Chapter 11, Problem 19P

To determine

Radius of the asteroid.

Answer to Problem 19P

Near about $3\text{km}$

Explanation of Solution

Introduction:

The escape velocity on the surface of the asteroid is

  $v_{e\text{ Asteroid}}=\sqrt{\frac{2G{M}_{\text{asteroid}}}{R_{\text{asteroid}}}}$

Here, $M_{\text{asteroid}}$ is the mass of asteroid and $R_{\text{asteroid}}$ is radius of asteroid.

Relation among mass, density and volume is ,

  $\begin{array}{l}{M}_{\text{ Asteroid}}={\rho }_{\text{ Asteroid}}{V}_{\text{ Asteroid}}\\ ={\rho }_{\text{ Asteroid}}\frac{4}{3}\pi R_{\text{ Asteroid}}^3\end{array}$

The second kinematic equation of motion, $v_{\text{f}}^{\text{2}}-v_{\text{i}}^{\text{2}}=2gh$

The escape velocity on the surface of the asteroid is

  $\begin{array}{l}{v}_{e\text{ Asteroid}}=\sqrt{\frac{2G{M}_{\text{asteroid}}}{R_{\text{asteroid}}}}\\ =\sqrt{\frac{2G{\rho }_{\text{ Asteroid}}\frac{4}{3}\pi R_{\text{ Asteroid}}^3}{R_{\text{asteroid}}}}\\ =\left(\sqrt{\frac{8}{3}\pi G{\rho }_{\text{asteroid}}}\right)R_{\text{asteroid}}\mathrm{....}\left(1\right)\\ \end{array}$

From the equation $\left(1\right)$ ,

The radius of the Asteroid is given by

  $R_{\text{asteroid}}=v_{e\text{ Asteroid}}\sqrt{\frac{3}{8\pi G{\rho }_{\text{asteroid}}}}$

From the kinematic equation of motion, $v_f^2-v_i^2=2gh$

Here ‘h’ is the height to which you can jump on the surface of the Earth.

As finally come to rest, the final speed of you is $v=0\text{ m/s}$

Therefore

  $\begin{array}{l}{v}_{\text{f}}^{\text{2}}-v_{\text{i}}^{\text{2}}=2gh\\ 0-v_{\text{i}}^{\text{2}}=2gh\\ v_{\text{i}}=\sqrt{2gh}\end{array}$

Let $v_i=v_{e\text{ asteroid}}$

Therefore, the radius of the asteroid

  $\begin{array}{l}{R}_{\text{asteroid}}=v_{e\text{ Asteroid}}\sqrt{\frac{3}{8\pi G{\rho }_{\text{asteroid}}}}\\ =\sqrt{2gh}\sqrt{\frac{3}{8\pi G{\rho }_{\text{asteroid}}}}\\ =\sqrt{\frac{3}{8\pi G{\rho }_{\text{asteroid}}}\left(gh\right)}\mathrm{........}\left(2\right)\end{array}$

Asssume that you can jump a height $h=0.75\text{ m}$ , on the surface of Earth.

Density of the asteroid is ${\rho }_{\text{Asteroid}}=\left(3.0{\text{ g/cm}}^3\right)\left(\frac{1\text{ kg}}{1000\text{ g}}\right)\left(\frac{{10}^6{\text{ cm}}^3}{1{\text{ m}}^3}\right)$

  $=3.0\times {10}^3{\text{ kg/m}}^3$

Universal gravitational constant is $G=6.673\times {10}^{-11}\text{ N}{\text{.m}}^2/{\text{kg}}^2$

By substituting all known values in the equation $\left(2\right)$ , we get

Radius of the asteroid is

  $\begin{array}{l}{R}_{\text{asteroid}}=\sqrt{\frac{3}{8\pi G{\rho }_{\text{asteroid}}}\left(2gh\right)}\\ =\sqrt{\frac{3}{8\left(3.14\right)\left(6.73\times {10}^{-11}\text{ N}{\text{.m}}^2/{\text{kg}}^2\right)\left(3.0\times {10}^3{\text{kg/m}}^3\right)}\left(2\right)\left(9.8{\text{m/s}}^2\right)\left(0.75\text{ m}\right)}\\ =\left(2.96\times {10}^3\text{ m}\right)\left(\frac{1\text{ km}}{1000\text{ m}}\right)\\ =\boxed{3.0\text{ km}}\end{array}$

Conclusion:

Radius of asteroid is $3\text{ km}$