Fuel is ejected from a rocket of mass \({ M }_{ 0 }\) at a velocity \(u\) \(m/s\) relative to Rocket. Fuel is ejecting at constant rate of \(\mu \quad Kg/s\). Initial Velocity rocket starts from rest from the surface of earth

( Neglect air and other resistance force in both Cases)

Q1). Assume gravity is constant , Find acceleration, velocity and Thrust as a function of time.

Q2). Assume gravity changes with height \(h\) from earth surface as \(g\left( h \right) =\frac { g }{ { \left( 1+\frac { h }{ { R }_{ e } } \right) }^{ 2 } } \), Find acceleration, Velocity and Thrust as a function of time.

## Comments

Sort by:

TopNewestUse impulse momentum theoerem in both cases Consider any arbitrary moment in time t and distance y from surface of earth Let us consider that rocket as my system with mass at that point in spacetime as m In time dt let dm mass be ejected absolute velocity of dm is u-v where u is the relative velocity and v is the absolute velocity of rocket at that moment Impulse due to weight is mgdt which is the change in momentum of the system that is rocket After this we form a differentia equation which is dificult to solve in case 2 I am not able to solve the equation In case 1 V= uln(m■\m) -g\p(m■-m) Where m■ is initia mass p is dm\dt – Akash Yadav · 1 month, 1 week ago

Log in to reply

Is acceleration \(\frac{\mu u}{M_{0} -\mu t}\) ? – Harsh Shrivastava · 1 year, 3 months ago

Log in to reply

– Rishabh Deep Singh · 1 year, 3 months ago

How do you find it ?? I don't know the answer.Log in to reply

– Harsh Shrivastava · 1 year, 3 months ago

Use conservation of momentum and a=dv/dt.Log in to reply

We can't apply conservation of momentum as there is gravity and also mass is changing with Time. – Rishabh Deep Singh · 1 year, 3 months ago

Log in to reply