Question

A rocket, which is in deep space and initially at rest relative to an inertial reference frame, has a mass of 60.4 × 105 kg, of which 3.51 × 105 kg is fuel. The rocket engine is then fired for 380 s, during which fuel is consumed at the rate of 490 kg/s. The speed of the exhaust products relative to the rocket is 2.36 km/s. (a) What is the rocket's thrust? After the 380 s firing, what are (b) the mass and (c) the speed of the rocket?

Answer 1

Answer:

a The rocket's thrust is $T = 1.15*10^6N$

b The mass of the rocket $M_{new} =5.85*10^{6}kg$

c  The speed of the rocket $v =75.4 m/s$

Explanation:

From the question we are told that

            The mass of Rocket  is  $m_R = 60.4 *10^{5} \ kg$

             The mass of fuel is $m_F = 3.51 kg$

            The duration the rocket engine was fired $t = 380s$  

            The rate at  which fuel was consumed $R=\frac{\Delta M}{\Delta t} = 490 kg/s$

  The speed of the exhaust product is  $s = 2.36 km/s$ $=2.36 *1000 = 2360 m/s$

Generally Thrust can be  represented as

           $T = s * R$

   Substituting values

                 $T =2360*490 = 1.15*10^6N$

  The mass of fuel consumed in 380 s would  be

                 Rate of  fuel consumption * The the time taken (The duration the rocket engine was fired)

               =  490 × 380 = 186200 kg

  So the new mass of rocket would be

              $M_{new} = Old \ mass \ of \ rocket - mass \ of \ fuel \ consumed$

                        $= 60.4*10^5 -186200$

                        $=5.85*10^6 kg$

Generally velocity of a rocket can be mathematically represented as

                        $v = v_E *ln [\frac{M_i}{M_f} ]$

where $v_E$ is the speed of the exhaust

           $M_i$ is the initial mass of the rocket

         $M_f$ is the final mass of the rocket

   $v = 2360 * ln[\frac{60.4*10^5}{5.85*10^6} ]$

     $=2368 * ln (1.0325)$

   $=75.4 m/s$