Newtonian gravity and differential equations

1. Nov 20, 2015

Chuckstabler

Hello,

So i know that the gravitational acceleration experienced by a body is -GM/||d||^2 * dhat, where dhat is the current displacement unit vector, which has a magnitude of 1. the magnitude of a vector is equal to to the square root of the sum of its squared components. This will be a 2d case. Assuming that some constant times the x and y displacement components will result in a displacement vector with a magnitude of 1, we get c^2(x^2+y^2) = sqrt(1) = 1, so c = 1/(x^2+y^2)^0.5, so our displacement unit vectors x component is x/(x^2+y^2)^0.5, and its y component is y/(x^2+y^2)^0.5. The magnitude of our displacement ||d|| is (x^2+y^2)^0.5, so ||d||^2 = x^2+y^2, so multiplying -gm/||d||^2 by the displacement unit vector I get an x component of -xgm/(x^2+y^2)^1.5, and a y component of -ygm/(x^2+y^2)^1.5

So thats the acceleration vector in cartesian coordinates. Breaking them down into two coupled non linear second order differential equations i get...
y" = -gmy/(y^2+x^2)^1.5
x" = -gmx/(y^2+x^2)^1.5

I'm well aware that we can find solutions in spherical coordinates that give us radius as a function of theta for the first formula (newtonian gravity), but can we find solutions giving us radius as a function of time in spherical coordinates? Or x and y displacement as a function of time in cartesian coordinates using the above coupled differential equations? I know that we can do so for a one dimensional case, by finding velocity as a function of distance and then taking advantage of the implicit solution that Time = the integral of (1/velocity(distance)) dv, but can it be done in two dimensions or more? or must a time independent solution suffice (radius(theta) via the well known conic solutions in spherical coordinates).

2. Nov 25, 2015

Staff: Mentor

It is possible to write down integrals that give the position as function of time, but I'm not aware of closed solutions for those integrals.