Chapter 7, Problem 21PQ

(a)

To determine

The net gravitational force on the particle due to the two spheres.

Answer to Problem 21PQ

The net gravitational force on the particle due to the two spheres is $\underset{\_}{4.27\times {10}^{-8}\text{N towards 750 kg sphere}}$.

Explanation of Solution

Write the equation for the net force on the particle.

  $F_{net}=F_1-F_2$                                                                                                           (I)

Here, $F_{net}$ is the net force, $F_1$ is the force due to first sphere and $F_2$ is the force due to the second force.

Write the equation for the force due to first sphere.

  $F_1=\frac{Gm{M}_1}{r_1^2}$                                                                                                           (II)

Here, $G$ is the gravitational constant, $m$ is the mass of the particle, $M_1$ is the mass of the first sphere and $r_1$ is the separation distance

Write the equation of the separation distance.

  $r_1=\frac{r}{2}$                                                                                                                   (III)

Here, $r$ is the distance between two spheres.

Write the equation for the force due to first sphere.

  $F_2=\frac{Gm{M}_2}{r_2^2}$                                                                                                        (IV)

Here, $M_2$ is the mass of the second sphere, and $r_2$ is the separation distance.

Write the equation of the separation distance.

  $r_2=\frac{r}{2}$                                                                                                                    (V)

Here, $r$ is the distance between two spheres.

Rewrite the equation of net force from equation (I).

  $F_{net}=\frac{Gm{M}_1}{{\left(r/2\right)}^2}-\frac{Gm{M}_2}{{\left(r/2\right)}^2}$                                                                                      (VI)

Conclusion:

Substitute $6.67\times {10}^{-11}{\text{N-m}}^{\text{2}}{\text{/kg}}^{\text{2}}$ for $G$, $10.0\text{kg}$ for $m$, $750\text{kg}$ for $M_1$, $350\text{kg}$ for $M_2$ and $5.00\text{m}$ for $r$ in equation (VI) to find $F_{net}$.

  $\begin{array}{c}{F}_{net}=\left[\frac{\left(6.67\times {10}^{-11}{\text{N-m}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)\left(10.0\text{kg}\right)\left(750\text{kg}\right)}{{\left(5.00\text{m}/2\right)}^2}\right]-\left[\frac{\left(6.67\times {10}^{-11}{\text{N-m}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)\left(10.0\text{kg}\right)\left(350\text{kg}\right)}{{\left(5.00\text{m}/2\right)}^2}\right]\\ =\frac{\left(6.67\times {10}^{-11}{\text{N-m}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)\left(10.0\text{kg}\right)}{{\left(5.00\text{m}/2\right)}^2}\left[\left(750\text{kg}\right)-\left(350\text{kg}\right)\right]\\ =4.27\times {10}^{-8}\text{N}\end{array}$

Thus, the net gravitational force on the particle due to the two spheres is $\underset{\_}{4.27\times {10}^{-8}\text{N towards 750 kg sphere}}$.

(b)

To determine

The position at which the net gravitational force on the particle will be zero.

Answer to Problem 21PQ

The position at which the net gravitational force on the particle will be zero is $\underset{\_}{2.97\text{m}}$.

Explanation of Solution

Considering the distance is $d$ from $M_1$.

Write the equation of net force from equation (VI).

  $F_{net}=\frac{Gm{M}_1}{d^2}-\frac{Gm{M}_1}{{\left(\left(5.00\text{m}\right)-d\right)}^2}$                                                                        (VII)

Here, $d$ is the point at which the net force is zero.

Conclusion:

Substitute $0$ for $F_{net}$, $750\text{kg}$ for $M_1$ and $350\text{kg}$ for $M_2$  in equation (VII).

  $0=\frac{Gm{M}_1}{d^2}-\frac{Gm{M}_1}{{\left(\left(5.00\text{m}\right)-d\right)}^2}$

Rearrange the equation to find $d$.

  $\begin{array}{c}0=400d^2-7500d+18750\\ d=2.97\text{m, 15}\text{.77m}\end{array}$

The value $15.77\text{m}$ is not valid.

Thus, the position at which the net gravitational force on the particle will be zero is $\underset{\_}{2.97\text{m}}$.