Chapter 7, Problem 91A

a.

To determine

Speed of cannonball.

Answer to Problem 91A

Speed of cannonball $v=1.674\times {10}^3\text{m/s}$ .

Explanation of Solution

Given:

Moon’s mass $m=7.3\times {10}^{22}\text{kg}$

Radius of Moon $r=1.738\times {10}^6\text{m}$

Formula used:

Velocity is given as,

  $v=\sqrt{\frac{Gm}{r}}$

Calculation:

The gravitational constant is,

  $G=6.67\times {10}^{-11}{\text{N-m}}^2/{\text{kg}}^{\text{2}}$

Substitute the values to find the vlocity,

  $\begin{array}{c}v=\sqrt{\frac{6.67\times {10}^{-11}\times 7.3\times {10}^{22}}{1.738\times {10}^6}}\\ =1.674\times {10}^3\text{m/s}\end{array}$

Conclusion:

Therefore, the speed of cannonball is $1.674\times {10}^3\text{m/s}$ .

b.

To determine

Time taken by cannonball to return to cannon.

Answer to Problem 91A

Time taken by cannonball to return to cannon is $T=6.77\times {10}^3\text{s}$ .

Explanation of Solution

Given:

Moon’s mass $m=7.3\times {10}^{22}\text{kg}$

Radius of Moon $r=1.738\times {10}^6\text{m}$

Formula used:

The time period is given as,

  $T=2\pi \sqrt{\frac{r^3}{Gm}}$

Calculation:

Substitute the given values, to find the time period T .

  $\begin{array}{c}T=2\times 3.14\times \sqrt{\frac{{\left(1.738\times {10}^6\right)}^3}{6.67\times {10}^{-11}\times 7.3\times {10}^{22}}}\\ =6.77\times {10}^3\text{s}\end{array}$

Conclusion:

Therefore, time taken by ball is $6.77\times {10}^3\text{s}$ .