Chapter 11, Problem 74P

(a)

To determine

To Calculate:The magnitude of gravitational force due to $M_1{\text{ and M}}_2$ at $x=3a$ .

Answer to Problem 74P

  $\left|\overrightarrow{\text{F}}\right|=\frac{Gm}{a^2}\left[\frac{M_1}{9}+\frac{M_2}{4.84}\right]$

Explanation of Solution

Given data:

Radius of $M_1$ is $R_2=2a$

  $R_1=a$

Center has shifted, $x=0.8a$

Formula Used:

Gravitational field:

  $\overrightarrow{g}=-\frac{G{m}_1}{r^2}\widehat{r}$

Where, G is the gravitational constant, m is the mass and r is the distance of the point where field is calculated.

The gravitational force is a force of attraction between any two bodies having mass and separated by a distance d . Mathematically, the gravitational force can be represented by:

  $\overrightarrow{F}=M\overrightarrow{g}$

  $\overrightarrow{F}=\frac{GmM}{r^2}\widehat{r}$

Where, G is the gravitational constant, m and M are the masses and r is the distance between them.

Calculation:

The magnitude of force due to $M_1$ at $r=3a$

  $\left|{\overrightarrow{\text{F}}}_1\right|=\frac{G{M}_{1}m}{9a^2}$

Again, due to $M_2$ at $r=3a-0.8a=2.2a$

Force, $\left|{\overrightarrow{\text{F}}}_2\right|=\frac{G{M}_{2}m}{{\left(2.2a\right)}^2}=\frac{G{M}_{2}m}{4.84a^2}$

Hence, net force

  $\left|\overrightarrow{F}\right|=\left|\overrightarrow{F_1}\right|+\left|\overrightarrow{F_2}\right|$

At $r=3a$

  $\left|\overrightarrow{\text{F}}\right|=\frac{Gm}{a^2}\left[\frac{M_1}{9}+\frac{M_2}{4.84}\right]$

Conclusion:

Thus, the magnitude of gravitational force due to $M_1{\text{ and M}}_2$ at $x=3a$ is $\left|\overrightarrow{\text{F}}\right|=\frac{Gm}{a^2}\left[\frac{M_1}{9}+\frac{M_2}{4.84}\right]$ .

(b)

To determine

The magnitude of gravitational force due to $M_1{\text{ and M}}_2$ at $x=1.9a$ .

Answer to Problem 74P

  $\left|\overrightarrow{\text{F}}\right|=\frac{G{M}_{2}m}{1.21a^2}$

Explanation of Solution

Given data:

Radius of $M_1$ is $R_2=2a$

  $R_1=a$

Center has shifted, $x=0.8a$

Formula Used:

Gravitational field:

  $\overrightarrow{g}=-\frac{G{m}_1}{r^2}\widehat{r}$

Where, G is the gravitational constant, m is the mass and r is the distance of the point where field is calculated.

The gravitational force is a force of attraction between any two bodies having mass and separated by a distance d . Mathematically, the gravitational force can be represented by:

  $\overrightarrow{F}=M\overrightarrow{g}$

  $\overrightarrow{F}=\frac{GmM}{r^2}\widehat{r}$

Where, G is the gravitational constant, m and M are the masses and r is the distance between them.

Calculation:

The value of $\left|\overrightarrow{F_1}\right|$ , due to ${\text{M}}_1$ at $r=1.9a<\left(R_2=2a\right)$

Hence, $\left|\overrightarrow{{\text{F}}_1}\right|=0$

The value of $\left|\overrightarrow{{\text{F}}_{\text{2}}}\right|$ ; due to ${\text{M}}_2$ at $r=1.9a-0.8a=1.1a$

  $\left|\overrightarrow{F_2}\right|=\frac{G{M}_{2}m}{{\left(1.1a\right)}^2}$

Hence, net force $\left|\overrightarrow{F}\right|=\left|\overrightarrow{F_2}\right|+\left|\overrightarrow{F_1}\right|$

  $\left|\overrightarrow{\text{F}}\right|=\frac{G{M}_{2}m}{1.21a^2}$

Conclusion:

Thus, the magnitude of gravitational force due to $M_1{\text{ and M}}_2$ at $x=1.9a$ is $\left|\overrightarrow{\text{F}}\right|=\frac{G{M}_{2}m}{1.21a^2}$ .

(c)

To determine

The magnitude of gravitational force due to $M_1{\text{ and M}}_2$ at $x=0.9a$ .

Answer to Problem 74P

  $\left|\overrightarrow{F}\right|=0$

Explanation of Solution

Given data:

Radius of $M_1$ is $R_2=2a$

  $R_1=a$

Center has shifted, $x=0.8a$

Formula Used:

Gravitational field:

  $\overrightarrow{g}=-\frac{G{m}_1}{r^2}\widehat{r}$

Where, G is the gravitational constant, m is the mass and r is the distance of the point where field is calculated.

The gravitational force is a force of attraction between any two bodies having mass and separated by a distance d . Mathematically, the gravitational force can be represented by:

  $\overrightarrow{F}=M\overrightarrow{g}$

  $\overrightarrow{F}=\frac{GmM}{r^2}\widehat{r}$

Where, G is the gravitational constant, m and M are the masses and r is the distance between them.

Calculation:

The value of $\left|\overrightarrow{F_2}\right|$ , due to $M_2$ , at $r=0.9a-0.8a=0.1a<<R_2$

  $\left|\overrightarrow{{\text{F}}_{\text{2}}}\right|=0$

Again, the value of $\left|{\overrightarrow{\text{F}}}_1\right|$ , due to $M_1$ , at $r=0.9a<R_1$

So, $\left|\overrightarrow{{\text{F}}_1}\right|=0$

Hence, net force

  $\begin{array}{c}\left|\overrightarrow{F}\right|=\left|\overrightarrow{F_2}\right|+\left|\overrightarrow{F_1}\right|\\ =0+0\end{array}$

  $\left|\overrightarrow{F}\right|=0$

Conclusion:

Thus, the magnitude of gravitational force due to $M_1{\text{ and M}}_2$ at $x=0.9a$ is $\left|\overrightarrow{F}\right|=0$ .