# B Rocket science.

1. Mar 18, 2016

### Biker

I was just scrambling and thought of rockets. So I came up with an equation for (hovering).
Note: This might be totally wrong.
Okay so we have a rocket with fuel with a mass of M. There is gravity acting on it so I need a force to balance things out.

F - Mg = 0
F = Mg
What I have learned is every force come in pairs
So F = -Fthrust.
and F thrust = mair * a
so -(m(air)*a) = (m-mair)*g
^^ Because the rocket loses mass over time.
Lets call the rate of mass ejected R
-(R * t * a) = (M-R*t)*g
a = dv/t
a = (-Ve - v)/t ( Ve( velocity of thrust) is down and v(velocity of the rocket) is upward, I assumed up is positive and down is negative)
As it is hovering then the v of the rocket is equal to 0.
so a = -Ve/t
Now substitute that in.
-(R * t *-Ve/t) = (M-R*t)*g
R * Ve = (M-R*t)*g

Hmm I hope this is right :/

2. Mar 18, 2016

### Guneykan Ozgul

If you over simplify a hovering rocket, your equations seems correct to me however there are some issues I want to to comment on.
If you do your calculations for an infinetisimal time interval it will be more general because in your notation you assumed that acceleration of air is constant and R is constant (in reality probably not). If you use dv and dt instead of v and t it will be more difficult to solve but more general.
Anyway for a simple object ejecting mass to hover in the air, your way of thinking seems correct.