In the figure, a particle of mass$m_1=0.67\text{kg}$is a distance$d=23\text{cm}$from one end of a uniform rod with length$L=3.0\text{m}$and mass$M=5.0\text{kg}$. What is the magnitude of the gravitational forceon the particle from the rod?

1. Mass of the particle,$m=0.67\text{\hspace{0.33em}kg}$
2. Distance of particle from the end of the rod,$d=0.23\text{\hspace{0.33em}m}$
3. Length of the rod,$L=3.0\text{\hspace{0.33em}m}$
4. Mass of the rod,$M=5.0\text{\hspace{0.33em}kg}$
5. Gravitational constant,$G=6.67\times {10}^{-11}\text{\hspace{0.33em}N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}$

Newton’s law of gravitation states that any particle in the universe attracts any other particle with a gravitational force whose magnitude is

$F=G\frac{m_1{m}_2}{r^2}$

Here, $m_1$and $m_2$are masses of the particles and$r$is their separation and is the gravitational constant.

We are using the concept of Newton’s law of gravitation. We can write the small element of the rod in terms of density and length. Using the small element, we can write the equation of force and integrate it between the given limits to find the force.

Formula:

Gravitational Force, $F=\frac{GMm}{r^2}$ ...(i)

We consider the small differential element of rod of mass $dm$and thickness$dr$ which is at distance$r$ from$m_1$.

Using equation (i), the gravitational force between $m$and $dm$is given by,

$dF=\frac{Gmdm}{r^2}$

We can write change in mass in terms of density and change in length.

So, we have,

$dm=\left(\frac{M}{L}\right)dr$

Substituting this we get,

$dF=\frac{Gm\left(\frac{M}{L}\right)dr}{r^2}$

Integrating above equation fromwe get,

$\begin{array}{c}F=\int dF\\ =\int \frac{Gm\left(\frac{M}{L}\right)dr}{r^2}\\ =\left(\frac{GmM}{L}\right){\int }_d^{L+d}\frac{dr}{r^2}\\ =-\left(\frac{GmM}{L}\right)\left(\frac{1}{L+d}-\left(\frac{1}{d}\right)\right)\\ =\left(\frac{GmM}{d\left(L+d\right)}\right)\end{array}$

Substituting all the values we get,

$\begin{array}{c}F=\frac{6.67\times {10}^{-11}\text{\hspace{0.33em}N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}\times 0.67\text{\hspace{0.33em}kg}\times 5.0\text{kg}}{\left(0.23\text{\hspace{0.33em}m}\right)\left(3.0\text{\hspace{0.33em}m}+0.23\text{\hspace{0.33em}m}\right)}\\ =3.0\times {10}^{-10}\text{\hspace{0.33em}N}\end{array}$

Hence, the force is$3.0\times {10}^{-10}\text{\hspace{0.33em}N}$