91Ó°ÊÓ

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:

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

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

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

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

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

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

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,$r=R_E+h=6.37×{10}^6+7×{10}^5=7.07×{10}^{6}m$.

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

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

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

$v_2=6.05×{10}^3\frac{m}{s}$

So, the additional speed required is

$\begin{array}{l}{V}_a=V−v_2\\ =1.46×{10}^3\\ â‰\dots 1.5×{10}^3\frac{m}{s}\end{array}$

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

$V_a=1.5×{10}^3\frac{m}{s}$