recenty i read the rocket equation, derivation of, however i think i have a slight confusion with signs(adsbygoogle = window.adsbygoogle || []).push({});

suppost initially a rocket has

mass= [tex]M[/tex]

velocity= [tex]\overrightarrow{v}[/tex]

then at a time dt later,

mass of rocket= [tex]M-dM[/tex]

velocity of rocket= [tex]\overrightarrow {v} +d\overrightarrow {v} [/tex]

mass of ejacted gas= [tex]dM[/tex]

velocity of gas= [tex]\overrightarrow{u}[/tex]

using conservation of momentum

[tex]\overrightarrow{v}M=(M-dM)(\overrightarrow{v}+d\overrightarrow{v})+\overrightarrow{u}dM[/tex]

[tex](\overrightarrow{u}-\overrightarrow{v})dM+Md\overrightarrow{v}=0[/tex]

but [tex](\overrightarrow{u}-\overrightarrow{v})[/tex]=velocity of gas relative to rocket

let [tex](\overrightarrow{u}-\overrightarrow{v})=\overrightarrow{U}[/tex]which is a constant

[tex]\overrightarrow{U}dM+Md\overrightarrow{v}=0[/tex]

[tex]-\int_{M_0}^{M}\frac{dM}{M}=\frac{1}{\overrightarrow{U}}\int_{\overrightarrow{v}_0}^{\overrightarrow{v}}d\overrightarrow{v}[/tex]

now [tex]-ln\frac{M}{M_0}=\frac{\overrightarrow{v}-\overrightarrow{v_0}}{\overrightarrow{U}}[/tex]

the problem is , when taking the velocity in the direciton rocket is travelling

[tex]\overrightarrow{U}<0[/tex]

[tex]-ln\frac{M}{M_0}>0[/tex]since [tex]\frac{M}{M_0}<1[/tex]

then

[tex]\overrightarrow{v}-\overrightarrow{v_0}<0[/tex] which is impossibe as the rocket is accelerating???

**Physics Forums - The Fusion of Science and Community**

# Rocket equation

Have something to add?

- Similar discussions for: Rocket equation

Loading...

**Physics Forums - The Fusion of Science and Community**