In his 1865 science fiction novelFrom the Earth to the Moon,Jules Verne described how three astronauts are shot to the Moonby means of a huge gun. According to Verne, the aluminum capsule containing the astronauts is accelerated by ignition of nitrocellulose to a speed of$\mathbf{11}\mathbf{}\mathbf{km}\mathbf{/}\mathbf{s}$along the gun barrel’s length of $\mathbf{220}\mathbf{m}$.

(a) In gunits, what is the average acceleration of the capsule and astronauts in the gun barrel?

(b) Is that acceleration tolerable

Or deadly to the astronauts?

A modern version of such gun-launched spacecraft (although without passengers) has been proposed. In this modern version,

called the SHARP (Super High Altitude Research Project) gun, ignition of methane and air shoves a piston along the gun’s tube, compressing hydrogen gas that then launches a rocket. During this launch, the rocket moves$\mathbf{3}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{km}$ and reaches a speed of $\mathbf{7}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{km}\mathbf{/}\mathbf{s}$. Once launched, the rocket can be fired to gain additional speed.

(c) In gunits, what would be the average acceleration of the rocket within the launcher?

(d) How much additional speed is needed (via the rocket engine) if the rocket is to orbit Earth at an altitude of $\mathbf{700}\mathbf{}\mathbf{km}$?

The final speed of capsule is$v_c=11000\frac{m}{s}$

The length of gun barrel is$x=220m$

The final speed of rocket is$v_r=7000\frac{m}{s}$

The distance covered by rocket is

$\begin{array}{l}d=3.5km\\ =3500m\end{array}$

Using the kinematic equation of motion, we can find the average acceleration of the capsule and astronauts. Then from the formula for velocity obtained from the equation of equilibrium of gravitational force and centripetal force, we can find the velocity required for the rocket to orbit around the earth. Then, using the law of conservation of energy, we can find the velocity obtained by the rocket within the launcher. Taking the difference between them, we can find the additional speed needed for the rocket to orbit around the earth at an altitude of 700 km.

Formulae:

$$\begin{gathered} \overrightarrow{F}=\frac{GMm}{r^2}\widehat{r} \\ {v}^2=v_0^2+2ax \\ \frac{1}{2}m{v}_2^2-\frac{GMm}{r_2}=\frac{1}{2}m{v}_1^2 \end{gathered}$$

We have the kinematic equation

$v^2=v_0^2+2ax$

Rearranging this equation for acceleration and dividing by 9.8 for ‘g’ units, we get

$\begin{array}{l}a=\frac{v^2-v_0^2}{9.8\left(2x\right)}\\ =\frac{{11000}^2-0}{10(2\times 220)}\\ =2.75\times {10}^{5}m/s^2\\ =2.8\times {10}^{4}g\end{array}$

The average acceleration of the capsule and astronauts in g unit is $a=2.8\times {10}^{4}g$.

The acceleration is deadly enough to kill the passengers, as we know that humans cannot sustain an acceleration of more than approximately 10g.

Using the kinematic equation of part a, we get

$\begin{array}{l}a=\frac{v^2-v_0^2}{9.8\left(2x\right)}\\ =\frac{{7000}^2-0}{9.8(2\times 3500)}\\ =7000m/s^2\\ =714g\end{array}$

The average acceleration of the rocket within the launcher in g units is $a=714g$.

Let us calculate the velocity required for the rocket to orbit around the earth at an altitude of 700 km. So,$r=R_E+h=6.37\times {10}^6+7\times {10}^5=7.07\times {10}^{6}m$.

We can get the equation for velocity from the equation of equilibrium of gravitational force and centripetal force as

$\begin{array}{l}v=\sqrt{\frac{GM}{r}}\\ =\sqrt{\frac{(6.67\times {10}^{-11})(5.98\times {10}^{24})}{7.07\times {10}^6}}\\ =7.51\times {10}^3\frac{m}{s}\end{array}$

Now, the velocity obtained by the rocket within the launcher can be calculated using the conservation of energy equation.

$\frac{1}{2}m{v}_2^2-\frac{GMm}{r_2}=\frac{1}{2}m{v}_1^2$

Solving this equation for velocity at an altitude 700 km, we get

$v_2=6.05\times {10}^3\frac{m}{s}$

So, the additional speed required is

$\begin{array}{l}{V}_a=V-v_2\\ =1.46\times {10}^3\\ \cong 1.5\times {10}^3\frac{m}{s}\end{array}$

The additional speed needed for the rocket to orbit around the earth at an altitude 700 km

$V_a=1.5\times {10}^3\frac{m}{s}$