Chapter 2, Problem 46PQ

In Example 2.6, we considered a simple model for a rocket launched from the surface of the Earth. A better expression for the rocket’s position measured from the center of the Earth is given by

$\stackrel{\to }{y}(t)={\left(R_{\oplus }^{3/2}+3\sqrt{\frac{g}{2}}{R}_{\oplus }t\right)}^{2/3}\hat{j}$

where $R_{\oplus }$ is the radius of the Earth (6.38 × 10⁶ m) and g is the constant acceleration of an object in free fall near the Earth’s surface (9.81 m/s²).

a. Derive expressions for ${\stackrel{\to }{v}}_y(t)$ and ${\stackrel{\to }{a}}_y(t)$.

b. Plot y(t), v_y(t), and a_y(t). (A spreadsheet program would be helpful.)

c. When will the rocket be at $y=4R_{\oplus }$?

d. What are ${\stackrel{\to }{v}}_y$ and ${\stackrel{\to }{a}}_y$ when $y=4R_{\oplus }$?

To determine

The expressions for ${\overrightarrow{\text{v}}}_y\left(t\right)$ and ${\overrightarrow{\text{a}}}_y\left(t\right)$.

Answer to Problem 46PQ

The expression for ${\overrightarrow{\text{v}}}_y\left(t\right)$ is ${\overrightarrow{\text{v}}}_y\left(t\right)=2\sqrt{\frac{g}{2}}{R}_{\oplus }{\left(R_{\oplus }^{3/2}+3\sqrt{\frac{g}{2}}{R}_{\oplus }t\right)}^{-1/3}\widehat{\text{j}}$ and the expression for ${\overrightarrow{\text{a}}}_y\left(t\right)$ is ${\overrightarrow{\text{a}}}_y\left(t\right)=-\left(g{R}_{\oplus }^2\right){\left(R_{\oplus }^{3/2}+3\sqrt{\frac{g}{2}}{R}_{\oplus }t\right)}^{-4/3}\widehat{\text{j}}$.

Explanation of Solution

Write the given expression for the position vector.

  ${\overrightarrow{\text{y}}}_y\left(t\right)={\left(R_{\oplus }^{3/2}+3\sqrt{\frac{g}{2}}{R}_{\oplus }t\right)}^{2/3}\widehat{\text{j}}$                                                                                  (I)

Here, $\overrightarrow{\text{y}}\left(t\right)$ is the position vector, $R_{\oplus }$ is the radius of the Earth, $g$ is the acceleration due to gravity, and $t$ is the time.

Velocity is the time derivative of position vector. Write the equation for velocity.

  ${\overrightarrow{\text{v}}}_y\left(t\right)=\frac{d\overrightarrow{\text{y}}\left(t\right)}{dt}$                                                                                                   (II)

Here, ${\overrightarrow{\text{v}}}_y\left(t\right)$ is the velocity.

Acceleration is the time derivative of velocity. Write the expression for acceleration.

  ${\overrightarrow{\text{a}}}_y\left(t\right)=\frac{d{\overrightarrow{\text{v}}}_y\left(t\right)}{dt}$                                                                                                (III)

Here, ${\overrightarrow{\text{a}}}_y\left(t\right)$ is the acceleration.

Conclusion:

Put equation (I) in equation (II).

  $\begin{array}{c}{\overrightarrow{\text{v}}}_y\left(t\right)=\frac{d\left[{\left(R_{\oplus }^{3/2}+3\sqrt{\frac{g}{2}}{R}_{\oplus }t\right)}^{2/3}\widehat{\text{j}}\right]}{dt}\\ =\frac{2}{3}{\left(R_{\oplus }^{3/2}+3\sqrt{\frac{g}{2}}{R}_{\oplus }t\right)}^{-1/3}3\sqrt{\frac{g}{2}}{R}_{\oplus }\widehat{\text{j}}\\ =2\sqrt{\frac{g}{2}}{R}_{\oplus }{\left(R_{\oplus }^{3/2}+3\sqrt{\frac{g}{2}}{R}_{\oplus }t\right)}^{-1/3}\widehat{\text{j}}\end{array}$                                                               (IV)

Put equation (IV) in equation (III).

