Rocket Propulsion: Understanding the Equation Behind It

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SUMMARY

The discussion centers on the ideal rocket equation, expressed as v = vex * ln(m0/m(t)), which describes the relationship between a rocket's velocity and its mass over time. Participants clarify that while the equation may suggest decreasing acceleration due to the logarithmic function, differentiation reveals that acceleration actually increases as mass decreases, particularly when fuel is burned at a constant rate. This principle explains why rockets, such as the Space Shuttle, must throttle down towards the end of their launch to prevent excessive acceleration on astronauts and equipment. The equation emphasizes that adding fuel has a diminishing return on final velocity, necessitating significant fuel addition for marginal velocity increases.

PREREQUISITES
  • Understanding of the ideal rocket equation and its components
  • Basic knowledge of calculus, specifically differentiation
  • Familiarity with concepts of momentum and acceleration
  • Awareness of rocket propulsion principles and fuel dynamics
NEXT STEPS
  • Study the derivation of the ideal rocket equation in detail
  • Learn about the implications of mass flow rate on rocket acceleration
  • Explore the throttling mechanisms used in rocket launches
  • Investigate the relationship between fuel mass and final velocity in rocket design
USEFUL FOR

Aerospace engineers, physics students, and anyone interested in rocket propulsion and the dynamics of mass loss during flight will benefit from this discussion.

aaaa202
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Look at the attached picture, which is taken from my textbook. I don't understand the equation:

v = vex * ln(m0/m(t))
If the rocket looses mass at a constant rate, wouldn't that equation then say, that the acceleration decreases as time goes. Since the curve of ln(>1) flattens out. That doesn't match my intution behind conservation of momentum which says that the lighter you are pushing something away from you, the greather acceleration you will attain.
 

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aaaa202 said:
Look at the attached picture, which is taken from my textbook. I don't understand the equation:

v = vex * ln(m0/m(t))
If the rocket looses mass at a constant rate, wouldn't that equation then say, that the acceleration decreases as time goes. Since the curve of ln(>1) flattens out. That doesn't match my intution behind conservation of momentum which says that the lighter you are pushing something away from you, the greather acceleration you will attain.

Did you mean to say acceleration increases as time goes?
 
aaaa202 said:
I don't understand the equation:

v = vex * ln(m0/m(t))
If the rocket looses mass at a constant rate, wouldn't that equation then say, that the acceleration decreases as time goes.
That equation says nothing about acceleration, at least not directly. Differentiating it does say something about acceleration:
\frac{dv}{dt} =<br /> v_e\left(\frac 1{m_0/m(t)}\right)\left(\frac {-m_0}{m(t)^2}\frac{dm(t)}{dt}\right)<br /> = -v_e\frac{\dot m}{m(t)}<br />
So assuming a constant burn rate, acceleration does increase as mass decreases. This is the reason that the Shuttle had to be throttled down near the end of the launch lest the astronauts and equipment be exposed to excessive acceleration.

What the ideal rocket equation (that's the name for this equation) says directly is that adding fuel has a small effect on final velocity. If you want to have a slightly faster final velocity when the fuel is depleted you need to add a considerable amount of fuel at launch. The relationship is not linear.
 

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