Find the final burnout velocity of a two-stage rocket which has a payload mass of 100 kg. Specific impulse of both the stages is 290s. First stage gets separated from the second stage at the instant of burnout. Rocket has the following design characteristics:
Find the final burnout velocity of a two-stage rocket which has a payload mass of 100 kg. Specific impulse of both the stages is 290s. First stage gets separated from the second stage at the instant of burnout. Rocket has the following design characteristics:
Stages: First stage Second stage
Propellant mass: 8000 kg 6000 kg
Structural mass : 900 kg 700 kg
Burnout velocity: It is the velocity of a rocket at the instant of burnout of its propellant. There is a depletion of propellants. Burnout velocity is given as
\[{V_b} = {g_0}{I_{sp}}\ln \left( {\frac{{{M_i}}}{{{M_f}}}} \right)\]
Burnout velocity for the first stage is
\[{V_{b1}} = {g_0}{I_{sp}}\ln \left( {\frac{{{M_L} + {M_{s1}} + {M_{s2}} + {M_{p1}} + {M_{p2}}}}{{{M_L} + {M_{s1}} + {M_{s2}} + {M_{p2}}}}} \right)\]
\[ \Rightarrow {V_{b1}} = \left( {9.8} \right)\left( {290} \right)\ln \left( {\frac{{100 + 900 + 700 + 8000 + 6000}}{{100 + 900 + 700 + 6000}}} \right) = 2024.76\,m/s\]
Increase in velocity with the second stage is,
\[{V_{b2}} – {V_{b1}} = {g_0}{I_{sp}}\ln \left( {\frac{{{M_L} + {M_{s2}} + {M_{p2}}}}{{{M_L} + {M_{s2}}}}} \right)\]
\[ \Rightarrow {V_{b2}} – {V_{b1}} = \left( {9.8} \right)\left( {290} \right)\ln \left( {\frac{{100 + 700 + 6000}}{{100 + 700}}} \right) = 6082.07\,m/s\]
Therefore,
\[{V_{b2}} – {V_{b1}} = 6082.07 \Rightarrow {V_{b2}} = 6082.07 + 2024.76 = 8106.83\,m/s\]
