Chapter 5, Problem 70P

(a)

To determine

The acceleration due to gravity at the surface of the Vesta.

Answer to Problem 70P

The acceleration due to gravity at the surface of the Vesta is $0.293{\text{m/s}}^2$.

Explanation of Solution

Write the expression from the kinematics equation of motion.

  $\Delta y=v_{0}t+\frac{1}{2}a{t}^2$        (I)

Here, $\Delta y$ is the height of the rock drops, $v_0$ is the initial speed of the rock, $t$ is the time taken by the rocks strikes on the ground, and $a$ is the acceleration.

Substitute $0$ for $v_0$ in above relation and rewrite for acceleration.

  $\begin{array}{c}\Delta y=\left(0\right)t+\frac{1}{2}a{t}^2\\ \Delta y=\frac{1}{2}a{t}^2\\ a=\frac{2\Delta y}{t^2}\end{array}$

Conclusion:

Substitute $1.5\text{m}$ for $\Delta y$ and $3.2\text{s}$ for $t$ in above relation to find acceleration..

  $\begin{array}{c}a=\frac{2\left(1.5\text{m}\right)}{{\left(3.2\text{s}\right)}^2}\\ =0.293{\text{m/s}}^2\end{array}$

Therefore, the acceleration due to gravity at the surface of the Vesta is $0.293{\text{m/s}}^2$.

(b)

To determine

The mass of the Vesta.

Answer to Problem 70P

The mass of the Vesta is $3.2\times {10}^{14}\text{kg}$.

Explanation of Solution

Write the expression for acceleration due to gravity.

  $g=\frac{GM}{r^2}$        (II)

Here, $M$ is the mass of the Vesta, $G$ is the gravitational constant, $r$ is the radius of the surface of Vesta, and $g$ is the acceleration due to gravity.

Rewrite the above expression for mass.

  $M=\frac{g{r}^2}{G}$        (III)

Conclusion:

Substitute $0.293{\text{m/s}}^2$ for $g$, $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2$ for $G$, and $270\text{m}$ for $r$ in equation (III).

  $\begin{array}{c}M=\frac{\left(0.293{\text{m/s}}^2\right){\left(270\text{m}\right)}^2}{\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\right)}\\ =3.2\times {10}^{14}\text{kg}\end{array}$

Therefore, the mass of the Vesta is $3.2\times {10}^{14}\text{kg}$.

(c)

To determine

The height of the astronaut could jump on Vesta planet.

Answer to Problem 70P

The height of the astronaut could jump on Vesta planet is $27.4\text{m}$.

Explanation of Solution

Write the expression from the equation of motion.

  $v^2=v_0^2-2g\Delta y$        (IV)

Here, $v$ is the final speed, $v_0$ is the initial speed, $\Delta y$ is the height, and $g$ is the acceleration due to gravity.

Rewrite the above expression for initial speed by substituting $0$ for $v$.

  $v_{0\text{Earth}}^2=2g_{\text{Earth}}\Delta y_{\text{Earth}}$        (V)

Here, $v_{0\text{Earth}}$ is the initial speed of the object on Earth, $\Delta y_{\text{Earth}}$ is the height of the rock drops on Earth surface, and $g_{\text{Earth}}$ is the acceleration due to gravity on Earth surface.

Similarly the initial speed of the object on Vesta surface,

  $v_{0\text{Vesta}}^2=2g_{\text{Vesta}}\Delta y_{\text{Vesta}}$        (VI)

Here, $v_{0\text{Vesta}}$ is the initial speed of the object on Vesta, $\Delta y_{\text{Vesta}}$ is the height of the rock drops on Vesta surface, and $g_{\text{Vesta}}$ is the acceleration due to gravity on Vesta surface.

Conclusion:

Compare the equations (V) and (VI).

  $\begin{array}{c}{v}_{0\text{Earth}}^2=v_{0\text{Vesta}}^2\\ 2g_{\text{Earth}}\Delta y_{\text{Earth}}=2g_{\text{Vesta}}\Delta y_{\text{Vesta}}\\ g_{\text{Earth}}\Delta y_{\text{Earth}}=g_{\text{Vesta}}\Delta y_{\text{Vesta}}\\ \Delta y_{\text{Vesta}}=\left(\frac{g_{\text{Earth}}}{g_{\text{Vesta}}}\right)\Delta y_{\text{Earth}}\end{array}$

Substitute $0.293{\text{m/s}}^2$ for $g_{\text{Vesta}}$, $9.80{\text{m/s}}^2$ for $g_{\text{Earth}}$, and $82\text{cm}$ for $\Delta y_{\text{Earth}}$ in above relation to find $\Delta y_{\text{Vesta}}$.

  $\begin{array}{c}\Delta y_{\text{Vesta}}=\left(\frac{9.80{\text{m/s}}^2}{0.293{\text{m/s}}^2}\right)\left(82\text{cm}\right)\left(\frac{0.01\text{m}}{1\text{cm}}\right)\\ =27.4\text{m}\end{array}$

Therefore, the height of the astronaut could jump on Vesta planet is $27.4\text{m}$.