Chapter 11, Problem 9P

(a)

To determine

The free-fall acceleration on the surface of the satellite.

Answer to Problem 9P

The free-fall acceleration on the surface of the satellite is $\underset{\_}{0.0761\text{m}/{\text{s}}^2}$.

Explanation of Solution

Write the expression for the force on the test object.

    $F_g=m_{\text{obj}}g$        (I)

Here, $F_g$ is the force on the test object in the field, $m_{\text{obj}}$ is the mass of the test object, $g$ is the free-fall acceleration.

Write the expression for the gravitational force.

    $F_G=\frac{G{M}_{\text{miranda}}{m}_{\text{obj}}}{r_{\text{miranda}}^2}$        (II)

Here, $F_G$ is the gravitational force, $G$ is the gravitational constant, $M_{\text{miranda}}$ is the mass of the satellite, $r_{\text{miranda}}$ is the radius of the satellite.

Equate equation (I) and (II) to solve for $g$.

    $\begin{array}{c}{m}_{\text{obj}}g=\frac{G{M}_{\text{miranda}}{m}_{\text{obj}}}{r_{\text{miranda}}^2}\\ g=\frac{G{M}_{\text{miranda}}}{r_{\text{miranda}}^2}\end{array}$        (III)

Conclusion:

Substitute $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2$ for $G$, $6.68\times {10}^{19}\text{kg}$ for $M_{\text{miranda}}$, $242\text{km}$ for $r_{\text{miranda}}$ in equation (III) to find $g$.

    $\begin{array}{c}g=\frac{\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\left(6.68\times {10}^{19}\text{kg}\right)}{\left(242\text{km}\times \frac{{10}^3\text{m}}{1\text{km}}\right)}\\ =0.0761\text{m}/{\text{s}}^2\end{array}$

Therefore, the free-fall acceleration on the surface of the satellite is $\underset{\_}{0.0761\text{m}/{\text{s}}^2}$.

(b)

To determine

The time taken by the athlete to climb the cliff vertically.

Answer to Problem 9P

The time taken by the athlete to climb the cliff vertically is $\underset{\_}{363\text{s}}$.

Explanation of Solution

Write the expression for the equation of motion in vertical direction.

    $s_y=v_{yi}t+\frac{1}{2}{a}_y{t}^2$        (IV)

Here, $s_y$ is the vertical displacement from the surface of the satellite, $v_{yi}$ is the initial velocity in vertical direction, $t$ is the time, $a_y$ is the acceleration in the vertical direction.

Set the acceleration in the vertical direction, $a_y$ as the free-fall acceleration on the surface.

    $a_y=-g$        (V)

Use equation (V) in (IV) to solve for $s_y$.

    $s_y=v_{yi}t+\frac{1}{2}\left(-g\right)t^2$        (VI)

If the object starts from the rest, the equation (VI) becomes,

    $s_y=\frac{1}{2}\left(-g\right)t^2$        (VII)

Use equation (VII) to solve for $t$.

    $t={\left(\frac{2s_y}{-g}\right)}^{1/2}$        (VIII)

Conclusion:

Substitute $-5.00\text{km}$ for $s_y$, $0$ for $v_{yi}$, $0.0761\text{m}/{\text{s}}^2$ for $g$ in equation (VIII) to find $t$.

    $\begin{array}{c}t={\left(\frac{2\left(-5.00\text{km}\times \frac{{10}^3\text{m}}{1\text{km}}\right)}{-0.0761\text{m}/{\text{s}}^2}\right)}^{1/2}\\ =363\text{s}\end{array}$

Therefore, the time taken by the athlete to climb the cliff vertically is $\underset{\_}{363\text{s}}$.

(c)

To determine

The distance covered by the athlete from the base of the vertical cliff to the icy surface of the satellite.

Answer to Problem 9P

The distance covered by the athlete from the base of the vertical cliff to the icy surface of the satellite is $\underset{\_}{3.08\times {10}^3\text{m}}$.

Explanation of Solution

Write the expression for the equation of motion in horizontal direction

    $s_x=v_{xi}t+\frac{1}{2}{a}_x{t}^2$        (IX)

Here, $s_x$ is the displacement in the horizontal direction, $v_{xi}$ is the initial horizontal velocity, $a_x$ is the acceleration in the horizontal direction.

Conclusion:

Substitute $8.5\text{m}/\text{s}$ for $v_{xi}$, $363\text{s}$ for $t$, $0$ for $a_x$ in equation (IX) to find $s_x$.

    $\begin{array}{c}{s}_x=\left(8.5\text{m}/\text{s}\right)\left(363\text{s}\right)+0\\ =3.08\times {10}^3\text{m}\end{array}$

Therefore, the distance covered by the athlete from the base of the vertical cliff to the icy surface of the satellite is $\underset{\_}{3.08\times {10}^3\text{m}}$.

(d)

To determine

The vector impact velocity of the athlete in climbing the cliff.

Answer to Problem 9P

The vector impact velocity of the athlete in climbing the cliff is $\underset{\_}{28.9\text{m}/\text{s}\text{at }72.9°\text{below the horizontal}}$.

Explanation of Solution

Write the expression for the velocity vector.

    ${\overrightarrow{v}}_f=\left(v_x\widehat{\text{i}}\text{+}{v}_y\widehat{j}\right)$        (X)

Here, is the final velocity vector, ${\overrightarrow{v}}_x$ is the velocity vector in horizontal direction, ${\overrightarrow{v}}_y$ is the velocity vector in the vertical direction.

Write the expression for $v_{yf}$.

    $v_y=v_{yi}+a_{y}t$        (XI)

Here, $v_y$ is the final velocity in the vertical direction.

Write the expression for the magnitude of the ${\overrightarrow{v}}_f$.

    ${\text{v}}_f=\sqrt{v_x^2+v_y^2}$        (XII)

Write the expression for the angle of the cliff making with the $X$ axis.

    $\varphi ={\tan }^{-1}\left(\frac{v_y}{v_x}\right)$        (XIII)

Here, $\varphi$ is the angle of the cliff making with the $X$ axis.

Conclusion:

Substitute $0$ for $v_{yi}$, $-0.0761\text{m}/{\text{s}}^2$ for $a$, $363\text{s}$ for $t$ in equation (XI) to find $v_y$.

    $\begin{array}{c}{v}_y=0-\left(0.0761\text{m}/{\text{s}}^2\right)\left(363\text{s}\right)\\ =-27.6\text{m}/\text{s}\end{array}$

Substitute $8.50\text{m}/\text{s}$ for $v_x$, $-27.6\text{m}/\text{s}$ for $v_y$ in equation (XII) to find $v_f$.

    $\begin{array}{c}{v}_f=\sqrt{{\left(8.50\text{m}/\text{s}\right)}^2+{\left(-27.6\text{m}/\text{s}\right)}^2}\\ =28.88\text{m}/\text{s}\\ \approx 28.9\text{m}/\text{s}\end{array}$

Substitute $27.6\text{m}/\text{s}$ for $v_y$, $8.50\text{m}/\text{s}$ for $v_x$ in equation (XIII) to solve for $\varphi$.

    $\begin{array}{c}\varphi ={\tan }^{-1}\left(\frac{27.6\text{m}/\text{s}}{8.50\text{m}/\text{s}}\right)\\ =72.9°\end{array}$

Therefore, the vector impact velocity of the athlete in climbing the cliff is $\underset{\_}{28.9\text{m}/\text{s}\text{at }72.9°\text{below the horizontal}}$.