Question

A 6100 kgrocket is set for vertical firing from the ground. If the exhaust speed is 1200 m/s, how much gas must be ejected each second if the thrust (a) is to equal the magnitude of the gravitational force on the rocket and (b) is to give the rocket an initial upward acceleration of $\mathbf{21}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$?

Short answer

a) Therate of consumptionis 50 kg/s

b) The rate of consumption with the additional effect due to gravity is$1.6\times {10}^2\mathrm{kg}/\mathrm{s}$

Step-by-step solution

Listing the given quantities

The value ofexhaust speed is 1200 m/s

Mass of the rocket is 6100 kg

Initial upward acceleration of${\text{21m/s}}^2$

Understanding the concept of launching the rocket

To successfully launched a rocket from the ground, fuel is consumed at a rate that results in a thrust big enough to overcome the gravitational force

Formula:

The thrust is given by T =Rvrel

Where, R is the rate of the fuel consumption

is the speed of the exhaust gas relative to the rocket

 Calculation of the rate of consumption

(a)

$$\begin{gathered} T=Rvrel \\ R=\frac{T}{Vrel} \\ =\frac{Mg}{Vrel} \\ =\frac{(6100\mathrm{kg})(9.8\mathrm{m}/{\mathrm{s}}^2)}{1200\mathrm{m}/\mathrm{s}} \\ =49.8\mathrm{kg}/\mathrm{s} \\ \approx 50\mathrm{kg}/\mathrm{s} \end{gathered}$$

The rate of consumption is 50 kg/s

Calculation of the rate of consumption with the additional effect due to gravity

(b)

$$\begin{gathered} (Rvrel-Mg=Ma) \\ R=\frac{M(g+a)}{Vrel} \\ =\frac{\left(6100\mathrm{kg}\right)(9.8\mathrm{m}/{\mathrm{s}}^2+21\mathrm{m}/{\mathrm{s}}^2)}{1200\mathrm{m}/\mathrm{s}} \\ =56.6\mathrm{kg}/\mathrm{s} \\ \approx 1.6\times {10}^2\mathrm{kg}/\mathrm{s} \end{gathered}$$

The rate of consumption with the additional effect due to gravity is$1.6\times {10}^2\mathrm{kg}/\mathrm{s}$