Chapter 8, Problem 74PQ

(a)

To determine

The maximum height of the ball using conservation of energy.

Answer to Problem 74PQ

The maximum height of the ball is $1.26\times {10}^4\text{ m}$ .

Explanation of Solution

Write the expression for the conservation of energy for the situation.

  $K_i+U_{gi}=K_f+U_{gf}$                                                                                                   (I)

Here, $K_i$ is the initial kinetic energy of the ball, $U_{gi}$ is the initial gravitational potential energy of the ball, $K_f$ is the final kinetic energy of the ball and $U_{gf}$ is the gravitational potential energy of the ball at maximum height.

Assume that the gravitational potential energy is zero on Mimas’s surface. The kinetic energy of the ball will be zero at the maximum height.

Write the expression for $U_{gi}$ and $K_f$ .

  $\begin{array}{c}{U}_{gi}=0\\ K_f=0\end{array}$                                                                                                                      (II)

Write the expression for $K_i$ .

  $K_i=\frac{1}{2}m{v}_i^2$                                                                                                              (III)

Here, $m$ is the mass of the ball and $v_i$ is the speed of the ball.

Write the expression for $U_{gf}$ .

  $U_{gf}=mg{h}_{\text{max}}$                                                                                                           (IV)

Here, $g$ is the acceleration due to gravity and $h_{\text{max}}$ is the maximum height of the ball.

Put equations (II) to (IV) in equation (I) and rewrite it for $h_{\text{max}}$ .

  $\begin{array}{c}\frac{1}{2}m{v}_i^2+0=0+mg{h}_{\text{max}}\\ g{h}_{\text{max}}=\frac{1}{2}{v}_i^2\\ h_{\text{max}}=\frac{v_i^2}{2g}\end{array}$                                                                                         (V)

Conclusion:

It is given that the speed of the ball is $40\text{ m/s}$ and the value of $g$ on Mimas is $0.0636{\text{ m/s}}^2$ .

Substitute $40\text{ m/s}$ for $v_i$ and $0.0636{\text{ m/s}}^2$ for $g$ in equation (V) to find $h_{\text{max}}$ .

  $\begin{array}{c}{h}_{\text{max}}=\frac{{\left(40\text{ m/s}\right)}^2}{2\left(0.0636{\text{ m/s}}^2\right)}\\ =1.26\times {10}^4\text{ m}\end{array}$

Therefore, the maximum height of the ball is $1.26\times {10}^4\text{ m}$ .

(b)

To determine

The maximum height of the ball using universal gravitation.

Answer to Problem 74PQ

The maximum height of the ball using universal gravitation is $1.34\times {10}^4\text{ m}$ .

Explanation of Solution

Write the equation for the initial gravitational potential energy of the ball using universal gravitation.

  $U_{gi}=-\frac{G{M}_{m}m}{R_m}$                                                                                                      (VI)

Here, $G$ is the universal gravitation constant, $M_m$ is the mass of Mimas and $R_m$ is the radius of Mimas.

Write the equation for the gravitational potential energy of the ball at maximum height using universal gravitation.

  $U_{gf}=-\frac{G{M}_{m}m}{r}$                                                                                                      (VII)

Here, $r$ is the distance from the center of the Mimas to the maximum height of the ball.

The final kinetic energy of the ball is zero.

  $K_f=0$                                                                                                               (VIII)

Put equations (III), (VI), (VII) and (VIII) in equation (I) and rewrite it for $r$ .

    $\begin{array}{c}\frac{1}{2}m{v}_i^2+\left(-\frac{G{M}_{m}m}{R_m}\right)=0+\left(-\frac{G{M}_{m}m}{r}\right)\\ -\frac{G{M}_m}{r}=\frac{1}{2}{v}_i^2-\frac{G{M}_m}{R_m}\\ \frac{1}{r}=\frac{1}{R_m}-\frac{v_i^2}{2G{M}_m}\\ r=\frac{1}{\frac{1}{R_m}-\frac{v_i^2}{2G{M}_m}}\end{array}$                                                                   (IX)

Write the expression for the height of the ball above the surface of Mimas.

  $h_{\text{max}}=r-R_m$                                                                                                             (X)

Conclusion:

Given that the value of $R_m$ is $1.98\times {10}^5\text{ m}$ and the value of $M_m$ is $3.75\times {10}^{19}\text{ kg}$ . The value of $G$ is $6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^2/{\text{kg}}^2$ .

Substitute $1.98\times {10}^5\text{ m}$ for $R_m$, $40\text{ m/s}$ for $v_i$, $6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^2/{\text{kg}}^2$ for $G$ and $3.75\times {10}^{19}\text{ kg}$ for $M_m$ in equation (IX) to find $r$ .

  $\begin{array}{c}r=\frac{1}{\frac{1}{1.98\times {10}^5\text{ m}}-\frac{{\left(40\text{ m/s}\right)}^2}{2\left(6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\left(3.75\times {10}^{19}\text{ kg}\right)}}\\ =2.11\times {10}^5\text{ m}\end{array}$

Substitute $2.11\times {10}^5\text{ m}$ for $r$ and $1.98\times {10}^5\text{ m}$ for $R_m$ in equation (X) to find $h_{\text{max}}$ .

  $\begin{array}{c}{h}_{\text{max}}=2.11\times {10}^5\text{ m}-1.98\times {10}^5\text{ m}\\ =1.34\times {10}^4\text{ m}\end{array}$

Therefore, the maximum height of the ball using universal gravitation is $1.34\times {10}^4\text{ m}$ .

(c)

To determine

The difference in result of part (a) with that of part (b) as percent and whether the estimate is too high or low.

Answer to Problem 74PQ

The percent difference of estimate of part (a) with the result of part (b) is $6.0\text{\%}$ and the estimate is lower than the more accurate value of part (b).

Explanation of Solution

The value obtained for $h_{\text{max}}$ in part (a) is $1.26\times {10}^4\text{ m}$ and that obtained in part (b) is $1.34\times {10}^4\text{ m}$ .

Calculate the percent difference between the two values.

  $\frac{1.34\times {10}^4\text{ m}-1.26\times {10}^4\text{ m}}{1.34\times {10}^4\text{ m}}\times 100=6.0\text{ \%}$

Conclusion:

The result obtained in part (a) is lower than the more accurate value in part (b) by $6.0\text{\%}$ . This difference cannot be considered as negligible since the maximum height is a significant fraction of the radius of the Mimas.

Therefore, the percent difference of estimate of part (a) with the result of part (b) is $6.0\text{\%}$ and the estimate is lower than the more accurate value of part (b).