Chapter 8, Problem 34A

In the previous chapter, rocket propulsion was explained in terms of Newton’s third law. That is, the force that propels a rocket is from the exhaust gases pushing against the rocket, the reaction to the force the rocket exerts on the exhaust gases. Explain rocket propulsion in terms of momentum conservation.

To determine

To explain:The rocket propulsion in terms of momentum conservation.

Explanation of Solution

Introduction:

One of the best examples of Newton’s laws of motion is the Rocket motion. Rocket motion is the change in mass system, where mass of the system is continuously changing. Due to burning of fuel,rocket loses its mass in its flight.The exhaust gas provides acceleration to the rocket in order to attain high velocity.

Suppose rocket is projected vertically by burning fuel. Suppose m be the mass of the rocket at time t . m is the mass of the rocket which is the sum of the masses of vehicle and fuel. Velocity acquired by the rocket at time t is v . When mass of the rocket decrease due to burning of the fuel is Δm within the time interval of Δt . Due to the reduction in mass, the velocity of the rocket increases by an amount Δv . Velocity of the exhaust gasses with respect to the rocket is u .

Then the relativevelocity of exhaust gas with respect to the stationary frame on earth

( ν+u+Δν ).

Thus the law of conservation of momentum gives,

  $\therefore mv=(m-\Delta m)(v+\Delta v)+\Delta m(v+u+\Delta v)$

After solving we get,

  $\therefore m\Delta v=-u\Delta m$

Negative sign indicate the decreased in mass.

Converting these into the exact differential form by dividing by Δt and taking limit as Δv →0

  $\therefore m\frac{dv}{dt}=-u\frac{dm}{dt}$

Correct the above equation by considering the constant gravitational field acting in the opposite direction to the velocity vector

  $\begin{array}{l}\therefore m\frac{dv}{dt}=-mg-u\frac{dm}{dt}\\ \therefore \frac{dv}{dt}=-g-\frac{u}{m}\frac{dm}{dt}\\ \therefore dv=-gdt-u\frac{dm}{m}\end{array}$

Integrating above equation we get velocity,

  $\therefore v=-gt-u\ln (m)+C$

Where, Cis the constant of integration.

Applying the initial conditions we get

  $\therefore v=-gt-u\ln (\frac{m}{m0})$

Where, m₀ = m_f + m_vis the initial mass of the rocket

This is the velocity of the rocket in vertical direction.

Conclusion: By considering the momentum conservation we can derive the expression for the velocity, displacement of the rocket at any time t .