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?
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?