Chapter 10, Problem 46QAP

To determine

(a)

The speed of the asteroid by which it strikes the moon's surface when it is falling with speed of $2.00\times {10}^{2}m/s$at the altitude of $1.00\times {10}^{3}m.$

Answer to Problem 46QAP

Speed of the asteroid by which it strikes the surface of moon is $1436m/s.$

Explanation of Solution

Given:

Mass of the moon = $M=7.35\times {10}^{22}kg$

Radius of the moon = $R=1.737\times {10}^6\text{}m$

Altitude of the asteroid = $h=1.00\times {10}^{3}km=1.00\times {10}^{6}m$

Mass of the asteroid = $m=1.00\times {10}^{2}kg$

Formula used:

Gravitation potential energy is defined as

  $\begin{array}{l}U=-\frac{GMm}{r}\\ r=\text{ distance between center to center of the two objects}\text{.}\end{array}$

Kinetic energy of an object is written as,

  $\begin{array}{l}K.E=\frac{1}{2}m{v}^2\\ \text{Here},v\text{is the speed of the object}\text{.}\end{array}$

Calculation:

Consider the speed at which asteroid strike to the surface of moon is $v_{surface}$

By conservation of mechanical energy,

  $\begin{array}{l}{(}_K={(}_K\\ \frac{1}{2}m{v}_{surface}{}^2-\frac{GMm}{R}=\frac{1}{2}m{v}^2-\frac{GMm}{(R+h)}\\ \frac{\cancel{m}{v}^2{}_{surface}}{2}=\frac{GM\cancel{m}}{R}-\frac{GM\cancel{m}}{(R+h)}+\frac{1}{2}\cancel{m}{v}^2\\ \frac{v^2{}_{surface}}{2}=GM\left(\frac{1}{R}-\frac{1}{(R+h)}\right)+\frac{1}{2}{v}^2\\ v_{surface}=\sqrt{2GM\left(\frac{1}{R}-\frac{1}{(R+h)}\right)+v^2}\\ v_{surface}=\sqrt{2GM\left(\frac{R+h-R}{R(R+h)}\right)+v^2}\\ v_{surface}=\sqrt{\frac{2GMh}{(R+h)R}+v^2}\end{array}$

Now, let's plug all these values,

  $\begin{array}{l}{v}_{surface}=\sqrt{\frac{2\times 6.67\times {10}^{-11}N{m}^2/k{g}^2\times 7.35\times {10}^{22}kg\times 1.00\times {10}^{6}m}{(1.737\times {10}^{6}m+1.00\times {10}^{6}m)(1.737\times {10}^{6}m)}+{(2.00\times {10}^{2}m/s)}^2}\\ v_{surface}=1436m/s\end{array}$

Conclusion:

Thus, the speed of the asteroid when it strikes to the surface of the moon is $1436m/s$.

To determine

(b)

The work done on asteroid by the moon when it strikes to the surface of the moon and stops there

Answer to Problem 46QAP

The work done on the asteroid during collision $-1.03\times {10}^{8}J.$

Explanation of Solution

Given:

Mass of the asteroid = $m=1.00\times {10}^{2}kg$

Speed of the asteroid just before colliding with the surface of moon = $v_{surface}=1436m/s$

Speed of the asteroid after stopping = $v_{stopping}=0$

Calculation:

Since, asteroid is coming to stop and its displacement inside the surface of moon is very small comparison to the radius of the moon. So, in this case, change in potential energy would be zero $(\Delta U=0).$

Thus, total work done on the asteroid by the moon consumes in the change (loss) in kinetic energy.

  $\begin{array}{l}W=\Delta K.E\\ W=\frac{1}{2}m{v}^2{}_{stopping}-\frac{1}{2}m{v}^2{}_{surface}\\ W=\frac{1}{2}m(0)-\frac{1}{2}\times 1.00\times {10}^{2}kg\times {(}^{1436}\\ W=-1.03\times {10}^{8}J\end{array}$

Conclusion:

Thus, the work done on the asteroid by the moon is $-1.03\times {10}^{8}J.$