Rocket Propulsion

The principle of conservation of momentum can be applied to study the motion of a rocket. This application is of special significance; as the rocket is a system, in which mass varies with time.

System of variable mass: Rocket

Usually, we assume that the total mass of a system remains constant. Sometimes, as in the case of a rocket, the mass does not remain constant. Most of the mass of a rocket on its launching pad is of fuel. The whole of the fuel is eventually burnt and ejected from the nozzle of the rocket engine. We handle the variable mass situation by applying Newton’s second law—not to the rocket alone, but to the rocket and its ejected combustion products taken together.

The mass of the combined system does not change as the rocket accelerates. In this way, we assume that external forces, e.g., gravitational forces and frictional forces of earth’s atmosphere, are negligibly small and do not affect the rocket’s motion.

Consider the flight of a rocket directed vertically upwards from the surface of earth. At $t = 0$, suppose

$ \displaystyle {{m}_{0}}$ $ \displaystyle =$ initial mass of the rocket including that of the fuel

$ \displaystyle {{v}_{0}}=$ initial velocity of the rocket

At any time $t$, suppose

$m = \text{mass of the rocket left}$,

$v = \text{velocity acquired by the rocket}$.

As the exhaust gases are escaping, $m < m_0$ and $v > v_0$.

In a small interval of time $dt$,

suppose

$ \displaystyle dm=$ a small decrease in mass of the rocket $ \displaystyle =$ mass of the exhaust gases that escape

$ \displaystyle dv=$ corresponding small increase in velocity of the rocket

$ \displaystyle {{v}_{g}}=$ velocity of exhaust gases w.r.t earth.

The velocity $v_g$ of exhaust gases is taken as negative, because these gases move in a direction opposite to that of the rocket.

According to the principle of conservation of linear momentum, the linear momentum of mass $m$ of the rocket at instant $t$ must be equal to the vector sum of linear momentum of rocket mass $(m – dm)$ moving with velocity $(v + dv)$ and linear momentum of mass $(dm)$ of exhaust gases moving downwards with velocity $(v – v_g)$. Therefore,

$$mv = (m – dm)(v + dv) + dm(v – v_g) \quad \text{…(1)}$$

or

$ \displaystyle mv$ $ \displaystyle =mv+(dm)(v-{{v}_{g}})-(dm)v$ …(as $ \displaystyle dm\text{ }dv$ is negligible)

As $ \displaystyle dm\text{ and }dv$, both are small, their product is negligibly small.

$$m \, dv = dm(v + v_g) \quad \text{…(2)}$$

When the rocket has gone far away from earth, the velocity of exhaust gases with respect to earth becomes irrelevant. We take the relative velocity of exhaust gases with respect to the rocket as $u$. As the gases and the rocket are moving in opposite directions, therefore:

$$v + v_g = -u \quad \text{…(3)}$$

The minus sign is again for the downward direction of motion of exhaust gases, in contrast to the upward direction of motion of the rocket. From (2), we have:

$$
m \, dv = -u \, dm \quad \text{…(4)}
$$

or

$$
dv = -\frac{u \, dm}{m}
$$

Now, at $t = 0; \, m = m_0$

$$
\text{and } v = v_0
$$

At $t = t; \, m = m \text{ and } v = v$

Therefore, integrating both sides of (4) within proper limits, we obtain

$$
\int_{v_0}^{v} dv = – \int_{m_0}^{m} \frac{u \, dm}{m}
$$

As the velocity of exhaust gases $(u)$ is constant throughout the firing of the rocket, therefore,

$$
\int_{v_0}^{v} dv = -\int_{m_0}^{m} \frac{u \, dm}{m}
$$

or

$$
[v]_{v_0}^{v} = -u \left[ \log_e m \right]_{m_0}^{m}
$$

$$
v – v_0 = -u \log_e m + u \log_e m_0
$$

$$
= -u \log_e \left( \frac{m}{m_0} \right) + u \log_e m_0
$$

or

$$
v = v_0 + u \log_e \left( \frac{m_0}{m} \right) \quad \text{…(5)}
$$

This equation gives us the velocity of the rocket at any time $t$, when its mass is $m$.

Usually, the initial velocity of the rocket at $t = 0$ is zero i.e., $v_0 = 0$.

