Question

Consider a rocket that is in deep space and at rest relative to an inertial reference frame. The rocket’s engine is to be fired for a certain interval. What must be the rocket’s mass ratio (ratio of initial to final mass) over that interval if the rocket’s original speed relative to the inertial frame is to be equal to (a) the exhaust speed (speed of the exhaust products relative to the rocket) and (b)2.0times the exhaust speed?

Short answer

a) The rocket’s mass ratio when the rocket’s original speed is equal to the exhaust speed,$\frac{M_i}{M_f}=2.7$

b) The rocket’s mass ratio when the rocket’s original speed is 2.0 times the exhaust speed, $\frac{M_i}{M_f}=.47$

Step-by-step solution

Listing the given quantities

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

Understanding the concept of law of conservation of momentum

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

Formula:

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

Explanation

We have, the second rocket equation as,

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

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

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

(a) Calculation of the rocket’s mass ratio when the rocket’s original speed is equal to the exhaust speed

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

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

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

 Step 5: (b) Calculation of the rocket’s mass ratio when the rocket’s original speed is equal to double the exhaust speed

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

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

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