Chapter 11, Problem 64P

(a)

To determine

The gravitational force on the particle of mass $m$ due to the five objects

Answer to Problem 64P

The gravitational force on the particle of mass $m$ due to the five particles is $9.7\times {10}^{-8}\text{N}\widehat{j}$.

Explanation of Solution

Given data:

Following figure shows the distribution of masses on the semi-circular arc of radius $R$

Formula Used:

The expression for the gravitational force between the masses $m\text{ and }M$ is

  $F_1=\frac{GMm}{R^2}\widehat{i}$

Here, $G$ is the gravitational constant, $M$ is the mass of the one object, $m$ is the mass of the other object and $R$ is the radius between the two objects. Here, $\widehat{i}$ represents the gravitational force along x -direction.

Calculation:

Radius of the arc is $10\text{cm}$ . Convert the distance from cm into $m$ .

  $\begin{array}{c}R=\left(\text{10}\text{.0}\text{cm}\right)\left(\frac{\text{1}\text{m}}{{\text{10}}^{\text{2}}\text{cm}}\right)\\ {\text{=10×10}}^{\text{-2}}\text{m}\end{array}$

Choose the unit vector $\widehat{i}\text{ and }\widehat{j}$ along the positive x -axis and positive y -axis respectively.

The gravitational force between the masses $m\text{ and }M$ which makes an angle ${45}^{\text{o}}$ with the positive x -axis is:

  $F_2=\left(\frac{GMm}{R^2}\right)\cos {45}^{\text{o}}\widehat{i}+\left(\frac{GMm}{R^2}\right)\sin {45}^{\text{o}}\widehat{j}$

The gravitational force between the masses $m\text{ and }M$ which is on positive y-axis is:

  $F_3=\frac{GMm}{R^2}\widehat{j}$

The gravitational force between the masses $m\text{ and }M$ which makes an angle ${135}^{\text{o}}$ with the positive x -axis is:

  $F_4=\left(\frac{GMm}{R^2}\right)\cos {135}^{\text{o}}\widehat{i}+\left(\frac{GMm}{R^2}\right)\sin {135}^{\text{o}}\widehat{j}$

The gravitational force between the masses $m\text{ and }M$ which is on lest side x -axis is,

  $F_5=-\frac{GMm}{R^2}\widehat{i}$

Hence, net force on the particle of mass $m$ is,

  $\begin{array}{c}F=F_1+F_2+F_3+F_4+F_5\\ =\frac{\text{GMm}}{{\text{R}}^{\text{2}}}\text{<mover accent="true"> <mo>i</mo> <mo>^</mo></mover> +}\left(\frac{\text{GMm}}{{\text{R}}^{\text{2}}}\right){\text{cos45}}^{\text{o}}\text{<mover accent="true"> <mo>i</mo> <mo>^</mo></mover> +}\left(\frac{\text{GMm}}{{\text{R}}^{\text{2}}}\right){\text{sin45}}^{\text{o}}\text{<mover accent="true"> <mo>j</mo> <mo>^</mo></mover> +}\frac{\text{GMm}}{{\text{R}}^{\text{2}}}\text{<mover accent="true"> <mo>j</mo> <mo>^</mo></mover> -}\left(\frac{\text{GMm}}{{\text{R}}^{\text{2}}}\right){\text{cos45}}^{\text{o}}\text{<mover accent="true"> <mo>i</mo> <mo>^</mo></mover> +}\left(\frac{\text{GMm}}{{\text{R}}^{\text{2}}}\right){\text{sin45}}^{\text{o}}\text{<mover accent="true"> <mo>j</mo> <mo>^</mo></mover> -}\frac{\text{GMm}}{{\text{R}}^{\text{2}}}\text{<mover accent="true"> <mo>i</mo> <mo>^</mo></mover>}\\ \text{=}\left(\text{2}\left(\frac{\text{GMm}}{{\text{R}}^{\text{2}}}\right){\text{sin45}}^{\text{o}}\text{+}\frac{\text{GMm}}{{\text{R}}^{\text{2}}}\right)\text{<mover accent="true"> <mo>j</mo> <mo>^</mo></mover> N}\\ \text{=}\frac{\text{GMm}}{{\text{R}}^{\text{2}}}\left({\text{2sin45}}^{\text{o}}\text{+1}\right)\text{<mover accent="true"> <mo>j</mo> <mo>^</mo></mover> N}\end{array}$

Substitute the values and solve:

  $\begin{array}{c}\text{F=}\frac{\left(\text{6}{\text{.67×10}}^{\text{-11}}\text{N}{\text{.m}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)\left(\text{3}\text{.0}\text{kg}\right)\left(\text{2}\text{.0}\text{kg}\right)}{{\left(\text{0}\text{.1}\text{m}\right)}^{\text{2}}}\left({\text{2sin45}}^{\text{o}}\text{+1}\right)\text{<mover accent="true"> <mo>j</mo> <mo>^</mo></mover> N}\\ \text{=}\text{9}{\text{.7×10}}^{\text{-8}}\text{<mover accent="true"> <mo>j</mo> <mo>^</mo></mover> N}\end{array}$

Conclusion:

Therefore, the gravitational force on the particle of mass $m$ due to the five particles is $9.7\times {10}^{-8}\text{N}\text{<mover accent="true"> <mo>j</mo> <mo>^</mo></mover>}$ .

(b)

To determine

The gravitational field at center of curvature of the arc.

Answer to Problem 64P

The gravitational field at center of curvature of the arc is $4.85\times {10}^{-8}\text{N/kg}\text{<mover accent="true"> <mo>j</mo> <mo>^</mo></mover>}$ .

Explanation of Solution

Given data:

Following figure shows the distribution of masses on the semi-circular arc of radius $R$

Formula Used:

Gravitational field at a point is:

  $\overrightarrow{g}=\frac{F_g}{m}$

Here, net gravitational force is $F_g$ and mass is $m$

Calculation:

The gravitational field at the center of curvature of the arc can be determined as follows.

  $\overrightarrow{g}=\frac{F_g}{m}$

Substitute $9.7\times {10}^{-8}N\widehat{j}\text{ for }{F}_g\text{ and 2}\text{.0}\text{kg for }m$

  $\begin{array}{c}g=\frac{9.7\times {10}^{-8}\text{N}\widehat{j}}{\text{2}\text{.0}\text{kg}}\\ =4.85\times {10}^{-8}\text{N/kg}\text{j}\end{array}$

Conclusion:

The gravitational field at center of curvature of the arc is $4.85\times {10}^{-8}\text{N/kg}\text{<mover accent="true"> <mo>j</mo> <mo>^</mo></mover>}$ .