It is possible for the velocity of a rocket to be greater than the exhaust velocity of the gases it ejects. When that is the case, the gas velocity and gas momentum are in the same direction as that of the rocket. How is the rocket still able to obtain thrust by ejecting the gases?

A moving body's motion is measured as the product of its mass and velocity.

i.e.\(P = m.V\)

Equation between escape velocity and velocity of the rocket is given by;

\({\rm{v = }}{{\rm{v}}_{\rm{e}}}{\rm{ln}}\dfrac{{{{\rm{m}}_{\rm{0}}}}}{{{{\rm{m}}_{\rm{r}}}}}\)

Where\({{\rm{m}}_{\rm{r}}}\)is the mass of the rocket andis the mass of the gas,

Now if\({\rm{v > }}{{\rm{v}}_{\rm{e}}}\)then,

\(\begin{aligned}{{\rm{ln}}\dfrac{{{{\rm{m}}_{\rm{o}}}}}{{{{\rm{m}}_{\rm{r}}}}}{\rm{ > 1}}}\\{\dfrac{{{{\rm{m}}_{\rm{o}}}}}{{{{\rm{m}}_{\rm{r}}}}}{\rm{ > 0}}}\end{aligned}\)

Hence when compared to the rocket, mass of gas will not all be negligible.

But in this case, the direction of the rocket is same as the velocity and momentum of the gas.

Now finding out the acceleration

\({\rm{a = }}\dfrac{{{{\rm{v}}_{\rm{e}}}}}{{\rm{m}}}\dfrac{{{\rm{\Delta m}}}}{{{\rm{\Delta t}}}}{\rm{ - g}}\)

Where \({\rm{m}}\)is the mass, \(\dfrac{{{\rm{\Delta m}}}}{{{\rm{\Delta t}}}}\)is the change in mass per second and \({\rm{g}}\)is the acceleration due to gravity.

Now, let us consider if the mass is increased then \(\dfrac{{{\rm{\Delta m}}}}{{{\rm{\Delta t}}}}\)will also be increased. Now the acceleration of the rocket will be very high. To obtain the thrust the above acceleration will be needed to gain the velocity.