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December 23rd, 2018, 01:58 PM  #1 
Newbie Joined: Dec 2018 From: Colombia Posts: 6 Thanks: 0  Varying Gravity and Air Resistance in projectile motion
Salutations, I have been trying to approach a case about projectile motion considering variation of gravity acceleration and air resistance: A spherical baseball with mass "m" is hit with inclination angle $\theta$ and launching velocity $v_0$, then, the wind has a drag force equals to $F=kv$ and according the acceleration of gravity force is varying in function of height. So, analzying the gravity in function of height, I got this: $$ mg=\frac{GM_Tm}{\left(R+y\right)^2}\\\ \\ g=\frac{GM_T}{R^2 \left(1+\frac{y}{R}\right)^2}\\\ g=\frac{g_+}{\left(1+\frac{y}{R}\right)^2}$$ Then, regarding the gravitational varying acceleration according height of the ball is considered insignificant above the radius of the Earth, which I considered to apply binomial expansion $((1+x)^n=1+nx)$: $$ \Gamma=g\left(1+\frac{y}{R}\right)^{2}\\\ \Gamma=g\left(1\frac{2y}{R}\right)$$ After that, analyzing the applied forces to the ball when rises up, I got this equation: $$ma=m\Gammakv\\\ a=\Gamma\frac{k}{m}v$$ $$\frac{dv}{dt}=g\left(1\frac{2y}{R}\right)\frac{k}{m}v$$ So, when the ball is falling down, I consider this model: $$\frac{dv}{dt}=\frac{k}{m}v\Gamma\\ \frac{dv}{dt}=\frac{k}{m}vg\left(1\frac{2y}{R}\right)$$ The objective of the modelling is finding maximum height, total flight time of the ball and maximum horizontal displacement Finally, my doubt is: Are the mathematical model well posed for rising and falling down of the ball? This is just academic curiosity, and it's the first time that I approach varying gravity and air resistance in projectile motion, and I'm not sure if the varying gravity is well applied in the models. So, I would like any guidance or starting steps or explanations to find the solutions because it's an interesting case of projectile motion. Thanks for your attention. 
December 23rd, 2018, 09:43 PM  #2  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,162 Thanks: 879 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
$\displaystyle \dfrac{d^2y}{dt^2} = \dfrac{k}{m} \cdot \dfrac{dy}{dt}  g \left ( 1  \dfrac{2y}{R} \right )$ Once you solve that the rest falls out by the usual derivative methods. Dan  

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air, gravity, motion, projectile, resistance, varying 
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