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May 19th, 2016, 03:28 AM   #1
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Acceleration components of gravity vector

Hello,

for "fun", I thought I'd have a go at modelling the orbit of planets around the sun in Excel. I'm trying to complete this worksheet I found online:

Building Planetary Orbits

I've stumbled at the first hurdle, though; the solution requires me to devise equations for the $\displaystyle x$ and $\displaystyle y$ components of acceleration of a planet orbiting the Sun, from the equation for vector gravitational force:

$\displaystyle \vec{F_g}=-G\frac{M.m}{r^2}\hat{r}$

...where $\displaystyle G$ is a gravitational constant, $\displaystyle M$ is the mass of the Sun, $\displaystyle m$ is the mass of the planet, $\displaystyle \hat{r}$ is the unit vector between the Sun and the planet and $\displaystyle r$ is the radius of the planets orbit around the Sun.

I've come up with a solution (because I had another model that did something similar, so I just copied that) but I don't understand why it works:

$\displaystyle a_x=\vec{F_g}\frac{x}{r}$

$\displaystyle a_y=\vec{F_g}\frac{y}{r}$

...where $\displaystyle x$ and $\displaystyle y$ are the planets position in 2D space - I have a known starting position.

So...why do these two equations give me acceleration? I appeciate that this might be a very elementary maths question and, before anybody suggests it, I have an introductory book on orbital mechanics on order from Amazon

Any help would be appreciated, though,

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May 19th, 2016, 06:18 AM   #2
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What you have, $\displaystyle a_x= \vec{F_g}\frac{x}{r}$ and $\displaystyle a_y= \vec{F_g}\frac{y}{r}$, are clearly incorrect for two different reasons. I presume you know that "force equals mass times acceleration" so that "acceleration equals force divided by mass". Since force is proportional to mass, acceleration due to gravity does not depend upon mass. But M is still part of $\displaystyle \vec{F_g}$ so that your components of acceleration are proportional to mass. Second, since both $\displaystyle a_x$ and $\displaystyle a_y$ are numbers times $\displaystyle \vec{F_g}$ you have both components pointing parallel to the force vector not parallel to the x and y axes as they should be.
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May 19th, 2016, 06:33 AM   #3
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Well, that's a very definite $\displaystyle WRONG!$ then - thanks for clearing that up

Can you explain why they work? I use them to model the path of a planet around the Sun and the orbit is as expected...

More helpful, though, would be a suggestion as to what the formulae for $\displaystyle a_x$ and $\displaystyle a_y$ should be...

Thanks for your thoughts on this
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May 19th, 2016, 08:07 AM   #4
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OK...

I've had a sit and a think about this (plus I read all of this - page 19 is very helpful!).

I've revised my calculations as follows:

Because $\displaystyle r=\sqrt{r_x^2+r_y^2}$ and $\displaystyle \hat{r}=\frac{\vec{r}}{r}$, we can derive $\displaystyle \hat{r_x}=\frac{r_x}{r}$ and $\displaystyle \hat{r_x}=\frac{r_x}{r}$

So $\displaystyle \vec{F_g}=\frac{G.M.m}{r^2}$ can be expanded into its x/y components as follows:

$\displaystyle F_g(x)=\frac{-r_x}{r}.F_g$

and

$\displaystyle F_g(y)=\frac{-r_y}{r}.F_g$

where $\displaystyle r_x$ and $\displaystyle r_y$ are the $\displaystyle x,y$ co-ordinates of the planet.

Thus, as we know $\displaystyle \vec{a}=\frac{\vec{F_g}}{M}$;

$\displaystyle a_x=\frac{F_g(x)}{M}$

and

$\displaystyle a_y=\frac{F_g(y)}{M}$

I've substituted these equations into my model and they also return the orbits expected - hooray!

The reason my model worked was because the mass $\displaystyle M$ I'm using is set to 1 for simplicity (i.e. the Sun=1 Solar Mass), so $\displaystyle \frac{F_g}{1}=F_g$ in this case...

I'm off to read the rest of this Newtonian Mechanics paper, anyway...

Thanks

Last edited by eigenvexed; May 19th, 2016 at 08:10 AM. Reason: A sort of sudden epiphany...
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May 19th, 2016, 09:40 AM   #5
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sorry, but $a \ne \dfrac{F_g}{M}$

$a = \dfrac{F_g}{m}$, where $m$ represents the planetary mass.

$F_g = -\dfrac{GMm}{r^2} \hat{r}$

$ma_r = -\dfrac{GMm}{r^2}$

$a_r = -\dfrac{GM}{r^2}$

$a_x = a_r \cos{\theta} = -\dfrac{GM \cos{\theta}}{r^2}$

$a_y = a_r \sin{\theta} = -\dfrac{GM \sin{\theta}}{r^2}$

where $\theta$ is the angle between positive x-axis and the orbital radius
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May 19th, 2016, 10:51 PM   #6
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Ah, Ok - to be honest, I found the use of $\displaystyle M$ and $\displaystyle m$ a bit ambiguous in what I was reading - thanks for clearing that up.

The original equation for the gravitational force vector is:

$\displaystyle \vec{F_g}=G\frac{M.m}{r^2}$

I've read that, where $\displaystyle M$ is significantly greater than $\displaystyle m$, a gravitational field around the larger mass can be defined as $\displaystyle g=-\frac{GM}{r^2}\hat{r}$ which is where the equations I'm using then derive from - perhaps this is why they work, despite my transposing $\displaystyle M$ and $\displaystyle m$...?

The equations...

$\displaystyle a_x=a_rcos\theta$

and

$\displaystyle a_y=a_rsin\theta$

...were my first thought on how to approach this, too - the problem I found was that there was no easy way of calculating $\displaystyle \theta$...how do you do it?

Thanks for your help
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May 20th, 2016, 01:26 AM   #7
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you don't need to calculate $\theta$ ...

$a_x=a_r\cos{\theta}=a_r \cdot \dfrac{x}{r} = -\dfrac{GM}{r^2} \cdot \dfrac{x}{r} = -\dfrac{GMx}{r^3}$

similarly, $a_y=-\dfrac{GMy}{r^3}$
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