Question

Would this system of differential equations accurately model earth's trajectory around the sun, given that the sun's mass is M, the gravitational constant is g, and the sun is modeled to be stationary at origin (0,0)?

Answer 1

\(\Huge \frac{d^2x}{dt^2}=\frac{-gMx}{x^2+y^2}\)
\(\Huge \frac{d^2y}{dt^2}=\frac{-gMy}{x^2+y^2}\)

Answer 2

with initial conditions x(0)=1.49*10^11 (distance from earth to sun), y(0)=0
y'(0)=30000 (speed of earth's orbit on average in m/s)
x'(0)=0
gM=(gravitational constant g muliplied by mass of the sun)=6.67e-11*1.989e30=1.327e20

Answer 3

wouldn't the second deriv model the acceleration?

Answer 4

yeah, taking F=gmM/r^2, I got the acceleration of earth, a=F/m or gM/r^2
since r is just sqrt(x^2+y^2) I got r^2=(x^2+y^2), and I took component forms of x and y for each of the equations

Answer 5

supposedly if I solve for x(t) and y(t) numerically or analytically I could get some sort of trajectory, correct?

Answer 6

I would just attempt to integrate wrt t and come up with the vector that way... I think

Answer 7

well, solving the system is not too much an issue for me, but I need to make sure that I've come up with something without any major mistakes

Answer 8

well F=Ma to derive a seems fine the substitution for gM/r^2 not sure, you seem to be missing the mass of the earth which would modify the force if I recall correctly. The x and y seems applicable here though

Answer 9

I thought the mass of the earth cancelled out with the force because ma=Mmg/r^2 so a=Mg/r^2. Does that look right to you?

Answer 10

I'm rusty on physics, I need the eq for the force, and isn't that the law for gravitation? wouldn't that be the acceleration?

Answer 11

oh, I think newton's law of gravitation is that F=gmM/r^2, and newton's second law is F=ma. so I combined the two to get a (of earth)=F/m=gM/r^2, there's also an acceleration for the sun but I think it's negligible.

Answer 12

btw use capital G for the constant. Now are you certain on that law of gravitation?

Answer 13

yeah; http://en.wikipedia.org/wiki/Newton's_law_of_universal_gravitation

Answer 14

yea i just checked it too

Answer 15

hmm

Answer 16

I feel like that x in the numerator doesn't belong there, but I don't know what I should do with it.

Answer 17

k so ma=GmM/r^2
a=GM/r^2
r^2=GM/a
x^2+y^2= some number
implies a perfect circle no?

Answer 18

well, then the path is the eq listed above. isn't it?

Answer 19

Wait, I'm mistaken entirely. My assumption is more so based upon the numbers that I'm using than for the system of equations here. I mean, I'm assuming the average velocity of the earth is around 30km/s, and that the distance the earth is from the sun is some constant, but really I'm trying to reproduce the orbital results (I should see some ellipses)

Answer 20

right, but using your method we get circles

Answer 21

yeah - I intend to take data like the distance 147,098,291 km as the perihelion of the orbit, and 30,300m/s as the speed at perihelion

Answer 22

since the earth's orbit is elliptical I'm hoping I can see some sort of overlapping ellipses after solving the system.

Answer 23

I think I'll play around with this a bit more and see if I get anything conclusive. Thanks for your time :)

Answer 24

got it, my final input was
NDSolve[{x''[t] == -grav*x[t]/(((x[t])^2 + (y[t])^2)^(3/2)),
y''[t] == -grav*y[t]/(((x[t])^2 + (y[t])^2)^(3/2)),
x[0] == 147098291000, y[0] == 0, x'[0] == 0, y'[0] == 30300 }, {x,
y}, {t, 100000000}] with grav = 1.327*10^20

Answer 25

nice! good luck, sorry i wasn't much help

Answer 26

tuns out the system I was supposed to be using was
\(\Huge \frac{d^2x}{dt^2}=\frac{-gMx}{(x^2+y^2)^{\frac{3}{2}}}\)
\(\Huge \frac{d^2y}{dt^2}=\frac{-gMy}{(x^2+y^2)^\frac{3}{2}}\)

Answer 27

yucky