**Mass of rocket** = (M – R.t); R.t is defined as rate at which the rocket loose it’s mass within a given time frame

**Mass of Exhaust** = R.t (amount of mass that has been expelled within a given time frame)

Now comes a bit complicated part, to calculate the acceleration of exhaust. Acceleration is defined as the rate of change of velocity. Exhaust velocity goes from V (before it is expelled) to Ve (after it is expelled). This diference is in negative as the acceleration is downward direction. So the acceleration of exhaust velocity is calculated by

**(-Ve – V)/t**

Now putting up all the values in toatl force equation, we can derive the equation which tells us how the whole system moves

**-mg = (m-R.t). a + (R.t). (-Ve – v)/t**

Above equation can be further reduced as “t” will get cancelled out

**-mg = (m-R.t).a + R.(-Ve – V) –** Eq. (1)

The above equation shows how the whole rocket plus exhaust system moves.

**What do you think should be the minimum exhaust velocity for the rocket to just hover?**

When the rocket is hovering then it’s velocity and acceleration are zero. In this case the Eq. (1) reduces to

**-mg = R.Ve – **Eq. (2)

But here still there are two variables to find the force required – “R” – rate at which fuel is coming out & “Ve” – Exhaust Velocity

Let’s assume that the rocket exhaust has an area of “a”. So the volume of exhaust coming out can be expressed as

**Vol = a.Ve**

Now we need to relate the volume and rate “R”. Rate is expressed as kilogram of fuel coming out within a given time frame. 1 Kg is 1 L and there are 1000 litres in 1 cubic meter

So, **R = 1000.Vol**

Now the Eq. (2) reduces to

**-mg = (1000.Vol). Ve**

Which further reduces down to