  $\begin{array}{c}{\overrightarrow{\text{a}}}_y\left(t\right)=\frac{d\left[2\sqrt{\frac{g}{2}}{R}_{\oplus }{\left(R_{\oplus }^{3/2}+3\sqrt{\frac{g}{2}}{R}_{\oplus }t\right)}^{-1/3}\widehat{\text{j}}\right]}{dt}\\ =2\sqrt{\frac{g}{2}}{R}_{\oplus }\left(-\frac{1}{3}\right){\left(R_{\oplus }^{3/2}+3\sqrt{\frac{g}{2}}{R}_{\oplus }t\right)}^{-4/3}3\sqrt{\frac{g}{2}}{R}_{\oplus }\widehat{\text{j}}\\ =-\left(g{R}_{\oplus }^2\right){\left(R_{\oplus }^{3/2}+3\sqrt{\frac{g}{2}}{R}_{\oplus }t\right)}^{-4/3}\widehat{\text{j}}\end{array}$                                              (V)

Therefore, the expression for ${\overrightarrow{\text{v}}}_y\left(t\right)$ is ${\overrightarrow{\text{v}}}_y\left(t\right)=2\sqrt{\frac{g}{2}}{R}_{\oplus }{\left(R_{\oplus }^{3/2}+3\sqrt{\frac{g}{2}}{R}_{\oplus }t\right)}^{-1/3}\widehat{\text{j}}$ and the expression for ${\overrightarrow{\text{a}}}_y\left(t\right)$ is ${\overrightarrow{\text{a}}}_y\left(t\right)=-\left(g{R}_{\oplus }^2\right){\left(R_{\oplus }^{3/2}+3\sqrt{\frac{g}{2}}{R}_{\oplus }t\right)}^{-4/3}\widehat{\text{j}}$ .

To determine

Plots of $\overrightarrow{\text{y}}\left(t\right)$, ${\overrightarrow{\text{v}}}_y\left(t\right)$ and ${\overrightarrow{\text{a}}}_y\left(t\right)$ .

Answer to Problem 46PQ

The plot of $\overrightarrow{\text{y}}\left(t\right)$ is

The plot of ${\overrightarrow{\text{v}}}_y\left(t\right)$ is

And the plot of ${\overrightarrow{\text{a}}}_y\left(t\right)$ is

Explanation of Solution

The graph of position versus time of an object gives the position of the object at different instant of time. The slope of the position versus time graph gives the magnitude of the velocity of the object. In velocity versus time graph of an object, its velocity at different instants of time is plotted. The slope of this graph gives the magnitude of acceleration of the object. In acceleration versus time graph, acceleration is plotted as a function of time.

The plot of $\overrightarrow{\text{y}}\left(t\right)$ is shown in figure 1.

The plot of ${\overrightarrow{\text{v}}}_y\left(t\right)$ is shown in figure 2.

From the figure it is clear that the rocket has maximum velocity when it starts its motion and the velocity decreases with time. The graph has negative slope implying the acceleration is negative.

The plot of ${\overrightarrow{\text{a}}}_y\left(t\right)$ is shown in figure 3.

From the figure, it is clear that the rocket has negative acceleration.

To determine

The time at which the rocket will be at $y=4R_{\oplus }$ .

Answer to Problem 46PQ

The time at which the rocket will be at $y=4R_{\oplus }$ is $2661\text{ s}$.

Explanation of Solution

Equation (I) can be used to determine the time at which the rocket will be at $y=4R_{\oplus }$ .

Substitute $4R_{\oplus }$ for $y$ in equation (I).

  $4R_{\oplus }={\left(R_{\oplus }^{3/2}+3\sqrt{\frac{g}{2}}{R}_{\oplus }t\right)}^{2/3}$

Take the power $\left(3/2\right)$ of the above equation and rewrite it for $t$ .

