Question

A rocket of mass m = 20 kg has M = 180kg fuel. The uniform exhaust velocity of the fuel is $v_r$ = 1.6 km/s. (i) Calculate the minimum rate of consumption of fuel so that the rocket may rise from the ground. (ii) Also calculate the maximum vertical speed gained by the rocket when the rate of consumption of fuel $\mu$ is (g = 10 m/s²) & (ln 10 = 2.30)

Answer 1

The minimum rate of consumption of fuel is 1.25 kg/s. The maximum vertical speed gained is 2780 m/s for $\mu$ = 2 kg/s and 3590 m/s for $\mu$ = 20 kg/s.

Answer 2

Part (i): Minimum rate of consumption of fuel so that the rocket may rise from the ground.

For the rocket to lift off from the ground, the upward thrust force must be at least equal to the initial weight of the rocket. The thrust force ($F_t$) is given by the product of the rate of fuel consumption ($\mu = -\frac{dm}{dt}$) and the exhaust velocity relative to the rocket ($v_r$). $F_t = \mu v_r$

The initial total mass of the rocket is $M_0 = m_{rocket} + m_{fuel} = 20 \text{ kg} + 180 \text{ kg} = 200 \text{ kg}$. The initial gravitational force is $F_g = M_0 g$.

For lift-off, $F_t \ge F_g$. The minimum rate of fuel consumption ($\mu_{min}$) occurs when $F_t = F_g$. $\mu_{min} v_r = M_0 g$ $\mu_{min} = \frac{M_0 g}{v_r}$

Given: $M_0 = 200 \text{ kg}$, $g = 10 \text{ m/s}^2$, $v_r = 1.6 \text{ km/s} = 1600 \text{ m/s}$. $\mu_{min} = \frac{200 \text{ kg} \times 10 \text{ m/s}^2}{1600 \text{ m/s}} = \frac{2000}{1600} \text{ kg/s} = \frac{20}{16} \text{ kg/s} = \frac{5}{4} \text{ kg/s} = 1.25 \text{ kg/s}$.

The minimum rate of consumption of fuel is 1.25 kg/s.

Part (ii): Maximum vertical speed gained by the rocket when the rate of consumption of fuel $\mu$ is given.

The equation of motion for a rocket moving vertically upwards under gravity is: $m \frac{dv}{dt} = F_t - mg$ $m \frac{dv}{dt} = \mu v_r - mg$ where $m(t)$ is the instantaneous mass, $v(t)$ is the instantaneous velocity, $\mu = -\frac{dm}{dt}$ is the constant rate of fuel consumption, and $v_r$ is the exhaust velocity. Rearranging and integrating from $t=0$ (velocity $v=0$, mass $M_0$) to time $T$ when fuel is exhausted (velocity $v_f$, mass $m_{rocket}$): $dv = \left(\frac{\mu v_r}{m} - g\right) dt$ Since $\mu = -\frac{dm}{dt}$ is constant, $dt = -\frac{dm}{\mu}$. Also, $m(t) = M_0 - \mu t$, so $t = \frac{M_0 - m}{\mu}$. The burn time is $T = \frac{M_0 - m_{rocket}}{\mu}$.

Integrating the equation of motion: $\int_{0}^{v_f} dv = \int_{0}^{T} \frac{\mu v_r}{M_0 - \mu t} dt - \int_{0}^{T} g dt$ $v_f = \mu v_r \int_{0}^{T} \frac{dt}{M_0 - \mu t} - g \int_{0}^{T} dt$ For the first integral, let $u = M_0 - \mu t$, $du = -\mu dt$. When $t=0$, $u=M_0$. When $t=T$, $u=M_0-\mu T = m_{rocket}$. $\int_{0}^{T} \frac{dt}{M_0 - \mu t} = \int_{M_0}^{m_{rocket}} \frac{-du/\mu}{u} = -\frac{1}{\mu} [\ln|u|]_{M_0}^{m_{rocket}} = -\frac{1}{\mu} (\ln m_{rocket} - \ln M_0) = \frac{1}{\mu} \ln\left(\frac{M_0}{m_{rocket}}\right)$. The second integral is $\int_{0}^{T} g dt = gT$.

So, $v_f = \mu v_r \left(\frac{1}{\mu} \ln\left(\frac{M_0}{m_{rocket}}\right)\right) - gT = v_r \ln\left(\frac{M_0}{m_{rocket}}\right) - gT$. The burn time $T = \frac{m_{fuel}}{\mu} = \frac{M_0 - m_{rocket}}{\mu}$. $v_f = v_r \ln\left(\frac{M_0}{m_{rocket}}\right) - g \frac{M_0 - m_{rocket}}{\mu}$.

Given: $m_{rocket} = 20 \text{ kg}$, $M_0 = 200 \text{ kg}$, $v_r = 1600 \text{ m/s}$, $g = 10 \text{ m/s}^2$, $\ln 10 = 2.30$. $\frac{M_0}{m_{rocket}} = \frac{200}{20} = 10$. $v_r \ln\left(\frac{M_0}{m_{rocket}}\right) = 1600 \ln(10) = 1600 \times 2.30 = 3680 \text{ m/s}$. $M_0 - m_{rocket} = 200 - 20 = 180 \text{ kg}$.

The formula for the maximum vertical speed (at the end of the burn) is $v_f = 3680 - 10 \times \frac{180}{\mu}$.

(a) $\mu = 2 \text{ kg/s}$ $v_f = 3680 - 10 \times \frac{180}{2} = 3680 - 10 \times 90 = 3680 - 900 = 2780 \text{ m/s}$.

(b) $\mu = 20 \text{ kg/s}$ $v_f = 3680 - 10 \times \frac{180}{20} = 3680 - 10 \times 9 = 3680 - 90 = 3590 \text{ m/s}$.

The maximum vertical speed gained is 2780 m/s for $\mu = 2$ kg/s and 3590 m/s for $\mu = 20$ kg/s.