Chapter 11, Problem 65P

(a)

To determine

To determine: The magnitude of the relative acceleration as a function of $m$.

Answer to Problem 65P

Answer: The magnitude of the relative acceleration as a function of $m$ is $2.77\left(1+\frac{m}{5.98\times {10}^{24}\text{kg}}\right)\text{m}/{\text{s}}^{\text{2}}$.

Explanation of Solution

Explanation:

Given information:

A object of mass $m$ is distance from the Earth’s center is $1.20\times {10}^7\text{m}$.

Figure I

Formula to calculate the relative acceleration is,

$a_{\text{rel}}=a_1-a_2$ (I)

$a_1$ is the acceleration of the object having mass $m$.

$a_2$ is the acceleration of the Earth.

Formula to calculate the gravitational force exerted by the object on the Earth is,

$F_{12}=\frac{G{M}_{\text{E}}m}{r^2}$ (II)

$M_{\text{E}}$ is the mass of the Earth.

$m$ is the mass of the object.

$r$ is the distance of object having mass $m$ from the Earth center.

By Newton’s law the force exerted by the object is,

$F_{12}=m{a}_1$ (III)

From equation (II) and equation (III) is,

$\begin{array}{c}m{a}_1=\frac{G{M}_{\text{E}}m}{r^2}\\ a_1=\frac{G{M}_{\text{E}}}{r^2}\end{array}$

The forces $F_{12}$ and $F_{21}$ both are gravitational force equal in magnitude and opposite in nature.

$F_{12}=-F_{21}$

$F_{21}$ is the force exerted by Earth on the object having mass $m$.

Substitute $\frac{G{M}_{\text{E}}m}{r^2}$ for $F_{12}$.

$\begin{array}{c}\frac{G{M}_{\text{E}}m}{r^2}=-F_{21}\\ F_{21}=-\frac{G{M}_{\text{E}}m}{r^2}\end{array}$ (IV)

By Newton’s law the force exerted by the Earth is,

$F_{21}=M_{\text{E}}{a}_2$ (V)

From equation (IV) and equation (V) is,

$\begin{array}{c}{M}_{\text{E}}{a}_2=-\frac{G{M}_{\text{E}}m}{r^2}\\ a_2=-\frac{Gm}{r^2}\end{array}$

Substitute $-\frac{Gm}{r^2}$ for $a_2$ and $\frac{G{M}_{\text{E}}}{r^2}$ for $a_1$ in equation (I).

$a_{\text{rel}}=\frac{G{M}_{\text{E}}}{r^2}-\left(-\frac{Gm}{r^2}\right)$

Substitute $5.972\times {10}^{24}\text{kg}$ for $M_{\text{E}}$, $6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}/{\text{kg}}^{\text{2}}$ for $G$ and $1.20\times {10}^7\text{m}$ for $r$ to find $a_{\text{rel}}$.

$\begin{array}{c}{a}_{\text{rel}}=\frac{6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}/{\text{kg}}^{\text{2}}\times 5.972\times {10}^{24}\text{kg}}{{\left(1.20\times {10}^7\text{m}\right)}^2}-\left(-\frac{6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}/{\text{kg}}^{\text{2}}\times m}{{\left(1.20\times {10}^7\text{m}\right)}^2}\right)\\ =2.77\left(1+\frac{m}{5.98\times {10}^{24}\text{kg}}\right)\text{m}/{\text{s}}^{\text{2}}\end{array}$ (VI)

Conclusion:

Therefore, the magnitude of the relative acceleration as a function of $m$ is $2.77\left(1+\frac{m}{5.98\times {10}^{24}\text{kg}}\right)\text{m}/{\text{s}}^{\text{2}}$.

(b)

To determine

To determine: The magnitude of the relative acceleration for $m=5.00\text{kg}$.

Answer to Problem 65P

Answer: The magnitude of the relative acceleration for $m=5.00\text{kg}$ is $2.77\text{m}/{\text{s}}^{\text{2}}$.

Explanation of Solution

Explanation:

Given information:

A object of mass $m$ is distance from the Earth’s center is $1.20\times {10}^7\text{m}$.

