Chapter 7, Problem 11Q

(a)

To determine

The mass of a hypothetical spherical asteroid with a diameter equal to $\text{2 km}$ and is composed of rocks with an average density of $2500\text{}{\text{ kg/m}}^{\text{3}}$.

Answer to Problem 11Q

Solution:

The mass of asteroid is $1.047\times {10}^{13}\text{ kg}$.

Explanation of Solution

Given data:

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

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

Formula used:

The mass of the asteroid can be calculated by the following expression:

$m=\rho \cdot V$

Here, $m$ represents the mass of the asteroid, $\rho$ represents the density of the asteroid, and $V$ represents the volume of the asteroid.

The expression for the volume of a sphere is:

$V=\frac{4}{3}\pi r^3$

Here, $r$ represents the radius.

Conversion formula from kilometer to the meter is:

1 km = 1000 m

Explanation:

Recall the expression for calculating the volume.

$V=\frac{4}{3}\pi r^3$

Substitute 1 km for $r$ and use the conversion formula.

$\begin{array}{l}V=\frac{4}{3}\pi {\left(1\text{ km}\times \left(\frac{1000\text{ m}}{1\text{ km}}\right)\right)}^3\\ V=\frac{4}{3}\pi {\left(1000\text{ m}\right)}^3\end{array}$

Recall the expression of mass.

$m=\rho \cdot V$

Substitute $\frac{4}{3}\pi {\left(1000\text{ m}\right)}^3$ for $V$ and $2500\text{}{\text{ kg/m}}^{\text{3}}$ for $\rho$.

$\begin{array}{c}m=\rho \cdot V\\ =\rho \cdot \left(\frac{4}{3}\pi r^3\right)\\ =\left(2500\text{}{\text{ kg/m}}^{\text{3}}\right)\left(\frac{4}{3}\pi {\left(1000\text{ m}\right)}^3\right)\\ =1.047\times {10}^{13}\text{ kg}\end{array}$

Conclusion:

Thus, the mass of asteroid is $1.047\times {10}^{13}\text{ kg}$.

(b)

To determine

The escape velocity to escape from the surface of an asteroid, if the diameter of the spherical asteroid is 2 km and is composed of rocks with an average density of $2500\text{}{\text{ kg/m}}^{\text{3}}$.

Answer to Problem 11Q

Solution:

The escape velocity of the asteroid is 1.18 m/s.

Explanation of Solution

Given data:

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

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

Formula used:

The expression for escape speed required to escape from the surface is:

$v_{\text{escape}}=\sqrt{\frac{2GM}{R}}$

Here, $v_{\text{escape}}$ represents the escape speed, $G$ represents the universal gravitational constant, $M$ represents the mass of the body, and $R$ represents the radius of the body.

The expression for calculating the radius is:

$r=\left(\frac{\text{d }}{2}\right)$

Conversion formula from kilometer to the meter is:

1 km = 1000 m

Explanation:

Refer the sub-part (a) for the value of mass that is $1.047\times {10}^{13}\text{ kg}$.

Consider the value of $G$ as $6.67\times {10}^{-11}{\text{Nm}}^2/{\text{kg}}^2$.

Recall the expression for calculating the radius.

$\text{Radius}=\left(\frac{\text{Diameter }}{2}\right)$

Substitute $2\text{ km}$ for diameter and also use the conversion formula.

$\begin{array}{c}\text{Radius}=\left(\frac{2\text{ km}}{2}\right)\times \left(\frac{1000\text{ m}}{1\text{ km}}\right)\\ =1000\text{ m}\end{array}$

Recall the expression of escape velocity.

$v_{\text{escape}}=\sqrt{\frac{2GM}{R}}$

Substitute $6.67\times {10}^{-11}{\text{Nm}}^2/{\text{kg}}^2$ for $G$, $1.047\times {10}^{13}\text{ kg}$ for $M$, and $1000\text{ m}$ for $R$.

$\begin{array}{c}{v}_{\text{escape}}=\sqrt{\frac{2\left(6.67\times {10}^{-11}{\text{ Nm}}^{\text{2}}/{\text{kg}}^2\right)\left(1.047\times {10}^{13}\text{ kg}\right)}{\left(\text{1000 m}\right)}}\\ =\sqrt{13.966\times {10}^{-1}}\\ =1.18\text{ m/s}\end{array}$

Conclusion:

Thus, the velocity to escape this asteroid is given as $1.18\text{ m/s}$.

(c)

To determine

The situation of an astronaut, if he decided to go for a jog with the speed 3m/s on an asteroid. If the diameter of the spherical asteroid is 2 km and is composed of rocks, with an average density of $2500\text{}{\text{ kg/m}}^{\text{3}}$.

Answer to Problem 11Q

Solution:

He would eventually leave the planet and float in the space.

Explanation of Solution

Introduction:

If a body attains a speed greater than the escape velocity for that surface, then it would leave the surface and acquire its position in space.

Explanation:

From sub-part (a), the value of escape speed for an asteroid is, 1.8 m/s.

The astronaut started jogging on the asteroid with the speed 3 m/s and this speed is greater than the escape velocity for the asteroid surface. So, the astronaut would eventually escape from the asteroid surface.

Conclusion:

Since the jog speed of the astronaut is greater than the escape velocity of the asteroid, he will eventually leave the planet and float in the space.