# Trouble with a Rocket Propulsion question (Variable Mass & Momentum)

• vparam

#### vparam

Homework Statement
A fully fueled rocket has a mass of 21,000 kg, of which 15,000 kg is fuel. The burned fuel is spewed out the rear at a rate of 190 kg/s with a speed of 2800 m/s relative to the rocket. If the rocket is fired vertically upward calculate its final velocity at burnout (all fuel used up). Ignore air resistance and assume g is a constant 9.80 m/s^2.
Relevant Equations
M * dv/dt = ∑F_ext + v_rel * dM/dt
I chose to set the upwards direction to be positive and dM/dt = R = 190 kg/s, so I can solve the problem in variable form and plug in. With the only external force being gravity, this gives

M(t) * dv/dt = -M(t) * g + v_rel * R

where M(t) is the remaining mass of the rocket. Rearranging this equation gives:

dv/dt = -((v_rel * R)/M(t)) - g

Since R is constant, M(t) = M_0 - R * t, where M_0 is the initial mass of the rocket. Plugging in gives:

dv/dt = -((v_rel * R)/(M_0 - R * t)) - g.

Solving as a separable differential equation, I arrived at the answer (assuming v = 0 at t = 0):

v(t) = -g * t + v_rel * ln(M_0 - R * t).

However, after plugging in values, I'm not able to get the correct answer. The solution instead has a different equation for v(t):

v(t) = -g * t + v_rel * ln(M/M_0).

Any help about where I could be going wrong with the physical setup or the math of this problem would be much appreciated. Thanks in advance!

dv/dt = -((v_rel * R)/(M_0 - R * t)) - g.

Solving as a separable differential equation, I arrived at the answer (assuming v = 0 at t = 0):

v(t) = -g * t + v_rel * ln(M_0 - R * t).
Note that your result for v(t) does not satisfy the initial condition v = 0 at t = 0.

When you integrated ## \dfrac {dt}{M_0 - Rt}## from ##t = 0## to ##t = t##, did you forget to evaluate at the lower limit ##t = 0##?

Last edited:
• vparam
Yes, it looks like that's where I went wrong. Thank you for your help!

• TSny