Name: Centrifugal Force
Uploaded: 2019-07-19
Duration: 3 min 19 s
Description: Have you seen the mud flying off from a tyre? How does it happen? Have you ever think of it? Let us see one such force cause the mud to move away from the ty...

Question

Write notes on rockets and jets. Obtain an expression for the acceleration of a rocket when gas, due to combustion of fuel in a rocket is ejected.

Accepted Answer

The operation of rockets and jets is based on the law of conservation of momentum. There is a combustion chamber in which fuel is burnt. The hot gases produced during combustion are expelled through the narrow opening at the back of the body, in the form of the jet at a very high speed. As there is no external force act on the system of (rocket+gas), the total momentum of the system is conserved. The momentum of gases in the backward direction is equal to the momentum of the rocket or jet plane in the forward direction. Hence, the jet or plane moves forward.

We assume that a rocket is launched from an inertial reference frame $F$ in the positive X-direction. The burnt-out gases are ejected in the negative X-direction. Let the mass of the rocket be $m_{o}$ , the exhausted velocity of the gas of the rocket will be $- u \hat{i}$ , the rate of ejection of the gases be $r$ . Hence, the mass of the rocket at the time $t$ , after the rocket is launched:

$m = m_{o} - r t$

Let the velocity of the rocket after the time $t$ relative to the frame $= v \hat{i}$

The momentum of the rocket after the time interval, $\vec{p} = m v \vec{i}$

Mass of the ejected gas in a time interval $d t$ , $d m = r d t$

Let the increase in velocity on the ejection of gas of mass $d m$ in this time interval be $d v \hat{i}$ . So the velocity of the rocket after this small time interval will be $(v + d v) \hat{i}$ . To apply the law of conservation of momentum for a system of two bodies, we must know the momentum of both the bodies in the same reference frame. For the system of (gas+rocket):

${\vec{V}}_{g R} = {\vec{V}}_{g F} - {\vec{V}}_{R F}$

$\Rightarrow {\vec{V}}_{g F} = - u \hat{i} + v \hat{i} = (v - u) \hat{i}$

Applying the law of conservation of momentum of the (rocket+gas):

$(m - d m) (v + d v) \hat{i} + d m (v - u) \hat{i}$ $- m v \hat{i} = 0$

$\Rightarrow (m d v - d m d v - u d m) \hat{i} = 0$

Neglecting the product of small quantities $d m d v$ and on differentiating with respect to $t$ , we get:

$m \frac{d v}{d t} \hat{i} = u \frac{d m}{d t} \hat{i}$

Acceleration of the rocket:

$\vec{a} = \frac{d v}{d t} \hat{i} = (\frac{r u}{m}) \hat{i} = (\frac{r u}{m_{o} - r t})$

Write notes on rockets and jets. Obtain an expression for the acceleration of a rocket when gas, due to combustion of fuel in a rocket is ejected.

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