Rocket Fired in Deep Space: Calculating Exhaust Gas Speed

[ElectricMile]

A rocket is fired in deep space, where gravity is negligible. In the first second it ejects (1/160) of its mass as exhaust gas and has an acceleration of 15.3 m/s^2.

What is the speed V(gas) of the exhaust gas relative to the rocket?

 

[ElectricMile]

can anyone help me out?

 

[Tide]

Does this equation look familiar to you?

$$v = v_e \ln \frac {M_0}{M}$$

 

[HallsofIvy]

More fundamentally, use conservation of momentum. Take the initial values of the mass and speed of the rocket to be M and v. The initial momentum is Mv. After ejecting (1/160)M, the rocket's mass is (159/160) M. Since it accelerated at 15.3 m/s² for 1 sec. , it gained 15.3 m/s: it's final speed is v+15.3 and it's final momentum is (159/160)M(v+15.3). The expelled gas has mass (1/160)M and, with velocity -V (opposite to the direction of the rocket) so its momentum is -(1/160)MV.

By conservation of momentum, we must have
(159/160)M(v+15.3)- (1/160)MV= Mv. I don't see any way of solving for V without knowing v (just as Tide's equation required v_e).

 

[Clausius2]

More fundamentally, use conservation of momentum. Take the initial values of the mass and speed of the rocket to be M and v. The initial momentum is Mv. After ejecting (1/160)M, the rocket's mass is (159/160) M. Since it accelerated at 15.3 m/s² for 1 sec. , it gained 15.3 m/s: it's final speed is v+15.3 and it's final momentum is (159/160)M(v+15.3). The expelled gas has mass (1/160)M and, with velocity -V (opposite to the direction of the rocket) so its momentum is -(1/160)MV.

By conservation of momentum, we must have
(159/160)M(v+15.3)- (1/160)MV= Mv. I don't see any way of solving for V without knowing v (just as Tide's equation required v_e).

Hey Hallsofivy! Now I've noticed you don't ask the famous adage "What have you done till now?" and "show me what you've done!".

Times are changing...(nostalgia... )