given by:

  $\begin{array}{l}dU=-\frac{G{m}_{0}dm}{x_0-x_s}\\ dU=-\frac{GM{m}_0}{L\left(x_0-x_s\right)}d{x}_s\end{array}$

Explanation of Solution

Given information :

Mass of point particle $=m_0$

Formula used :

Gravitational potential energy

  $U=-\frac{G{m}_{o}m}{r}$

U is gravitational potential energy, G is the gravitational constant, m is the mass of large body and $m_0$ is the mass of the other body.

Calculation:

Let $U=0$ at $x=\infty$ .

The potential energy of an element of the stick $dm$ and the point mass $m_0$ is given by the definition of gravitational potential energy.

  $U=-\frac{G{m}_{o}m}{r}$

Where r is the separation of $dm$ and $m_0$ .

  $dU=\frac{-G{m}_{0}dm}{x_0-x_s}$

  $dm=\lambda d{x}_o$

  $\lambda =\frac{M}{L}$

  $\begin{array}{l}dU=-\frac{G{m}_0\lambda d{x}_0}{x_0-x_s}\\ dU=-\frac{GM{m}_{0}d{x}_0}{L\left(x_0-x_s\right)}\end{array}$

Conclusion:

The potential energy shared by an element of the rod of mass $dm$ and the point particle of mass $m_0$ located at $x_0\ge \frac{1}{2}L$ is given by:

  $\begin{array}{l}dU=-\frac{G{m}_{0}dm}{x_0-x_s}\\ dU=-\frac{GM{m}_0}{L\left(x_0-x_s\right)}d{x}_s\end{array}$

(b)

To determine

ToIntegrate:The result had in part (a).

Answer to Problem 103P

  $U=-\frac{GM{m}_0}{L}\ln \left(\frac{x_0+L/2}{x_0-L/2}\right)$

Explanation of Solution

Given information:

Mass of point particle $=m_0$

Formula used:

Gravitational potential energy

  $U=-\frac{G{m}_{o}m}{r}$

U is gravitational potential energy, G is the gravitational constant, m is the mass of large body and $m_0$ is the mass of the other body.

Calculation:

Integrate $dU$ to find the total potential energy of the system:

  $\begin{array}{l}U=-\frac{GM{m}_0}{L}{\displaystyle \underset{-L/2}{\overset{L/2}{\int }}\frac{d{x}_s}{x_0-x_s}}\\ U=\frac{GM{m}_0}{L}\left[\ln \left(x_0-\frac{L}{2}\right)-\ln \left(x_0+\frac{L}{2}\right)\right]\\ U=-\frac{GM{m}_0}{L}\ln \left(\frac{x_0+L/2}{x_0-L/2}\right)\end{array}$

Conclusion:

The integration of $dU=-\frac{GM{m}_{0}dx}{L\left(x_0-x_s\right)}$ is, $U=-\frac{GM{m}_0}{L}\ln \left(\frac{x_0+L/2}{x_0-L/2}\right)$ .

(c)

To determine

To Calculate: The force on $m_0$ at a general point x using $F_x=-dU/dx$ and compare the result with $m_{0}g$ .

Answer to Problem 103P

  $F\left(x_0\right)=-\frac{Gm{m}_0}{x^2-L^2/4}$

Explanation of Solution

Given information:

Mass of point particle $=m_0$

Formula used:

Force on mass ${\text{ m}}_{\text{0}}$

  $F\left(x_0\right)=-\frac{dU}{d{x}_0}$

Where, $dU$ is the potential energy difference and $d{x}_o$ is the distance to mass ${\text{ m}}_{\text{0}}$ .

Calculation:

  $U=\frac{GM{m}_0}{L}\left[\ln \left(x_0-\frac{L}{2}\right)-\ln \left(x_0+\frac{L}{2}\right)\right]$

Because $x_0$ is a general point along the x axis:

  $\begin{array}{l}F\left(x_0\right)=-\frac{dU}{d{x}_0}\\ F\left(x_0\right)=\frac{GM{m}_0}{L}\left[\frac{1}{x_0+\frac{L}{2}}-\frac{1}{x_0-\frac{L}{2}}\right]\end{array}$

  $F\left(x_0\right)=-\frac{Gm{m}_0}{x_o{}^2-L^2/4}$

This answer and the answer given in Example 11-8 are the same.

Conclusion:

The force on $m_0$ at a general point x using $F_x=-dU/dx$ is, $F\left(x_0\right)=-\frac{Gm{m}_0}{x^2-L^2/4}$ .