Therefore, from (5),

$$
v = u \log_e \left( \frac{m_0}{m} \right) \quad \text{…(6)}
$$

Hence we conclude that the velocity of the rocket at any time is directly proportional to the exhaust speed $(u)$ of the gases and the natural log of the ratio of the initial mass of the rocket to its mass $(m)$ at that instant of time.

Note: As the exhaust gases escape, the residual mass $(m)$ of the rocket decreases with time. Therefore, from (36), the velocity $v$ of the rocket increases with time. Hence, the motion of the rocket is an accelerated motion.

For numerical problems, we may rewrite equation (6) as:

$$
v = 2.303 \, u \, \log_{10} \left( \frac{m_0}{m} \right)
$$

Burnt-out Speed of the Rocket is the speed attained by the rocket when the whole of the fuel of the rocket has been burnt.

Obviously, it is the maximum velocity acquired by the rocket. When the whole of the fuel of the rocket has been used up, the residual mass $(m_1)$ of the rocket is equal to the mass of the empty container of the fuel. From (6):

$ \displaystyle {{v}_{b}}=u{{\log }_{e}}\left( {\frac{{{{m}_{0}}}}{{{{m}_{1}}}}} \right)$ $ \displaystyle =2.303u{{\log }_{{10}}}\left( {\frac{{{{m}_{0}}}}{{{{m}_{1}}}}} \right)\quad …\text{(7)}$

Thrust on the rocket

It is the force with which the rocket moves upwards.

Dividing both sides of (4) by a small time interval $dt$, we get:

$$
m \frac{dv}{dt} = -u \frac{dm}{dt} \quad \text{…(8)}
$$

As $ \dfrac{dv}{dt} = a,$

acceleration of the rocket at time $t$,

$ \displaystyle mdv=ma=F$ $ \displaystyle =$ thrust on the rocket at time $ \displaystyle t$

From (2),

$$
F = -u \frac{dm}{dt} \quad \text{…(9)}
$$

The negative sign indicates that the thrust on the rocket is in a direction opposite to the direction of escaping gases.

Ignoring the negative sign, we find that the thrust on the rocket at any instant is the product of the velocity of escape of fuel gases w.r.t. the rocket and the rate of combustion of fuel at that instant.

We can calculate the instantaneous acceleration $(a)$ of the rocket at any time from equation (3), using the relation $a = F/m$. Here, $m$ is the mass of the rocket left at time $t$.

Note: If the effect of gravity on the motion of the rocket were taken into account, the upward thrust on the rocket would decrease by $(mg)$, the weight of the rocket. However, because of the very large values of $(u)$ and $\left(\dfrac{dm}{dt}\right)$, the effect of gravity is negligible.

Example:

A rocket motor consumes one quintal of fuel per second. The exhaust speed of gases w.r.t rocket is 5 km/s. Calculate the force exerted on the rocket. What is the velocity acquired by the rocket, when its mass reduces to $ \displaystyle {{\left( {\frac{1}{{100}}} \right)}^{{th}}}$ of its initial mass.

Solution:

$$\displaystyle \begin{array}{l}\text{Here, }\dfrac{{dm}}{{dt}}=1\text{ quintal/sec}\\=100\text{ kg/s}\\u=-5\text{ km/s}=-5\times {{10}^{3}}\text{ m/s}\\\text{As }F=-u\left( {\dfrac{{dm}}{{dt}}} \right)\\\therefore F=-\left( {-5\times {{{10}}^{3}}} \right)\left( {100} \right)=5\times {{10}^{5}}\text{ N}\end{array}$$

$ \displaystyle \begin{array}{*{20}{l}} {\dfrac{m}{{{{m}_{{}^\circ }}}}=\dfrac{1}{{100}}\text{ or }\dfrac{{{{m}_{{}^\circ }}}}{m}=100} \\ {\text{From }v=u{{{\log }}_{e}}\left( {\dfrac{{{{m}_{{}^\circ }}}}{m}} \right)} \\ \begin{array}{l}=5\times {{10}^{3}}{{\log }_{e}}100\\=\left( {5\times {{{10}}^{3}}} \right)\times 2.303{{\log }_{{10}}}100\end{array} \\ {=5\times {{{10}}^{3}}\times 2.303\times 2} \\ {=2.303\times {{{10}}^{4}}\text{ m/s}} \end{array}$