Question

A $5000\,kg$ rocket is set for a vertical firing. The exhaust speed is $800\,ms^{- 1}$ . To give an initial upward acceleration of $30\,ms^{- 2}$ , the amount of gas ejected (in $kg\,s^{- 1}$ ) to supply the needed thrust will be (Take $g=10\,ms^{- 2}$ )

Answer 1

Given: Mass of rocket, $\left( m \right)=5000\,kg$
Exhaust speed, $\left( v \right)=800\,ms^{- 1}$
Acceleration of rocket, $\left( a \right)=20\,ms^{- 2}$
Gravitational acceleration, $\left( g \right)=10\,ms^{- 2}$
We know that the upward force, $F_{\text{net}}=F_{\text{Thrust}}-W...\left(1\right)$
where, the terms are net upward force on the rocket, thrust force on the rocket and the weight of rocket, respectively.
We know that the thrust force on rocket is given by,
$F_{\text{Thrust}}=v\left(- \frac{d m}{d t}\right)...\left(2\right)$
where, $v$ and $\frac{d m}{d t}$ are the velocity of exhaust and change of mass of rocket, respectively. Negative sign is there because the mass of a rocket decreases due to ejection of gas.
From equation $\left(1\right)$ and $\left(2\right)$ , we can write
$\Rightarrow ma=v\left(- \frac{d m}{d t}\right)-mg\Rightarrow -\frac{d m}{d t}=\frac{m \left(a + g\right)}{v}$
Substituting the given values in the formula and solving we get,
$-\frac{d m}{d t}=\frac{5000 \left(30 + 10\right)}{800}=250\,kgs^{- 1}$