# Average acceleration and total acceleration of gravity

1. Nov 9, 2015

### Nick Heller

Say you have a gravitational acceleration between a spherical object of mass M and inner radius R and a much smaller spherical object of mass m (essentially point-like) and distance ro away (M >> m) with an acceleration on mass m at any point r given by
a = GM/r2
Knowing that the acceleration with respect to the distance between the two objects is not constant, how would you calculate the amount of time it would take for the object to hit the surface of the larger sphere (also assuming the larger sphere moves negligibly)? The smaller object moves from a distance ro away from the center of mass of the larger object (r = 0) to a distance R away from this center.
Solving this as a differential equation has gotten hairy. My other approach was to take the average force with respect to the distance from the center given by (the definite integral from ro to R of Fdr)/(R - ro) where F is the standard gravitational force. This comes from the equation for the average of any function. Once this average force is calculated, the acceleration is constant, so can it be inserted into the regular kinematic constant-acceleration equations to find the amount of time it would take to transverse the distance?
I am not a physicist (yet), so be merciful. Thanks!

2. Nov 9, 2015

### BvU

Hello Nick, welcome to PF !

Not sure what you mean with "I am not a physicist (yet)" -- but if you are going to be one, you should walk the walk and talk the talk. I.e. write out equations and then solve them. In your case something like $$\vec F = m\vec a$$ and if I understand you well, $$| \vec F | = {GM m\over r^2}$$ and you describe the free fall of an object with mass m that is let go from a distance $r_0$ above M and you want to describe $|r|$ as a function of time. So you get an equation like $${d^2r\over dt^2} = -{A\over r^2}$$with $r(t=0) = r_0$. Am I right so far ?

--

3. Nov 9, 2015

### Nick Heller

Yes, right so far

4. Nov 10, 2015

### PWiz

You cannot find a closed form solution for a function $r(t)$, but you can find $t(r)$.
First, observe that $a= v \frac{dv}{dx}$
Now assume that $dx=-dr$ (i.e. only the displacement of the smaller mass leads to a reduction in the radial distance between the two masses), so that $\int v dv = \frac{v^2}{2} = \int_{r_0}^r - \frac{GM}{r^2} dr$ where $r_0$ is the radial distance of the smaller mass from the larger mass at t=0.
You can then calculate t by substituting $v=\frac{dx}{dt} = \frac{-dr}{dt}$, and rearranging the differentials so that $$t = \int_{r_0}^{r} \frac{-dr}{\sqrt{2 \int_{r_0}^r - \frac{GM}{r^2} dr}}$$
You can then just substitute $r=R$ into the final expression you get to calculate the time the smaller mass takes to fall that radial distance. (Remember to compute the definite integral in the denominator first.)

5. Nov 10, 2015

### PeroK

Solving the differential equation involves a couple of tricks that are not easy to spot.

You can also get the answer from Kepler's laws. But, again, the trick is not easy to spot.

6. Nov 10, 2015

### nasu

By the way, the average of the force over distance will not give you the average acceleration.
You need the average over time.

You can use the average over distance to calculate the work, though. Which is the same as doing that integral.

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