Solving a Rocket Equation: Mass, Velocity & Height

AI Thread Summary
The discussion revolves around solving a rocket equation involving mass, velocity, and height. The user is attempting to derive the rocket's burn-out time, final velocity, and height during the boost phase, given specific parameters such as mass, fuel rate, and gravitational force. They have successfully calculated the burn-out time as 200 seconds but are confused about how to proceed with the changing mass in their differential equations. The user acknowledges their mistake in treating mass as constant and realizes that the mass can be expressed as M = M_0 - kt, which will help in further calculations. The conversation highlights the complexities of rocket motion equations and the importance of correctly accounting for changing variables.
LondonLady
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Im having this problem with a rocket equation. Ill state the problem then show what I've done

Let M = mass of rocket and fuel

M_f = mass of fuel

M_0 = rockets total initial mass (including fuel)

(this is given as 10^5 kg)

V_R = rockets velocity

A 10^5 kg rocket has a total weight-to-payload ratio of 10 to 1. i.e M_f = 9M_0

The rocket starts from rest, pointing vertically. Its exhaust gases are expelled at a rate k (=450 kg/s) with a constant velocity v_0(= 2,600 m/s). Assume the rocket experiences a constant gravitational force with g = 9.8 m/s^2


Work out the answers to the following questions using the above symbols together with k, v_0 and g. Put in numerical values only at the end.

(1) What is the burn-out time?

(2) What is the rockets final velocity, assuming it rises vertically during the boost phase?

(3) How high is the rocket at the end of its boost phase?


Ok, I've done (1) and am trying to do (2). I am trying to set up the differential equation of the rockets motion.

I said that at t = 0

Total momentum = 0

then at t = dt

dp_{tot} = (M - dM_f)(dV_R) - (v_0dM_f)

neglecting 2-nd order terms

\displaystyle{\frac{dp_{tot}}{dt} = M\frac{dV_R}{dt} - v_0\frac{dM_f}{dt}}

Then as the only external force on the rocket is gravity

\displaystyle{-(M - dM_f)g = M\frac{dv_R}{dt} - v_0\frac{dM_f}{dt}}

I also know that \displaystyle{\frac{dM_f}{dt} = -k

so

\displaystyle{-(M - dM_f)g = M\frac{dv_R}{dt} + v_0k}

Is this at all right so far?
 
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That looks good to me! Of course, you will drop the dM terms as it is infinitesimal.
 
Thankyou very much for your help Tide, I was a bit unsure about that dM_f bit.

I continued on but have got a bit confused again!

I know that the time for the burn out is 200 from the first part, so I did this

\displaystyle{-Mg = M\frac{dv_R}{dt} + v_0k}

\displaystyle{\frac{v_0k}{M} + g = -\frac{dv_R}{dt}}

Kind of wandering on I did this

\displaystyle{\int^{200}_0 \left(\frac{v_0k}{M} + g\right) dt = -\int^{v_f}_0 dv_R}

This is worrying me though because i know that M is not constant...

What have I done wrong and what is my next step?

Thankyou in advance
 
M is not constant but you know the rate at which mass is expelled: M = M_0 - kt
 
Silly me! Thankyou!
 
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