91Ó°ÊÓ

The initial velocity of rocket,$V_i=0\mathrm{m}/\mathrm{s}$

Here, we can use the second rocket equation to calculate the mass ratio in both cases.

Formula:

$v_f-v_i=v_{rel}×\mathrm{In}\left(\frac{M_i}{M_f}\right)$

We have, the second rocket equation as,

$v_f-v_i=v_{rel}×\mathrm{In}\left(\frac{M_i}{M_f}\right)$

Here,$v_f$is the original speed of rocket relative to the inertial frame of reference.

And$v_{rel}$is the exhaust speed of rocket

So, substituting the value of$v_i$and rearranging above equation for mass ratio, we get

$\frac{M_i}{M_f}=\exp \left(\frac{{\mathrm{v}}_{\mathrm{f}}}{{\mathrm{v}}_{\mathrm{rel}}}\right)$

When,$v_f=2×v_{rel}$ the above equation become,

role="math" localid="1661250600364" $$\begin{gathered} \frac{M_i}{M_f}=\exp \left(2\right) \\ â‰\dots 7.4 \end{gathered}$$

Hence, the rocket’s mass ratio when the rocket’s original speed is equal to the exhaust speed$\frac{M_i}{M_f=7.4}$

When $v_f=2×v_{rel}$the above equation become,

$$\begin{gathered} \frac{M_i}{M_f}=\exp \left(2\right) \\ â‰\dots 7.4 \end{gathered}$$

Hence, the rocket’s mass ratio when the rocket’s original speed is times the exhaust speed, $\frac{M_i}{M_f}=7.4$