A rocket that is in deep space and initially at rest relative to an inertial reference frame has a mass of $\mathbf{2}\mathbf{.}\mathbf{55}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{}\mathbf{kg}$, of which $\mathbf{1}\mathbf{.}\mathbf{81}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{}\mathbf{kg}$is fuel. The rocket engine is then fired for 250 swhile fuel is consumed at the rate of $\mathbf{480}\mathbf{}\mathbf{kg}\mathbf{/}\mathbf{s}$. The speed of the exhaust products relative to the rocket is. (a) What is the rocket’s thrust? After the 250 sfiring, what are (b) the mass and (c) the speed of the rocket?

The initial velocity of rocket is$V_i=0\mathrm{m}/\mathrm{s}$.

The initial mass of rocket,$M_i=2.55\times {10}^5\mathrm{kg}$

The total mass of fuel is,$M_{fi}=1.81\times {10}^5\mathrm{kg}$

The rate of fuel consumption is$R=\frac{dM}{dt}=480\mathrm{kg}/\mathrm{s}$.

The exhaust speed is calculated as$V_{rel}=3.27\mathrm{km}/\mathrm{s}=3.27\times {10}^3\mathrm{m}/\mathrm{s}$.

The time of burning fuel,$t=250\mathrm{s}$

Here, we can use the first rocket equation to calculate the thrust of the rocket and second rocket equations to calculate the speed of rocket. The mass of rocket can be calculated by finding the mass of fuel burnt out.

Formula:

$$\begin{gathered} V_f-V_i=V_{rel}\times \mathrm{In}\left(\frac{{\mathrm{M}}_{\mathrm{i}}}{{\mathrm{M}}_{\mathrm{f}}}\right) \\ \mathrm{Ma}={\mathrm{RV}}_{\mathrm{rel}} \end{gathered}$$

The first rocket equation is

$R{V}_{rel}=Ma$

Applying Newton’s second law of motion, the net force or the thrust of rocket is

$$\begin{gathered} T=R{V}_{rel} \\ =480\times 3.27\times {10}^3 \\ =1.57\times {10}^6\mathrm{N} \end{gathered}$$.

Hence, the thrust of the rocket,$T=1.57\times {10}^6\mathrm{N}$

The rate of fuel consumption is

$R=\frac{dM}{dt}480\mathrm{kg}/\mathrm{s}$

So, the mass of the fuel burned is given by,

$$\begin{gathered} M_{fuel}=R\times t \\ =480\times 250 \\ =1.20\times {10}^5\mathrm{kg} \end{gathered}$$

So, the mass of rocket after 250 s is

$$\begin{gathered} M_f=M_i-M_{fuel} \\ =2.55\times {10}^5-1.20\times {10}^5 \\ =1.35\times {10}^5\mathrm{kg} \end{gathered}$$

Hence, the final mass of rocket,$M_f=1.35\times {10}^5\mathrm{kg}$

The second rocket equation is

$V_f-V_i=V_{rel}\times \mathrm{In}\left(\frac{M_i}{M_f}\right)$

Substituting the known values in this equation, we get

$$\begin{gathered} V_f-0=3.27\times {10}^3\times \mathrm{In}\left(\frac{2.55\times {10}^5}{1.35\times {10}^5}\right) \\ =2.08\times {10}^3\mathrm{m}/\mathrm{s} \\ {\mathrm{V}}_{\mathrm{f}}=2.08\mathrm{km}/\mathrm{s} \end{gathered}$$

Hence, the speed of the rocket,${\mathrm{V}}_{\mathrm{f}}=2.08\mathrm{km}/\mathrm{s}$