Question

For a rocket propulsion velocity of exhaust gases relative to rocket is 2km/s.If mass of rocket system is 1000kg,then what is the rate of fuel consumption for a rocket to rise up with acceleration 4.9m/s^2?

Answer 1

Answer on Question #44912, Physics, Mechanics | Kinematics | Dynamics

For a rocket propulsion velocity of exhaust gases relative to rocket is $2\mathrm{km/s}$ . If mass of rocket system is $1000\mathrm{kg}$ , then what is the rate of fuel consumption for a rocket to rise up with acceleration $4.9\mathrm{m/s^2}$ ?

Solution.

According to the general equation of variable-mass motion

$$\vec {F} _ {e x t} + \vec {V} _ {r e l} \frac {d m}{d t} = m \frac {d \vec {V}}{d t}$$

where $F_{\text{ext}}$ is the net external force on the body, $V_{\text{rel}}$ is the relative velocity of the escaping or incoming mass with respect to the center of mass of the body, and $v$ is the velocity of the body (copypaste from Wikipedia)

In projection on the vertical axis:

$$- m g - V _ {r e l} \left(- \frac {d m}{d t}\right) = m \frac {d V}{d t}$$

The mass of the rocket is decreasing so $\frac{dm}{dt}$ term has negative sign.

$$\frac {d V}{d t} = a$$

So:

$${}_{rel} \frac{dm}{dt} = m(a + g)$$ $$\frac{dm}{dt} = \frac{m(a + g)}{_{rel}}$$

Numerically:

$$\frac{dm}{dt} = \frac{1000kg \cdot \left(4.9 \frac{m}{s^2} + 9.8 \frac{m}{s^2}\right)}{2000 \frac{m}{s}} = 7.35 \frac{kg}{s}$$

Answer:

$$\frac{dm}{dt} = 7.35 \frac{kg}{s}$$

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