  $\begin{array}{c}{\left(4R_{\oplus }\right)}^{3/2}=R_{\oplus }^{3/2}+3\sqrt{\frac{g}{2}}{R}_{\oplus }t\\ t=\frac{R_{\oplus }^{3/2}\left(4^{3/2}-1\right)}{3\sqrt{\frac{g}{2}}{R}_{\oplus }}\\ =\frac{\left(4^{3/2}-1\right)}{3}\sqrt{\frac{2R_{\oplus }}{g}}\end{array}$                                                                                    (VI)

Conclusion:

Given that the radius of the Earth is $6.38\times {10}^6\text{ m}$ and the acceleration due to gravity is $9.81{\text{ m/s}}^2$ .

Substitute $6.38\times {10}^6\text{ m}$ for $R_{\oplus }$ and $9.81{\text{ m/s}}^2$ for $g$ in equation (VI) to find $t$ .

  $\begin{array}{c}t=\frac{\left(4^{3/2}-1\right)}{3}\sqrt{\frac{2\left(6.38\times {10}^6\text{ m}\right)}{9.81{\text{ m/s}}^2}}\\ =2661\text{ s}\end{array}$

Therefore, the time at which the rocket will be at $y=4R_{\oplus }$ is $2661\text{ s}$.

To determine

The value of ${\overrightarrow{\text{v}}}_y$ and ${\overrightarrow{\text{a}}}_y$ when $y=4R_{\oplus }$ .

Answer to Problem 46PQ

The value of ${\overrightarrow{\text{v}}}_y$ when $y=4R_{\oplus }$ is $\left(5590\text{ m/s}\right)\widehat{\text{j}}$ and the value of ${\overrightarrow{\text{a}}}_y$ is $\left(-0.613{\text{ m/s}}^2\right)\widehat{\text{j}}$ .

Explanation of Solution

Equation (IV) can be used to determine the value of ${\overrightarrow{\text{v}}}_y$ and equation (V) can be used to determine the value of ${\overrightarrow{\text{a}}}_y$ .

Conclusion:

Substitute $2661\text{ s}$ for $t$, $6.38\times {10}^6\text{ m}$ for $R_{\oplus }$ and $9.81{\text{ m/s}}^2$ for $g$ in equation (IV) to find ${\overrightarrow{\text{v}}}_y$ .

  $\begin{array}{c}{\overrightarrow{\text{v}}}_y=2\sqrt{\frac{9.81{\text{ m/s}}^2}{2}}\left(6.38\times {10}^6\text{ m}\right){\left({\left(6.38\times {10}^6\text{ m}\right)}^{3/2}+3\sqrt{\frac{9.81{\text{ m/s}}^2}{2}}\left(6.38\times {10}^6\text{ m}\right)\left(2661\text{ s}\right)\right)}^{-1/3}\widehat{\text{j}}\\ \text{=}\left(5590\text{ m/s}\right)\widehat{\text{j}}\end{array}$

Substitute $2661\text{ s}$ for $t$, $6.38\times {10}^6\text{ m}$ for $R_{\oplus }$ and $9.81{\text{ m/s}}^2$ for $g$ in equation (V) to find ${\overrightarrow{\text{a}}}_y$ .

  $\begin{array}{c}{\overrightarrow{\text{a}}}_y\left(t\right)=-\left(\left(9.81{\text{ m/s}}^2\right){\left(6.38\times {10}^6\text{ m}\right)}^2\right){\left({\left(6.38\times {10}^6\text{ m}\right)}^{3/2}+3\sqrt{\frac{\left(9.81{\text{ m/s}}^2\right)}{2}}\left(6.38\times {10}^6\text{ m}\right)\left(2661\text{ s}\right)\right)}^{-4/3}\widehat{\text{j}}\\ =\left(-0.613{\text{ m/s}}^2\right)\widehat{\text{j}}\end{array}$

Therefore, the value of ${\overrightarrow{\text{v}}}_y$ when $y=4R_{\oplus }$ is $\left(5590\text{ m/s}\right)\widehat{\text{j}}$ and the value of ${\overrightarrow{\text{a}}}_y$ is $\left(-0.613{\text{ m/s}}^2\right)\widehat{\text{j}}$ .