Chapter 13, Problem 13.58AP

(a)

To determine

The maximum height gain by the space vehicle.

Answer to Problem 13.58AP

The maximum height gain by the space vehicle is $\frac{v_{\text{i}}^{2}R}{2\left(gR-\frac{1}{2}{v}_{\text{i}}^2\right)}$.

Explanation of Solution

The initial speed of the vehicle is $v_{\text{i}}$, the escape speed of space vehicle is $v_{\text{esc}}$ and the height of meteorite from the Earth’s surface is $h$.

Formula to calculate the maximum height gain by the space vehicle by the conservation of energy is,

    $K{E}_1+P{E}_1=K{E}_2+P{E}_2$        (1)

Here, $K{E}_1$ is the kinetic energy of the space vehicle at the Earth’s surface, $K{E}_2$ is the final kinetic energy of the space vehicle, $P{E}_1$ is the potential energy of the space vehicle at the Earth’s surface and $P{E}_2$ is final potential energy of the space vehicle.

 Formula to calculate the kinetic energy of the space vehicle at the Earth’s surface is,

    $K{E}_1=\frac{1}{2}m{v}_{\text{i}}{}^2$

Here, $m$ is the mass of the space vehicle and $v_{\text{i}}$ is the velocity of the space vehicle.

Formula to calculate the potential energy of the space vehicle at the Earth’s surface is,

    $P{E}_1=-\frac{GMm}{R}$

Here, $M$ is the mass of Earth, $m$ is the mass of the space vehicle, $R$ is the radius of the Earth and $G$ is the universal gravitational constant.

Formula to calculate the potential energy space vehicle at the altitude is,

  $P{E}_2=-\frac{GMm}{R+h_{\text{max}}}$

Here, $h_{\text{max}}$ is the maximum height gain by the space vehicle.

The final kinetic energy of the space vehicle is zero because the space vehicle is rest at that point.

Substitute $-\frac{GMm}{R}$ for $P{E}_1$, $-\frac{GMm}{R+h_{\text{max}}}$ for $P{E}_2$, $\frac{1}{2}m{v}_{\text{i}}{}^2$ for $K{E}_1$ and $0$ for $K{E}_2$ in equation (1) to find $h_{\text{max}}$.

    $\begin{array}{c}\frac{1}{2}m{v}_{\text{i}}{}^2+\left(-\frac{GMm}{R}\right)=0+\left(-\frac{GMm}{R+h_{\text{max}}}\right)\\ \frac{1}{2}m{v}_{\text{i}}{}^2=\frac{GMm}{R}-\frac{GMm}{R+h_{\text{max}}}\\ =GMm\left(\frac{1}{R}-\frac{1}{R+h_{\text{max}}}\right)\\ =GMm\left(\frac{h_{\text{max}}}{R\left(R+h_{\text{max}}\right)}\right)\end{array}$        (2)

Further solve the above expression.

    $\begin{array}{c}\frac{1}{2}m{v}_{\text{i}}^{2}R\left(R+h_{\text{max}}\right)=GMm{h}_{\text{max}}\\ \frac{1}{2}m{v}_{\text{i}}^2{R}^2+\frac{1}{2}m{v}_{\text{i}}^{2}R{h}_{\text{max}}=GMm{h}_{\text{max}}\\ h_{\text{max}}m\left(GM-\frac{1}{2}{v}_{\text{i}}^{2}R\right)=\frac{1}{2}m{v}_{\text{i}}^2{R}^2\\ h_{\text{max}}=\frac{v_{\text{i}}^2{R}^2}{2\left(GM-\frac{1}{2}{v}_{\text{i}}^{2}R\right)}\end{array}$

Write the expression for the acceleration due to gravity.

    $\begin{array}{c}g=\frac{GM}{R^2}\\ GM=g{R}^2\end{array}$

Here, $g$ is the acceleration due to gravity.

Substitute $g{R}^2$ for $GM$ in the above equation to find $h_{\text{max}}$.

    $\begin{array}{c}{h}_{\text{max}}=\frac{v_{\text{i}}^2{R}^2}{2\left(g{R}^2-\frac{1}{2}{v}_{\text{i}}^{2}R\right)}\\ =\frac{v_{\text{i}}^2{R}^2}{2R\left(gR-\frac{1}{2}{v}_{\text{i}}^2\right)}\\ =\frac{v_{\text{i}}^{2}R}{2\left(gR-\frac{1}{2}{v}_{\text{i}}^2\right)}\end{array}$

Conclusion:

Therefore, the maximum height gain by the space vehicle is $\frac{v_{\text{i}}^{2}R}{2\left(gR-\frac{1}{2}{v}_{\text{i}}^2\right)}$.

