according to newtons second law:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\vec{F}=\frac{d\vec{p}}{dt}=m\frac{d\vec{v}}{dt}+\vec{v}\frac{dm}{dt}[/tex](1)

ie force impelled on a body is equal to the rate of change of momentum

however when we use calculus to derive rocket equation we get:

[tex]m\frac{d\vec{v}}{dt}={\vec{v}_{gas\; relative \;to \;rocket}}\frac {dm}{dt}[/tex](2)

where [tex]\vec{v}[/tex] is the velocity of the rocket

my problem now is that many text book concludes that the net force on the rocket

[tex]\vec{F}=\vec{T}=m\frac{d\vec{v}}{dt}={\vec{v}_{gas\; relative \;to \;rocket}}\frac{dm}{dt}[/tex] where T is the THRUST

but when you apply (2) in (1) would not the force on rocket by gas be

[tex]\vec{F}=m\frac{d\vec{v}}{dt}+\vec{v}\frac{dm}{dt}=({\vec{v}_{gas\; relative \;to \;rocket}+\vec{v})\frac{dm}{dt}[/tex]?????

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Netwons second law and rocket equation

**Physics Forums | Science Articles, Homework Help, Discussion**