# A rocket in space has a payload of 200 kg and 6000 kg of fuel. If the fuel is expelled at 150 m/s at a constant rate, what is the final velocity?

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As the rocket is in space, an assumption can be made that there is no acceleration due to the gravitational pull of any object on it.

The rocket has a payload of 200 kg and 6000 kg of fuel. A rocket moves forward due to the fact that momentum has to be conserved. As the rocket fuel is gradually expelled backward at a high velocity, the rocket with the payload and the remaining fuel moves in the opposite direction with an equal momentum.

It is assumed that the rocket starts from rest; the final velocity after all the fuel has been consumed has to be determined.

The increase in momentum of the payload and the remaining fuel is equal to fuel being expelled but this is not constant for the entire flight as there is a reduction in the mass of the fuel. The resulting change in velocity can be determined using integration but an easier way is to use the Tsiolkovsky rocket equation which gives: V = Ve*ln((M + P)/P) where Ve is the velocity of the exhaust, V is the final velocity attained, M is the mass of the fuel and P is the mass of the payload.

Substituting the values given:

V = 150*ln((6000+200)/200)

=> V = 150*ln(6200/200)

=> V = 150*3.433

=> V = 515.09

The final velocity of the rocket after all the fuel has been consumed is equal to 515.09 m/s.

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