(b)

To determine

The speed of the meteorite to strike the Earth.

Answer to Problem 13.58AP

The speed of the meteorite to strike the Earth is $\sqrt{2gh\left(\frac{1}{\left(1+\frac{h}{R}\right)}\right)}$.

Explanation of Solution

The initial speed of the vehicle is $v_{\text{i}}$, the escape speed of space vehicle is $v_{\text{esc}}$ and the height of meteorite from the Earth’s surface is $h$.

 From equation (2), the expression for the speed is given as,

    $\begin{array}{c}\frac{1}{2}m{v}^2=GMm\left(\frac{h}{R\left(R+h\right)}\right)\\ v^2=\frac{2GMm}{m}\left(\frac{h}{R\left(R+h\right)}\right)\end{array}$

Here, $v$ is the speed of the meteorite to strike the Earth and $h$ is the height of meteorite from the Earth’s surface.

Further solve the above expression.

    $\begin{array}{c}{v}^2=\frac{2GMm}{m}\left(\frac{h}{R\left(R+h\right)}\right)\\ v=\sqrt{2GM\left(\frac{h}{R\left(R+h\right)}\right)}\end{array}$

Substitute $g{R}^2$ for $GM$ in the above equation to find $v$.

    $\begin{array}{c}v=\sqrt{2g{R}^2\left(\frac{h}{R\left(R+h\right)}\right)}\\ =\sqrt{2gR\left(\frac{h}{\left(R+h\right)}\right)}\\ =\sqrt{2gh\left(\frac{1}{\left(1+\frac{h}{R}\right)}\right)}\end{array}$

Conclusion:

Therefore, the speed of the meteorite to strike the Earth is $\sqrt{2gh\left(\frac{1}{\left(1+\frac{h}{R}\right)}\right)}$.

(c)

To determine

The result from part (a) is consistent with $h=\frac{u_{\text{i}}^2{\sin }^2{\theta }_{\text{i}}}{2g}$

Answer to Problem 13.58AP

The result from part (a) is consistent with $h=\frac{u_{\text{i}}^2{\sin }^2{\theta }_{\text{i}}}{2g}$.

Explanation of Solution

Consider a baseball is tossed up with an initial speed that is very small as compared to the escape speed.

    $v_{\text{i}}<v_{\text{esc}}$

Here, $v_{\text{esc}}$ is the escape speed of space vehicle.

As the initial speed that is very small. So the initial speed of the vehicle tends to be zero.

    $v_{\text{i}}\to 0$

From part (a), the maximum height gain by the space vehicle is,

    $h_{\text{max}}=\frac{v_{\text{i}}^{2}R}{2\left(gR-\frac{1}{2}{v}_{\text{i}}^2\right)}$

Substitute $0$ for $v_{\text{i}}$ in the above equation to find $h_{\text{max}}$.

    $\begin{array}{c}{h}_{\text{max}}=\frac{v_{\text{i}}^{2}R}{2\left(gR-0\right)}\\ =\frac{v_{\text{i}}^2}{2g}\end{array}$        (3)

Write the expression for the maximum height of the projectile motion of the baseball.

    $h=\frac{v_{\text{i}}^2{\sin }^2{\theta }_{\text{i}}}{2g}$

Here, $h$ is the maximum height of the projectile motion of the baseball and ${\theta }_{\text{i}}$ is the angle of the projectile motion of the baseball.

From maximum height of the projectile motion of the baseball, the value of angle of the projectile motion of the baseball should be $90°$.

Substitute $90°$ for ${\theta }_{\text{i}}$ in the above equation to find $h$.

    $\begin{array}{c}h=\frac{v_{\text{i}}^2{\sin }^{2}90°}{2g}\\ =\frac{v_{\text{i}}^2{\left(1\right)}^2}{2g}\\ =\frac{v_{\text{i}}^2}{2g}\end{array}$        (4)

From equations (3) and (4).

    $h_{\text{max}}=h$

So, the maximum height gain by the space vehicle is consistent with $h=\frac{u_{\text{i}}^2{\sin }^2{\theta }_{\text{i}}}{2g}$.

Conclusion:

Therefore, the result from part (a) is consistent with $h=\frac{u_{\text{i}}^2{\sin }^2{\theta }_{\text{i}}}{2g}$.