Two-Stage Rocket: Find Speed w/ Fuel Exhaustion & v_ex Req'd for 6 km/s

[faller217]

I was hoping someone could help me with this homework problem that has me stumped. Any assistance would be appreciated.


Suppose the first stage of a two-stage rocket has total mass 1.30×10^4 kg, of which 1.10×10^4 kg is fuel. The total mass of the second stage is 1000 kg, of which 700 kg is fuel. Assume that the relative speed v_ex of ejected material is constant, and ignore any effect of gravity. (The effect of gravity is small during the firing period if the rate of fuel consumption is large.)


A) Suppose the entire fuel supply carried by the two-stage rocket is utilized in a single-stage rocket with the same total mass of 1.40×10^4 kg. In terms of v_ex, what is the speed of the rocket, starting from rest, when its fuel is exhausted?

B)For the two-stage rocket, what is the speed when the fuel of the first stage is exhausted if the first stage carries the second stage with it to this point? This speed then becomes the initial speed of the second stage. At this point, the second stage separates from the first stage.

C)What is the final speed of the second stage?

D)What value of v_ex is required to give the second stage of the rocket a speed of 6.00 km/s?

 

[D H]

We don't do your homework for you. Please show some work. What are the relevant equations for this kind of problem?

 

[Moetasim]

A) To find the speed of the single-stage rocket, we can use the conservation of momentum equation: mv = (m + dm)(v + dv), where m is the initial mass of the rocket, dm is the change in mass due to fuel consumption, and v and dv are the initial and final velocities, respectively. In this case, v is 0 since the rocket starts from rest, and dm is the total fuel mass of 2.10×10^4 kg. Solving for v, we get v = v_ex ln(m / (m + dm)). Plugging in the given values, we get v = v_ex ln(1.30×10^4 / 1.40×10^4) = -0.074v_ex. Therefore, the speed of the single-stage rocket when its fuel is exhausted is -0.074v_ex.

B) When the fuel of the first stage is exhausted, the mass of the rocket becomes 2.70×10^3 kg (1.40×10^4 kg - 1.10×10^4 kg - 700 kg). Using the same conservation of momentum equation, we get v = v_ex ln(m / (m + dm)) = v_ex ln(2.70×10^3 / 1.40×10^4) = -0.993v_ex. This becomes the initial speed of the second stage when it separates from the first stage.

C) To find the final speed of the second stage, we can again use the conservation of momentum equation. The initial mass of the second stage is 1000 kg and the final mass is 300 kg (700 kg - 400 kg). The initial velocity is -0.993v_ex and the final velocity is v. Solving for v, we get v = 3.31v_ex. Therefore, the final speed of the second stage is 3.31v_ex.

D) To find the value of v_ex required to give the second stage a speed of 6.00 km/s, we can use the final speed equation from part C and set it equal to 6.00 km/s. This gives us 6.00 km/s = 3.31v_ex, so v_ex = 1.81 km/s. Therefore, the required value of v_ex is 1.81 km/s to give the second stage a speed of