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Radial fall in a Newtonian gravitational field

This is how Wikipedia defines Newton's law of Gravitation:

Every point mass attracts every single other point mass by a force pointing along the line intersecting both points. The force is proportional to the product of the two masses and inversely proportional to the square of the distance between them:



  • F is the force between the masses,
  • G is the gravitational constant,
  • m1 is the first mass,
  • m2 is the second mass, and
  • r is the distance between the centers of the masses.

Now, say 2 spheres, one the size of the earth and the other the size of a ping pong ball are placed say 10 km apart. There are no other forces acting on the system other than gravitation. If I've understood rightly, then both the ball & the earth sized sphere will be pulled towards each other with the same huge force.

My question is- how can we calculate when the 2 spheres will meet? And while calculating that shouldn't we consider the fact that the force is changing every moment because the distance between them $r$ changes every moment? How can I include this factor into my calculations?

Further, since the force is changing every moment, are the spheres undergoing acceleration or accelerated acceleration?? (Forgive me if my terminology is improper. I'm a beginner.) Is there any name for such forces & accelerations which change at a predictable rate as in this question?

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The time needed for two balls to collide by the action of gravity: It's an analytically solvable problem but too mathematically contrived one for a beginner. The resulting formula is pretty messy. One has to find the solution to the Kepler problem (differential equations) etc. Yes, the acceleration is accelerated because the force (and therefore acceleration) itself is increasing as the particles get closer to each other. –  Luboš Motl Jan 23 '12 at 8:00
@LubošMotl If you want to calculate the time at which they meet, an approximation with constant r will get you close enough. Since the radius of the earth plus 10km will not change too much here. –  Bernhard Jan 23 '12 at 8:14
@LubošMotl Can you give a more detailed answer? I'm familiar with differential equations (though I don't claim to be an expert) I'll try to grasp whatever I can. –  Green Noob Jan 23 '12 at 8:15
@Bernhard I just took 10 km as an example. What I really wanted to know was how to solve the problem considering the variation in distance. –  Green Noob Jan 23 '12 at 8:18
Have a look at physics.stackexchange.com/questions/19813/… as this describes a similar calculation. –  John Rennie Jan 23 '12 at 8:34
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marked as duplicate by Qmechanic, David Z Jan 23 '12 at 9:13

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2 Answers

Problems of this kind can be solved with methods of differential calculus.

Let's suppose the acceleration to be constant during some short interval of time. We can find new position of the spheres at the end of this interval and calculate new acceleration for the next period and so on.

This will give us an approximate solution. The shorter the interval we select the higher the accuracy of the approximation is. In the limit of infinitesimal interval we get the exact solution.

There is no special name for such forces because almost any force depends on the relative position of the interacting objects. But if the force depends only on the position (like gravity) it is called conservative because in that case one can introduce the potential energy and then the total energy of the system which is conserved.

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The term you're looking for is "non constant acceleration". When we do elementary free fall problems, we always assume the distance is so short that the acceleration due to gravity doesn't vary much over this distance, so we're in a constant acceleration scenario. However here you're not...

Working in the centre of mass frame, the motion is given assuming the reduced mass $\mu={m1m2\over (m1+m2)}$

Defining r as the distance between the masses, writing the equation $F=\mu a$ using an inverse square force $-{Gm1m2\over r^2}$and cancelling the $m1m2$ factors we get:

${d^2r\over dt^2}=-{GM\over r^2}$ where $M=m1+m2$

Denoting $v={dr\over dt}$

${dv\over dt}=-{GM\over r^2}$

$v{dv\over dt}=-{GM\over r^2}v$

$\int vdv = -GM\int {v\over r^2}dt$

${v^2\over 2} ={GM\over r}+c$ some constant $c$

${dr\over dt}=\sqrt[2]{{2GM\over r}+2c}$

I can't see a way to integrate this in closed form to get r as a function of t (which I'd want to invert to get t as a function of r which you need to answer your question). I think I'd switch to numerical integration here.

Alternatively, there are probably ways to compute the time approximately using the techniques that have been developed for solving Kepler's equations in two dimensions (where the resulting motion is an ellipse - here the straight line motion is just a degenerate elliptical motion.

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