Chapter 13, Problem 13.3P

A 200-kg object and a 500-kg object are separated by 4.00 m. (a) Find the net gravitational force exerted by these objects on a 50.0-kg object placed midway between them. (b) At what position (other than an infinitely remote one) can the 50.0-kg object be placed so as to experience a net force of zero from the other two objects?

To determine

 The net gravitational field exerted on the object.

Answer to Problem 13.3P

 The net gravitational field exerted on the object is $2.5\times {10}^{-7}\text{N}$.

Explanation of Solution

Given info: The mass of the first object is $200\text{kg}$, and the mass of second object is $500\text{kg}$ and the mass of the third object is $50.0\text{kg}$. The distance between the first and the second object is $4.00\text{m}$.

The universal gravitational constant is $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2$.

The third object is at the midpoint of the distance between the first object and the second object.

The formula for gravitational force is,

$F=G\frac{m_1{m}_2}{r^2}$ (1)

Here,

$m_1$ is mass of object one.

$m_2$ is mass of object two.

$r$ is the distance between objects.

$G$ is the universal gravitational constant.

For first object and third object

Substitute $F_1$ for $F$, $200\text{kg}$ for $m_1$, $50\text{kg}$ for $m_2$, $4.00\text{m}$ for $r$ and $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2$ for $G$ in the above expression.

$\begin{array}{c}{F}_1=\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\frac{\left(200\text{kg}\right)\left(50.0\text{kg}\right)}{{\left(2.00\text{m}\right)}^2}\\ =1.67\times {10}^{-7}\text{N}\end{array}$

The force by the first object on third object is $1.67\times {10}^{-7}\text{N}$.

For second object and third object

Substitute $F_2$ for $F$, $500\text{kg}$ for $m_1$, $50\text{kg}$ for $m_2$, $4.00\text{m}$ for $r$ and $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2$ for $G$ in the above expression.

$\begin{array}{c}{F}_2=\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\frac{\left(500\text{kg}\right)\left(50.0\text{kg}\right)}{{\left(4.00\text{m}\right)}^2}\\ =4.17\times {10}^{-7}\text{N}\end{array}$

The force by the second object on third object is $4.17\times {10}^{-7}\text{N}$.

The expression for the net force on the third object is,

$F_{\text{n}}=F_2-F_1$

The direction of the forces by first and second object on the third object is opposite to each other.

Substitute $4.17\times {10}^{-7}\text{N}$ for $F_2$ and $1.67\times {10}^{-7}\text{N}$ for $F_1$ in the above expression.

$\begin{array}{c}{F}_{\text{n}}=4.17\times {10}^{-7}\text{N}-1.67\times {10}^{-7}\text{N}\\ =2.5\times {10}^{-7}\text{N}\end{array}$

Conclusion:

Therefore, the net gravitational field exerted on the object is $2.5\times {10}^{-7}\text{N}$.

To determine

 The position at which the net force on the object is zero.

Answer to Problem 13.3P

 The net force on the object is zero at a distance of $\text{2}\text{.45}\text{m}$ from $500\text{kg}$ object.

Explanation of Solution

Given info: The mass of the first object is $200\text{kg}$, and the mass of second object is $500\text{kg}$ and the mass of the third object is $50.0\text{kg}$. The distance between the first and the second object is $4.00\text{m}$.

The expression for the distance of the third object from the first object is,

$y=4.00\text{m}-x$

Here,

$y$ is the distance of the third object from first object.

$x$ is the distance of the third object from second object.

For first object and third object

Substitute $F_1$ for $F$, $200\text{kg}$ for $m_1$, $50\text{kg}$ for $m_2$, $y$ for $r$ and $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2$ for $G$ in the above expression.

$F_1=\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\frac{\left(200\text{kg}\right)\left(50.0\text{kg}\right)}{y^2}$

Substitute $4.00\text{m}-x$ for $y$ in the above expression.

$F_1=\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\frac{\left(200\text{kg}\right)\left(50.0\text{kg}\right)}{{\left(4.00\text{m}-x\right)}^2}$

For second object and third object

Substitute $F_2$ for $F$, $500\text{kg}$ for $m_1$, $50\text{kg}$ for $m_2$, $x$ for $r$ and $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2$ for $G$ in the above expression.

$F_2=\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\frac{\left(500\text{kg}\right)\left(50.0\text{kg}\right)}{x^2}$

The expression for the net force on the third object is,

$F_{\text{n}}=F_2-F_1$

The direction of the forces by first and second object on the third object is opposite to each other.

Substitute $0$ for $F_{\text{n}}$, $\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\frac{\left(500\text{kg}\right)\left(50.0\text{kg}\right)}{x^2}$ for $F_2$ and $\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\frac{\left(200\text{kg}\right)\left(50.0\text{kg}\right)}{{\left(4.00\text{m}-x\right)}^2}$ for $F_1$ in the above expression.

$\begin{array}{l}0=\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\frac{\left(500\text{kg}\right)\left(50.0\text{kg}\right)}{x^2}\\ -\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\frac{\left(200\text{kg}\right)\left(50.0\text{kg}\right)}{{\left(4.00\text{m}-x\right)}^2}\\ \end{array}$

Rearrange the above equation for value of $x$.

$\begin{array}{c}x=\frac{4\sqrt{10}}{2+\sqrt{10}}\text{m}\\ =\text{2}\text{.45}\text{m}\end{array}$

Conclusion:

Therefore, the net force on the object is zero at a distance of $\text{2}\text{.45}\text{m}$ from $500\text{kg}$ object.