Integration of velocity and thrust angle equation

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SUMMARY

The discussion focuses on integrating the equation for rocket launch velocities, specifically the equation V \frac{d\gamma}{dt}=-g \cos \gamma. The user outlines their integration process, leading to the expression \ln \frac{(\tan t + \sec t)}{(\tan t_0 + \sec t_0)} = -\int_{t_0}^t \frac{g}{V} dt. However, they express confusion regarding the final result in the textbook, particularly the role of the angle \gamma and its relationship to the lift L and gravitational force g. The user seeks clarification on the physical interpretation of \gamma and additional context from the textbook.

PREREQUISITES
  • Understanding of calculus, specifically integration techniques.
  • Familiarity with physics concepts related to rocket dynamics.
  • Knowledge of trigonometric functions, particularly tan and sec.
  • Basic understanding of forces acting on a rocket, including lift and gravity.
NEXT STEPS
  • Review the derivation of the equation MV\frac{d\gamma}{dt}=F\sin (\alpha + \delta) - Mg \cos \gamma + L + \frac{MV^2}{r}\cos \gamma.
  • Study the physical significance of the angle \gamma in rocket motion.
  • Explore the relationship between lift L and gravitational force g in the context of rocket launches.
  • Investigate standard integrals involving trigonometric functions, particularly \sec \gamma.
USEFUL FOR

Students and professionals in aerospace engineering, physicists studying rocket dynamics, and anyone interested in the mathematical modeling of rocket launches.

tomwilliam2
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This is from a physics textbook, a chapter on rocket launch velocities, but really the question is how to integrate the first equation to get to the next.
The way I was approaching it was like this:
From
## V \frac{d\gamma}{dt}=-g \cos \gamma##
Integrating from ##t=0## to some ##t##:
##\int_{t_0}^t \frac{1}{\cos \gamma}\frac{d\gamma}{dt} dt=-\int_{t_0}^t \frac{g}{V} dt##
##\int_{t_0}^t \frac{1}{\cos \gamma}d\gamma =-\int_{t_0}^t \frac{g}{V} dt##
Then, using a standard integral for ##\sec \gamma##:
##[\ln (\tan \gamma + \sec \gamma)]_{t_0}^t = -\int_{t_0}^t \frac{g}{V} dt##
##\ln (\tan t + \sec t) - \ln (\tan t_0 + \sec t_0) = -\int_{t_0}^t \frac{g}{V} dt##
##\ln \frac{(\tan t + \sec t)}{(\tan t_0 + \sec t_0)} = -\int_{t_0}^t \frac{g}{V} dt##

I think this is correct so far, but I don't see how to get the result in the book. Not least because the book keeps the gamma term.
Can anyone help?
 

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Does the book tell you what ##\gamma## is ? Because you don't tell us !
And posting eqn (7.2) would help too
 
BvU said:
Does the book tell you what ##\gamma## is ? Because you don't tell us !
And posting eqn (7.2) would help too
Equation 7.2 is:
##MV\frac{d\gamma}{dt}=F\sin (\alpha + \delta) - Mg \cos \gamma + L + \frac{MV^2}{r}\cos \gamma##
Where g is the local gravitation, which varies with r, ##\gamma## is the angle between the local horizontal and the velocity vector, and L is the lift.
Most of these terms are left out in the simplified version I posted above, and I didn't think I needed anything but maths to integrate the first equation to get to the second...but maybe I do need some information or physical understanding to get to the final equation.
Thanks in advance for your help
 

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