Question

To illustrate the use of a multistage rocket consider the following:
(a) A certain rocket carries 60% of its initial mass as fuel. (That is, the mass of fuel is 0.6mo) What is the rocket's final speed, accelerating from rest in free space, if it burns all its fuel in a single stage? Express your answer as a multiple of v ex .
(b) Suppose instead it burns the fuel in two stages as follows: In the first stage it burns a mass 0.3m o of fuel. It then jettisons the first-stage fuel tank, which has a mass of 0.1m o , and thenburns the remaining 0.3m 0 of fuel. Find the final speed in this case, assuming the same value of vex throughout and compare.

Answer 1

Answer:

a

 $v_f = 0.916 v_{ex}$

b

 $v_f = 1.05 v_{ex}$

Explanation:

Considering question a

From the question we are told that

   The mass of the rocket is  $m_o$  

    The mass of the fuel which is in the rocket is  m = 0.6 M  

Generally given that the rocket burns all it fuel in a single stage , the final velocity of the rocket is mathematically represented as

          $v_f = v_{ex} ln [\frac{m_o}{m_k} ]$

Here $m_k$ is the mass  of the rocket without fuel which is mathematically evaluated as

        $m_k = m_o- m$

=>      $m_k = m_o -0.6m_o$

=>    $m_k = [1- 0.6] m_o$

=>      $v_f = v_{ex} ln [\frac{m_o}{[1 - 0.6]m_o} ]$    

=>      $v_f = 0.916 v_{ex}$

Considering question b

From the question we are told that

   The mass of the fuel it burn at the first stage is  $m = 0.3 m_o$

     The mass of the first stage fuel tank $m_1 = 0.1 m_o$

    The mass of the fuel at the second stage is  $m_f = 0.3m_o$

Generally the  velocity of the rocket at the first stage  is mathematically represented as

           $v_i = v_{ex} * ln [ \frac{m_o }{[1- m]m_o } ]$

=>         $v_i = v_{ex} * ln [ \frac{m_o }{[1- 0.3]m_o } ]$

=>         $v_i = v_{ex} * ln [ \frac{1 }{0.7 } ]$  

=>         $v_i =0.357 v_{ex}$

Generally the mass of the rocket after first stage is  

    $m_r = m_o - 0.3m_o -0.1m_o$

=>   $m_r = 0.6m_o$

Generally the final  velocity of the rocket at the second stage is  

     $v_f = v_i + v_{ex} * ln [\frac{m_r}{ m_f } ]$

=>   $v_f = v_i + v_{ex} * ln [\frac{0.6 m_o }{0.3mo } ]$

=>   $v_f = 0.357 v_{ex} + 0.693 v_{ex}$

=>   $v_f = 1.05 v_{ex}$