Motion due to gravity without neglecting varying distance

1. Mar 30, 2015

omberlo

Hello everyone. In an attempt to brush up my Calculus skills and apply them to a physics problem, I made up an excercise which I tried to solve, but got stuck in the process. I'll write my attempt at a solution and I'd be grateful if someone could correct my approach and maybe even help me finish solving it.

The idea is to calculate the motion of a mass m2 towards a mass m1 due to the gravitational force (Newtonian's formulation), but without neglecting the effect of the varying distance, and thus the varying gravitational force, as m2 moves towards m1.

This are the framework conditions of the problem:
- we have two masses m1 and m2 that are distant r0 at the beginning.
- Both masses have a velocity of 0 at the beginning.
- m1 >> m2 so we only consider the displacement of m2 towards m1 when defining the distance. We neglect the displacement of m1 towards m2.
- we use Newtonian's formulation of gravity, i.e. $F_G = \frac{Gm_1m_2}{r^2}$
- we use the standard formulation of velocity and not the relativistic one.

The goal is to calculate the displacement, velocity and acceleration of m2 as functions of time.
Again, this is mostly a calculus exercise, so please accept the above approximations, no matter how incorrect they may be.

This is how I tried to tackle the problem:

> The mass m2 is affected by an acceleration equal to $a = \frac{F_G}{m_2}$(1) due to gravity.

> Substituting FG with Newton's law and cancelling out the mass term yields: $a = \frac{Gm_1}{r^2}$(2)

> Now I want to calculate the velocity. Acceleration is the rate of change of velocity with respect to time, meaning that $a = \frac{dv}{dt}$(3)

> Substituting this expression of acceleration in equation (2) yields: $\frac{dv}{dt} = \frac{Gm_1}{r^2}$(4)

> Now we have to integrate both sides of the equation over time: $\int \frac{dv}{dt}\, dt = \int \frac{Gm_1}{r^2} \, dt$(5)

> On my first attempt, this yielded $v = \frac{Gm_1t}{r^2}$(6) . I ran an approximated calculation using excel, and noticed that this formula for the velocity was only correct for low velocities. As the time and velocity increased, the calculated velocity was distancing more and more from the correct value. I then realized that given the way I've defined the problem, r, the distance between the two masses, is a function of time, as it changes over time, and thus cannot be treated as a constant when performing the integral.

> The distance between the two masses is decreasing as m2 moves towards m1, and is equal to $r(s) = r_0 - s$(7), where s is the total displacement of m2 towards m1. Now I have an expression for r as a function of s, but I need one as a function of t in order to integrate it over time.

> The definition of velocity is $v = \frac{ds}{dt}$(8) and by integrating both terms over time we have $s(t) = \int v\,dt$(9). Substituting s from equation (9) in equation (7) we have $r(t) = r_0 - \int v\,dt$(10)

> Substituting (10) in (5) we have $\int \frac{dv}{dt}\, dt = \int \frac{Gm_1}{(r_0 - \int v\,dt)^2} \, dt$

Here I stopped. First of all, are the calculation steps correct up to this point? Secondly, is there a way to solve the above integral and conclude the exercise?

Any help is much appreciated.

2. Mar 30, 2015

AlephNumbers

I think that your steps are not incorrect thus far, and your methods solvent (so long as there are only two masses involved). But it certainly would be easier to use numerical approximation to solve this.

3. Mar 30, 2015

AlephNumbers

Oh jeeze. This problem is even more difficult than I thought. You might want to wait for someone who really knows what they are talking about.

4. Mar 30, 2015

A.T.

I'm sure this has been discussed here before. Try the search function. If I remember correctly, the simplest solution was by treating the trajectories as degenerated orbits and applying Keppler's Laws.

5. Mar 30, 2015

Tom_K

Looks OK but you may find it easier to use the chain rule:
dv/dt = (ds/dt) (dv/ds)
so:
v(dv/ds) = Gm/s^2
leaves a more manageable integral:
∫ v dv = ∫ Gm/s^2 ds
But still a lot of work to do after that!