My Math Forum Varying Gravity and Air Resistance in projectile motion

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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: $$ma=-m\Gamma-kv\\\ a=-\Gamma-\frac{k}{m}v$$ $$\frac{dv}{dt}=-g\left(1-\frac{2y}{R}\right)-\frac{k}{m}v$$ So, when the ball is falling down, I consider this model: $$\frac{dv}{dt}=\frac{k}{m}v-\Gamma\\ \frac{dv}{dt}=\frac{k}{m}v-g\left(1-\frac{2y}{R}\right)$$ 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
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 Originally Posted by ht1204 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: $$ma=-m\Gamma-kv\\\ a=-\Gamma-\frac{k}{m}v$$ $$\frac{dv}{dt}=-g\left(1-\frac{2y}{R}\right)-\frac{k}{m}v$$ So, when the ball is falling down, I consider this model: $$\frac{dv}{dt}=\frac{k}{m}v-\Gamma\\ \frac{dv}{dt}=\frac{k}{m}v-g\left(1-\frac{2y}{R}\right)$$ 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.
So far as I can see your derivation is almost correct. There's just one little problem: Your differential equation is based on "y", not "v." So, your free fall equation for example, the last one you wrote, you really have a second order equation:
$\displaystyle \dfrac{d^2y}{dt^2} = \dfrac{k}{m} \cdot \dfrac{dy}{dt} - g \left ( 1 - \dfrac{2y}{R} \right )$

Once you solve that the rest falls out by the usual derivative methods.

-Dan

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