 My Math Forum Varying Gravity and Air Resistance in projectile motion
 User Name Remember Me? Password

 Physics Physics Forum

 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
Math Team

Joined: May 2013
From: The Astral plane

Posts: 2,304
Thanks: 961

Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
 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 Tags air, gravity, motion, projectile, resistance, varying Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Elize Physics 3 March 18th, 2018 03:45 AM jeho Physics 1 November 23rd, 2016 04:42 PM sweer6 New Users 4 May 21st, 2014 10:00 PM edwinandrew Physics 0 February 26th, 2014 03:30 PM symmetry Algebra 1 June 19th, 2007 10:26 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top      