Chapter 11, Problem 68P

(a)

To determine

ToShow: The element produces a gravitational field at a point $x_0$ on the x axis in the region $x_0>L$ is given by $d{g}_0=-\frac{GM}{L{\left(x_0-x\right)}^2}dx$ .

Explanation of Solution

Given information :

Mass of the rod $=M$

Length of the rod $=L$

Length of an element of the rod $=dx$

Mass of an element of the rod $=dm$

Formula used :

Acceleration due to gravity for small mass element ( dm ) can be obtained by:

  $dg=-\frac{Gdm}{r^2}$

Calculation:

The gravitational field at P due to the element $dm$ is

  $d{\overrightarrow{g}}_x=-\frac{Gdm}{r^2}\widehat{i}$

  $dm=\frac{M}{L}dx$

Distance to the point on the axis:

  $r=x_0-x$

  $d{\overrightarrow{g}}_x=\left\{-\frac{GM}{L{\left(x_0-x\right)}^2}dx\right\}\widehat{i}$

(b)

To determine

ToIntegrate: The result over the length of the rod to find the total gravitational field at the point $x_0$ due to the rod.

Explanation of Solution

Given information:

Mass of the rod $=M$

Length of the rod $=L$

Length of an element of the rod $=dx$

Mass of an element of the rod $=dm$

Formula used:

The element produces a gravitational field at a point $x_0$ on the x axis in the region $x_0>L$ is given by $d{g}_0=-\frac{GM}{L{\left(x_0-x\right)}^2}dx$ .

Calculation:

  $\begin{array}{l}{\overrightarrow{g}}_x=-\frac{GM}{L}{\displaystyle \underset{0}{\overset{L}{\int }}\frac{dx}{{\left(x_0-x\right)}^2}}\widehat{i}\\ {\overrightarrow{g}}_x=\left\{-\frac{GM}{L}{\left[\frac{1}{\left(x_0-x\right)}\right]}_0^L\right\}\widehat{i}\\ {\overrightarrow{g}}_x=-\frac{GM}{x_o\left(x_0-L\right)}\widehat{i}\end{array}$

Conclusion:

The total gravitational field at the point $x_0$ due to the rod is: ${\overrightarrow{g}}_x=-\frac{GM}{x_o\left(x_0-L\right)}\widehat{i}$

(c)

To determine

To Calculate: The gravitational force on a point particle of mass $m_0$ at $x_0$ .

Explanation of Solution

Given information:

Mass of the rod $=M$

Length of the rod $=L$

Length of an element of the rod $=dx$

Mass of an element of the rod $=dm$

Formula used:

Force of gravity

  $F=mg$

Where, m is the mass and gis the acceleration due to gravity.

The total gravitational field at the point $x_0$ due to the rod is: ${\overrightarrow{g}}_x=-\frac{GM}{x_o\left(x_0-L\right)}\widehat{i}$

Calculation:

By using the definition of gravitational field and the result from part (b) to express ${\overrightarrow{F}}_g$ at $x=x_0$ is:

  $\begin{array}{c}{\overrightarrow{F}}_g=m_0{\overrightarrow{g}}_x\\ {\overrightarrow{F}}_g=-\frac{GM{m}_0}{x_0\left(x_0-L\right)}\widehat{i}\end{array}$

Conclusion:

The gravitational force on a point particle of mass $m_0$ at $x_0$ is ${\overrightarrow{F}}_g=-\frac{GM{m}_0}{x_0\left(x_0-L\right)}\widehat{i}$

(d)

To determine

To Show:For $x_0>>L$ , the field of the rod approximates the field of a point particle of mass $M$ at $x=0$ .

Explanation of Solution

Given information:

Mass of the rod $=M$

Length of the rod $=L$

Length of an element of the rod $=dx$

Mass of an element of the rod $=dm$

Formula used:

The total gravitational field at the point $x_0$ due to the rod is: ${\overrightarrow{g}}_x=-\frac{GM}{x_o\left(x_0-L\right)}\widehat{i}$

Calculations:

  ${\overrightarrow{g}}_x=-\frac{GM}{x_o\left(x_0-L\right)}\widehat{i}$

  ${\overrightarrow{g}}_x=-\frac{GM}{x_0^2\left(1-\frac{L}{x_0}\right)}\widehat{i}$

For $x_0>>L$ , the second term in parentheses is very small.

  ${\overrightarrow{g}}_x\approx -\frac{GM}{x_0^2}\widehat{i}$