I am not a physicist.

Suppose a body A is falling towards body B in a vacuum. We know that A's speed will increase. However, as A draws near to B, the force of gravity will increase so the rate at which A accelerates will increase. Also we can presume that B's motion is affected by A. So now we have multiple levels of acceleration.


I understand that, for practical purposes we can usually neglect smaller effects. My question is: In Nature do we ever get to the end of this apparently bottomless pit of derivatives?


A and B are of comparable but non-identical mass. They will therefore accelerate towards one another at differing rates.

When they get close enough, they can no longer be assumed to be a dimensionless point wrt gravity.

  • 2
    $\begingroup$ Note: the rate at which an object accelerates, that is, the derivative of the acceleration, is called jerk. $\endgroup$ – Massimo Ortolano Jan 4 at 17:54
  • $\begingroup$ Also we can presume that B's motion is affected by A In what way is 'B's motion affected by A'? $\endgroup$ – Gert Jan 4 at 17:58
  • $\begingroup$ [...] this apparently bottomless pit of derivatives? There is no such bottomless pit of derivatives in nature. $\endgroup$ – Gert Jan 4 at 17:59
  • $\begingroup$ @Gert - They are of comparable but not identical mass. Therefore they are each accelerating towards the other at different rates. Also, when they get close enough, they can no longer be treated as dimensionless points wrt gravity. $\endgroup$ – chasly - supports Monica Jan 4 at 18:04
  • 2
    $\begingroup$ The differential equation to be solved involves only second derivatives. The solution has nonzero derivatives of all orders. It is a standard homework problem in classical mechanics. $\endgroup$ – G. Smith Jan 4 at 18:34

In your question, you are tracking both ${\bf r}_A$ and ${\bf r}_B$ (the position vectors in a fixed coordinate system), with something like:

$$ m_a\ddot {\bf r}_A = km_am_b\frac{{\bf r}_A - {\bf r}_B}{||{\bf r}_A - {\bf r}_B||^3}$$

$$ m_b\ddot {\bf r}_B = km_am_b\frac{{\bf r}_B - {\bf r}_A}{||{\bf r}_A - {\bf r}_B||^3}$$

which is not an endless pit of derivates, but rather a endless cycle of second derivatives.

However, there is a coordinate transformation that helps. If you rewrite the equations in terms of:

$$ {\bf R} = \frac{m_a{\bf r}_A + m_b{\bf r}_B}{m_a+m_b}$$

$$ {\bf r} = {\bf r}_A - {\bf r}_B $$

you should find:

$$ \ddot {\bf R} = 0 $$


$$ \frac{m_am_b}{m_a + m_b}\ddot{\bf r} = km_am_b\frac{{\bf r}}{||r||^3}$$

which breaks the cycle.

The first coordinate is the evolution of the center-of-mass: since there are no external forces, it moves at constant velocity.

The other coordinate is just the separation, which works when you use the reduced mass ($\mu$), defined via:

$$ \frac 1 {\mu} = \frac 1 {m_a} + \frac 1 {m_b} $$

  • $\begingroup$ It had not occurred to me that the center-of-mass also moves (in the direction of falling). I had assumed that was impossible. $\endgroup$ – chasly - supports Monica Jan 4 at 18:48
  • $\begingroup$ The use of a single equation for $\mathbf{r}$ obscures the fact that it it's a 3-D vector, and so contains three variables whose second derivatives are all mixed together if you write out its components. Through further clever choices of coordinates you can get it down to an equivalent problem involving one function of $t$, but the cycle isn't "broken" quite yet when you write down the equation for $\mathbf{r}$ alone. $\endgroup$ – Michael Seifert Jan 4 at 20:24
  • $\begingroup$ P.S. I meant the centre of mass of the system not of B. $\endgroup$ – chasly - supports Monica Jan 4 at 22:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.