Rocket Propulsion

Impluse And Momentum of Class 11

Consider a rocket of mass M with fuel of mass Δm. Their common velocity is v relative to some inertial frame. When the rocket engines are fired, the gases expelled backward with an exhaust velocity = −vex relative to the rocket. This is a fixed quantity determined by the design of the engine and the type of fuel. If the rocket’s velocity changes to relative to the inertial frame, then the velocity of gas with respect to the frame will be

= (−vex+ v + Δv)

Applying the law of conservation of momentum, we get

(M + Δm) v = M(v + Δv) + Δm(-vex+ v + Δv)

After some cancellation we find

0 = MΔv + Δm(-vex + Δv)

If both Δv and Δm are small quantities relative to v and M, respectively, their product , ΔvΔm, is negligible in comparison with the other terms, and we are left with

Δv = vex(9.30)

Since an increase in the mass of the expelled gases corresponds exactly to the loss in mass of the rocket system, we have Δm = -ΔM. In the limit as ΔM → 0 equation (9.30) becomes

dv = -vex

On integrating both sides,

we find

vf– vi= vex ln(9.31)

Example 9.19

The mass of a rocket is 2.8 × 106 kg at the launch time. Of this, 2 × 106 kg is fuel. The exhaust speed is 2500 m/s and the fuel is ejected at the rate of 1.4 × 104 kg/s.

(a) Find the thrust of the rocket

(b) What is its initial acceleration at launch time? Ignore air resistance.

Solution

(a) The magnitude of the thrust is given by

Thrust = vex = (2500)(1.4 × 104) = 3.5 × 107 N

(b)Using equation

We get the acceleration

a =

= -g + vex= -9.8 + 12.5 = +2.7 m/s2

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