From equation (VI) the relative acceleration is,

$a_{\text{rel}}=2.77\left(1+\frac{m}{5.98\times {10}^{24}\text{kg}}\right)\text{m}/{\text{s}}^{\text{2}}$

Substitute $5.00\text{kg}$ for $m$ to find $a_{\text{rel}}$.

$\begin{array}{c}{a}_{\text{rel}}=2.77\left(1+\frac{5.00\text{kg}}{5.98\times {10}^{24}\text{kg}}\right)\text{m}/{\text{s}}^{\text{2}}\\ =2.77\text{m}/{\text{s}}^{\text{2}}\end{array}$

Conclusion:

Therefore, the magnitude of the relative acceleration for $m=5.00\text{kg}$ is $2.77\text{m}/{\text{s}}^{\text{2}}$.

(c)

To determine

To determine: The magnitude of the relative acceleration for $m=2000\text{kg}$.

Answer to Problem 65P

Answer: The magnitude of the relative acceleration for $m=2000\text{kg}$ is $2.77\text{m}/{\text{s}}^{\text{2}}$.

Explanation of Solution

Explanation:

Given information:

A object of mass $m$ is distance from the Earth’s center is $1.20\times {10}^7\text{m}$.

From equation (VI) the relative acceleration is,

$a_{\text{rel}}=2.77\left(1+\frac{m}{5.98\times {10}^{24}\text{kg}}\right)\text{m}/{\text{s}}^{\text{2}}$

Substitute $5.00\text{kg}$ for $m$ to find $a_{\text{rel}}$.

$\begin{array}{c}{a}_{\text{rel}}=2.77\left(1+\frac{2000\text{kg}}{5.98\times {10}^{24}\text{kg}}\right)\text{m}/{\text{s}}^{\text{2}}\\ =2.77\text{m}/{\text{s}}^{\text{2}}\end{array}$

Conclusion:

Therefore, the magnitude of the relative acceleration for $m=2000\text{kg}$ is $2.77\text{m}/{\text{s}}^{\text{2}}$.

(d)

To determine

To determine: The magnitude of the relative acceleration for $m=2.00\times {10}^{24}\text{kg}$.

Answer to Problem 65P

Answer: The magnitude of the relative acceleration for $m=2.00\times {10}^{24}\text{kg}$ is $3.7\text{m}/{\text{s}}^{\text{2}}$.

Explanation of Solution

Explanation:

Given information:

A object of mass $m$ is distance from the Earth’s center is $1.20\times {10}^7\text{m}$.

From equation (VI) the relative acceleration is,

$a_{\text{rel}}=2.77\left(1+\frac{m}{5.98\times {10}^{24}\text{kg}}\right)\text{m}/{\text{s}}^{\text{2}}$

Substitute $5.00\text{kg}$ for $m$ to find $a_{\text{rel}}$.

$\begin{array}{c}{a}_{\text{rel}}=2.77\left(1+\frac{2.00\times {10}^{24}\text{kg}}{5.98\times {10}^{24}\text{kg}}\right)\text{m}/{\text{s}}^{\text{2}}\\ =3.7\text{m}/{\text{s}}^{\text{2}}\end{array}$

Conclusion:

Therefore, the magnitude of the relative acceleration for $m=2.00\times {10}^{24}\text{kg}$ is $3.7\text{m}/{\text{s}}^{\text{2}}$.

(e)

To determine

To determine: The pattern of variation of relative acceleration with $m$.

Answer to Problem 65P

Answer: The relative acceleration is directly proportional to the mass $m$.

Explanation of Solution

Explanation:

Given information:

A object of mass $m$ is distance from the Earth’s center is $1.20\times {10}^7\text{m}$.

From equation (VI) the relative acceleration is,

$a_{\text{rel}}=2.77\left(1+\frac{m}{5.98\times {10}^{24}\text{kg}}\right)\text{m}/{\text{s}}^{\text{2}}$

This is the linear equation and shows the relative acceleration is directly proportional to the object having mass $m$. As $m$ increases to become comparable to the mass of the Earth, the acceleration increases and can become arbitrarily large.

Conclusion:

Therefore, the relative acceleration is directly proportional to the object having mass $m$.