Chapter 5, Problem 77P

(a)

To determine

The maximum gravitational force the tractor could exert on the asteroid.

Answer to Problem 77P

The maximum gravitational force the tractor could exert on the asteroid $\underset{\_}{1.1\text{N}}$

Explanation of Solution

The gravitational force of an object on the Earth’s surface is determined by Newton’s universal law of gravitation.

    $F_{\text{grav}}=\frac{G{M}_{\text{Asteroid}}{M}_{\text{spacecraft}}}{r_{\text{oit}}^2}$        (I)

Here, $F_{\text{grav}}$ is the gravitational force, $G$ is the gravitational constant, $M_{\text{spacecraft}}$ is the mass of the spacecraft, $M_{\text{Asteroid}}$ is the mass of the Asteroid.

The mass of the asteroid is given by the density multiplied by the volume.

    $\begin{array}{l}{M}_{\text{asteroid}}=\rho V\\ M_{\text{asteroid}}=2000\text{kg}/{\text{m}}^3\left(\frac{4}{3}\pi r^3\right)\end{array}$        (II)

Use equation (II) in (I),

    $F_{\text{grav}}=\frac{4}{3}\pi rG{M}_{\text{spaseraft}}\left(2000\text{kg}/{\text{m}}^3\right)$        (III)

Conclusion:

Substitute $100\text{m}$ for $r$, $\left(6.67\times {10}^{-11}{\text{N}\cdot {\text{m}}^2/\text{kg}}^2\right)$ for $G$, $2\times {10}^4\text{kg}$ for $M_{\text{spacecraft}}$, in equation (III) to find $F_{\text{grav}}$.

    $\begin{array}{c}{F}_{\text{grav}}=\frac{4}{3}\pi \left(100\text{m}\right)\left(6.67\times {10}^{-11}{\text{N}\cdot {\text{m}}^2/\text{kg}}^2\right)\left(2\times {10}^4\text{kg}\right)\left(2000\text{kg}/{\text{m}}^3\right)\\ =1.1\text{N}\end{array}$

Therefore, the maximum gravitational force the tractor could exert on the asteroid $\underset{\_}{1.1\text{N}}$

(b)

To determine

The acceleration of the Asteroid.

Answer to Problem 77P

The acceleration of the Asteroid is $\underset{\_}{1.3\times {10}^{-10}{\text{m}/\text{s}}^2}$.

Explanation of Solution

The expression for the acceleration of the asteroid is given by,

    $a=\frac{G{M}_{\text{spacecraft}}}{r^2}$        (IV)

Conclusion:

Substitute $\left(6.67\times {10}^{-11}{\text{N}\cdot {\text{m}}^2/\text{kg}}^2\right)$ for $G$, $2\times {10}^4\text{kg}$ for $M_{\text{spacecraft}}$, in equation (IV) to find $a$.

    $\begin{array}{c}a=\frac{\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\left(2\times {10}^4\text{kg}\right)}{{\left(100\text{m}\right)}^2}\\ =1.3\times {10}^{-10}{\text{m}/\text{s}}^2\end{array}$

Therefore, The acceleration of the Asteroid is $\underset{\_}{1.3\times {10}^{-10}{\text{m}/\text{s}}^2}$.

(C)

To determine

The deflection of the asteroid if the tractor stayed near the asteroid for 1 year.

Answer to Problem 77P

The deflection of the asteroid if the tractor stayed near the asteroid for 1 year is $\underset{\_}{6.5\times {10}^4\text{m}}$.

Explanation of Solution

The expression for the deflection is given by,

    $\Delta x=v_{0}t+\frac{1}{2}a{t}^2$        (V)

Here, $\Delta x$ is the deflection of the asteroid,

Assume the initial velocity as zero but the time taken is given by,

    $\frac{1\text{year}}{1}\frac{365\text{days}}{1\text{year}}\frac{24\text{hours}}{1\text{day}}\frac{60\text{min}}{1\text{hour}}\frac{60\text{sec}}{1\text{min}}=3.154\times {10}^7\text{s}$

Conclusion:

Substitute $1.3\times {10}^{-10}{\text{m}/\text{s}}^2$ for $a$, $3.154\times {10}^7\text{sec}$ for $t$ and $0$ for $v$ in equation (V) to find $\Delta x$.

    $\begin{array}{c}\Delta x=\frac{1}{2}\left(1.3\times {10}^{-10}\right){\left(3.154\times {10}^7\right)}^2\\ =6.5\times {10}^4\text{m}\end{array}$

Therefore, The deflection of the asteroid if the tractor stayed near the asteroid for 1 year is $\underset{\_}{6.5\times {10}^4\text{m}}$.

(d)

To determine

Whether the asteroid hits the Earth or not.

Answer to Problem 77P

Yes, the asteroid will hit the Earth.

Explanation of Solution

The expression for the deflection is given by,

    $\Delta x=v_{0}t+\frac{1}{2}a{t}^2$

Here, $\Delta x$ is the deflection of the asteroid,

Assume the initial velocity as zero equation (V) changes to,

    $\Delta x=\frac{1}{2}a{t}^2$        (VI)

Conclusion:

Substitute $1.3\times {10}^{-10}{\text{m}/\text{s}}^2$ for $a$, $6.37\times {10}^6\text{m}$ for $\Delta x$ in equation (VI),

    $\begin{array}{c}6.37\times {10}^6\text{m}=\frac{1}{2}\left(1.3\times {10}^{-10}\right)t^2\\ t=3.13\times {10}^8\text{s}\end{array}$

It is about 10 years,

The time taken by the asteroid to cover the distance to Earth at a constant velocity is,

    $\begin{array}{c}\Delta t=\frac{\Delta x}{v}\\ =\frac{6.0\times {10}^{11}\text{m}}{3\times {10}^4\text{m}/\text{s}}\\ =2.0\times {10}^7\text{s}\end{array}$

It is about $7.6\text{months}$. Thus asteroid will hit the earth about seven and a half months after it is detected.

Therefore, , the asteroid will hit the Earth.