Solving a Rocket Equation: Mass, Velocity & Height

Click For Summary

Homework Help Overview

The discussion revolves around a rocket equation involving mass, velocity, and height. The original poster presents a problem related to the dynamics of a rocket, including parameters such as mass of the rocket and fuel, exhaust velocity, and gravitational force. The goal is to determine the burn-out time, final velocity, and height of the rocket during its boost phase.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to set up a differential equation for the rocket's motion, questioning the treatment of mass as a variable. Participants discuss the implications of mass expulsion on the equations of motion and the correct interpretation of terms in the equations.

Discussion Status

Some participants provide feedback on the original poster's approach, confirming the validity of certain steps while also addressing concerns about the non-constant mass of the rocket. There is an ongoing exploration of how to correctly incorporate the mass change into the equations.

Contextual Notes

The original poster has established a specific burn-out time and is working within the constraints of the problem, including the given values for mass and exhaust parameters. There is an acknowledgment of the complexity introduced by the changing mass during the rocket's flight.

LondonLady
Messages
14
Reaction score
0
Im having this problem with a rocket equation. Ill state the problem then show what I've done

Let [tex]M[/tex] = mass of rocket and fuel

[tex]M_f[/tex] = mass of fuel

[tex]M_0[/tex] = rockets total initial mass (including fuel)

(this is given as [tex]10^5 kg[/tex])

[tex]V_R[/tex] = rockets velocity

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

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


Work out the answers to the following questions using the above symbols together with [tex]k, v_0[/tex] and [tex]g[/tex]. 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 [tex]t = 0[/tex]

Total momentum = 0

then at [tex]t = dt[/tex]

[tex]dp_{tot} = (M - dM_f)(dV_R) - (v_0dM_f)[/tex]

neglecting 2-nd order terms

[tex]\displaystyle{\frac{dp_{tot}}{dt} = M\frac{dV_R}{dt} - v_0\frac{dM_f}{dt}}[/tex]

Then as the only external force on the rocket is gravity

[tex]\displaystyle{-(M - dM_f)g = M\frac{dv_R}{dt} - v_0\frac{dM_f}{dt}}[/tex]

I also know that [tex]\displaystyle{\frac{dM_f}{dt} = -k[/tex]

so

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

Is this at all right so far?
 
Physics news on Phys.org
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 [tex]dM_f[/tex] bit.

I continued on but have got a bit confused again!

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

[tex]\displaystyle{-Mg = M\frac{dv_R}{dt} + v_0k}[/tex]

[tex]\displaystyle{\frac{v_0k}{M} + g = -\frac{dv_R}{dt}}[/tex]

Kind of wandering on I did this

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

This is worrying me though because i know that [tex]M[/tex] 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: [itex]M = M_0 - kt[/itex]
 
Silly me! Thankyou!
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 42 ·
2
Replies
42
Views
7K
Replies
2
Views
2K
  • · Replies 23 ·
Replies
23
Views
8K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
3
Views
2K
  • · Replies 30 ·
2
Replies
30
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
13
Views
2K