Chapter 7, Problem 33PQ

To determine

The magnitude and the direction of the gravitational field at the origin due to the three particles.

Answer to Problem 33PQ

The magnitude of the gravitational field at the origin due to the three particles is $g=\frac{Gm}{L^2}\left(\sqrt{2}+\frac{1}{2}\right)$ and it is directed at an angle $45°$ with the horizontal.

Explanation of Solution

Write the equation for the gravitational field at a point due to a particle.

  $\overrightarrow{\text{g}}\left(r\right)=-\frac{Gm}{r^2}\widehat{\text{r}}$                                                                                                           (I)

Here, $\overrightarrow{\text{g}}\left(r\right)$ is the gravitational field at the point, $G$ is the universal gravitation constant, $m$ is the mass of the particle, $r$ is the distance between the particle and the point where the field has to be determined and $\widehat{\text{r}}$ is a unit vector directed away from the particle.

The arrangement of the particles is shown in figure 1.

Refer to figure (I) and use equation (I) to find the field produced at the origin due to the particle 1.

  $\begin{array}{c}{\overrightarrow{\text{g}}}_1=-\frac{Gm}{L^2}\left(-\widehat{\text{j}}\right)\\ =\frac{Gm}{L^2}\widehat{\text{j}}\end{array}$                                                                                                        (II)

Here, ${\overrightarrow{\text{g}}}_1$ is the gravitational field at the origin due to the particle 1 and $L$ is the distance between the origin and the particle 1.

Refer to figure (I) and use equation (I) to find the field produced at the origin due to the particle 2.

  $\begin{array}{c}{\overrightarrow{\text{g}}}_2=-\frac{Gm}{L^2}\left(-\widehat{\text{i}}\right)\\ =\frac{Gm}{L^2}\widehat{\text{i}}\end{array}$                                                                                                       (III)

Here, ${\overrightarrow{\text{g}}}_2$ is the gravitational field at the origin due to the particle 2 and $L$ is the distance between the origin and the particle 2.

Use equation (I) to find the field produced at the origin due to the particle 3.

  ${\overrightarrow{\text{g}}}_3=-\frac{Gm}{r_3^2}{\widehat{\text{r}}}_3$                                                                                                           (IV)

Here, ${\overrightarrow{\text{g}}}_3$ is the gravitational field at the origin due to the particle 3 , $r_3$ is the distance between the origin and the particle 3 and ${\widehat{\text{r}}}_3$ is the unit vector pointing away from particle 3.

Find the distance between the particle 3 and origin.

  $\begin{array}{c}{r}_3=\sqrt{L^2+L^2}\\ =\sqrt{2L^2}\\ =\sqrt{2}L\end{array}$

Refer to figure 1 and write the expression for ${\widehat{\text{r}}}_3$ .

  $\begin{array}{c}{\widehat{\text{r}}}_3=-\cos 45°\widehat{\text{i}}-\sin 45°\widehat{\text{j}}\\ =-\frac{1}{\sqrt{2}}\left(\widehat{\text{i}}+\widehat{\text{j}}\right)\end{array}$

Put the above two equations in equation (IV).

  $\begin{array}{c}{\overrightarrow{\text{g}}}_3=-\frac{Gm}{{\left(\sqrt{2}L\right)}^2}\left(-\frac{1}{\sqrt{2}}\left(\widehat{\text{i}}+\widehat{\text{j}}\right)\right)\\ =\frac{Gm}{2\sqrt{2}{L}^2}\left(\widehat{\text{i}}+\widehat{\text{j}}\right)\end{array}$                                                                                  (V)

The gravitational field at the origin is the vector sum of the gravitational fields due to the three particles.

  $\overrightarrow{\text{g}}={\overrightarrow{\text{g}}}_1+{\overrightarrow{\text{g}}}_2+{\overrightarrow{\text{g}}}_3$

Here, $\overrightarrow{\text{g}}$ is the net gravitational field at the origin.

Put equations (II), (III) and (V) in the above equation.

  $\begin{array}{c}\overrightarrow{\text{g}}=\frac{Gm}{L^2}\widehat{\text{j}}+\frac{Gm}{L^2}\widehat{\text{i}}+\frac{Gm}{2\sqrt{2}{L}^2}\left(\widehat{\text{i}}+\widehat{\text{j}}\right)\\ =\frac{Gm}{L^2}\left(\widehat{\text{i}}+\widehat{\text{j}}\right)+\frac{Gm}{2\sqrt{2}{L}^2}\left(\widehat{\text{i}}+\widehat{\text{j}}\right)\\ =\frac{Gm}{L^2}\left(\widehat{\text{i}}+\widehat{\text{j}}\right)\left(1+\frac{1}{2\sqrt{2}}\right)\\ =\frac{Gm}{L^2}\left(1+\frac{1}{2\sqrt{2}}\right)\left(\widehat{\text{i}}+\widehat{\text{j}}\right)\end{array}$                                                                          (VI)

Conclusion:

Multiply and divide equation (VI) by $\sqrt{2}$ .

  $\begin{array}{c}\overrightarrow{\text{g}}=\frac{Gm}{L^2}\left(\sqrt{2}+\frac{1}{2}\right)\frac{\left(\widehat{\text{i}}+\widehat{\text{j}}\right)}{\sqrt{2}}\\ =\frac{Gm}{L^2}\left(\sqrt{2}+\frac{1}{2}\right)\left(\cos 45°\widehat{\text{i}}+\sin 45°\widehat{\text{j}}\right)\end{array}$

Therefore, the magnitude of the gravitational field at the origin due to the three particles is $g=\frac{Gm}{L^2}\left(\sqrt{2}+\frac{1}{2}\right)$ and it is directed at an angle $45°$ with the horizontal.