Chapter 7, Problem 11Q

To determine

(a)

The mass of a hypothetical spherical asteroid.

Answer to Problem 11Q

The mass of the hypothetical spherical asteroid is $1.04\times {10}^{13}\text{kg}$.

Explanation of Solution

Given:

The diameter of the asteroid is $2\text{km}$.

The average density of the asteroid is $2500\text{kg}/{\text{m}}^{\text{3}}$.

Concept used:

The density of a body is defined as the total mass of it present in the total volume of the substance in the region of space.

Write the expression for the density of the body.

$\rho =\frac{m}{V}$

Rearrange the above expression for $m$.

$m=\rho V$ ...... (1)

Here, $\rho$ is the average density of the body, $m$ is the total mass of the body and $V$ is the total volume of the body.

Write the expression for the volume of the spherical asteroid.

$V=\frac{4}{3}\pi {\left(\frac{D}{2}\right)}^3$

Here, $D$ is the diameter of the asteroid.

Substitute $\frac{4}{3}\pi {\left(\frac{D}{2}\right)}^3$ for $V$ in equation (1).

$m=\rho \left(\frac{4}{3}\pi {\left(\frac{D}{2}\right)}^3\right)$ ...... (2)

Calculation:

Substitute $2500\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ and $2\text{km}$ for $D$ in equation (2).

$\begin{array}{c}m=\left(2500\text{kg}/{\text{m}}^{\text{3}}\right)\left(\frac{4}{3}\pi {\left(\frac{2\text{km}\left(\frac{1000\text{m}}{1\text{km}}\right)}{2}\right)}^3\right)\\ =\left(2500\text{kg}/{\text{m}}^{\text{3}}\right)\left(\frac{4}{3}\pi {\left(1000\text{m}\right)}^3\right)\\ =1.04\times {10}^{13}\text{kg}\end{array}$

$$

Conclusion:

Thus, the mass of the hypothetical spherical asteroid is $1.04\times {10}^{13}\text{kg}$.

To determine

(b)

The speed required to escape from the surface of the asteroid.

Answer to Problem 11Q

The escape velocity of the asteroid is $1.17\text{m}/\text{s}$.

Explanation of Solution

Given:

The mass of the hypothetical spherical asteroid is $1.04\times {10}^{13}\text{kg}$.

The diameter of the asteroid is $2\text{km}$.

Concept used:

According to the law of conservation of mechanical energy, the total mechanical energy of the system is always a constant and therefore, the sum of the kinetic and the potential energy is also a constant, provided only conserved forces are applied.

Write the expression for conservation of mechanical energy as the sum of the kinetic and the potential energy.

${\left(K+U\right)}_i={\left(K+U\right)}_f$ ...... (3)

Here, $K$ is the kinetic energy of the body and $U$ is the potential energy of the body from the asteroid.

Write the expression for kinetic energy.

$K=\frac{1}{2}m{v}^2$

Here, $m$ is the mass of the escaping body and $v$ is the velocity of the escaping body.

Write the expression for the gravitational potential energy.

$U=\frac{-GMm}{R}$

Here, $G$ is the universal gravitational constant, $M$ is the mass of the asteroid, $m$ is the mass of the body which escapes from the surface of the asteroid, $R$ is the distance of the escaped body from the center of the massive asteroid.

Substitute $\frac{1}{2}m{v}^2$ for $K$ and $\frac{-GMm}{R}$ for $U$ in equation (3).

${\left(\frac{1}{2}m{v}^2+\left(\frac{-GMm}{R}\right)\right)}_i={\left(=\frac{1}{2}m{v}^2+\left(\frac{-GMm}{R}\right)\right)}_f$

Simplify the above expression.

$\frac{1}{2}m{v}_i^2+\left(\frac{-GMm}{R_i}\right)=\frac{1}{2}m{v}_f^2+\left(\frac{-GMm}{R_f}\right)$ ...... (4)

Here, $v_i$ is the initial velocity of the escaping body, $R_i$ is the initial distance of the escaping body from the center of the asteroid or the radius of the asteroid, $R_f$ is the final distance of the escaping body from the center of the asteroid, and $v_f$ is the final velocity of the escaping body.

As a body escapes with velocity, the only force on it is the gravitational force of the massive body and when it goes out of the gravitational pull, the force on it is nearly zero. Therefore, it is considered to be at infinity and final velocity $v_f$ becomes negligibly small and the distance $R_f$ of it from the center of the massive body is infinitely large.

Substitute $0\text{m/s}$ for $v_f$ and $\infty$ for $R_f$ in equation (4).

$\begin{array}{c}\frac{1}{2}m{v}_i^2+\left(\frac{-GMm}{R_i}\right)=\frac{1}{2}m{\left(0\text{m/s}\right)}^2+\left(\frac{-GMm}{\infty }\right)\\ \frac{1}{2}m{v}_i^2+\left(\frac{-GMm}{R_i}\right)=0+0\end{array}$

Simplify and rearrange the above expression for $v_i$.

$v_i=\sqrt{\frac{2GM}{R_i}}$ ...... (5) Write the relation between the diameter and the radius.

$R_i=\frac{D_i}{2}$

Here, $D_i$ is the diameter of the asteroid.

Substitute $\frac{D_i}{2}$ for $R_i$ and rearrange the equation (5) for $v_i$.

$v_i=\sqrt{\frac{4GM}{D_i}}$ ...... (6)

Calculation:

Substitute $6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}/{\text{kg}}^{\text{2}}$ for $G$, $1.04\times {10}^{13}\text{kg}$ for $M$ and $2\text{km}$ for $D_i$ in equation (6).

$\begin{array}{c}{v}_i=\sqrt{\frac{4\left(6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}/{\text{kg}}^{\text{2}}\right)\left(1.04\times {10}^{13}\text{kg}\right)}{2\text{km}\left(\frac{1000\text{m}}{1\text{km}}\right)}}\\ =\sqrt{\frac{4\left(6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}/{\text{kg}}^{\text{2}}\right)\left(1.04\times {10}^{13}\text{kg}\right)}{2000\text{m}}}\\ =1.17\text{m}/\text{s}\end{array}$

Conclusion:

Thus, the escape velocity of the asteroid is $1.17\text{m}/\text{s}$.

To determine

(c)

The condition if an astronaut starts to jog on the surface of the asteroid with a velocity of $3\text{m/s}$.

Answer to Problem 11Q

The astronaut will escape out from the gravitational field of the asteroid.

Explanation of Solution

Concept used:

The escape velocity of the asteroid is $1.17\text{m/s}$ and the astronaut is jogging with a velocity of $3\text{m/s}$ and that’s because he will escape out breaking the gravitational pull of the asteroid.

Conclusion:

Thus, the astronaut will escape out from the gravitational field of the asteroid.