Rocket in space

  • #1
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?
 

Answers and Replies

  • #2
can anyone help me out?
 
  • #3
Tide
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Does this equation look familiar to you?

[tex]v = v_e \ln \frac {M_0}{M}[/tex]
 
  • #4
HallsofIvy
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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 ve).
 
  • #5
Clausius2
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HallsofIvy said:
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 ve).
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... :frown: )
 

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