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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.

Question

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?

Considerations

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.

Qmechanic
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chasly - supports Monica
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    Note: the rate at which an object accelerates, that is, the derivative of the acceleration, is called jerk. – Massimo Ortolano Jan 04 '21 at 17:54
  • Also we can presume that B's motion is affected by A In what way is 'B's motion affected by A'? – Gert Jan 04 '21 at 17:58
  • [...] this apparently bottomless pit of derivatives? There is no such bottomless pit of derivatives in nature. – Gert Jan 04 '21 at 17:59
  • @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. – chasly - supports Monica Jan 04 '21 at 18:04
  • @chasly-supportsMonica Correct but in what way does this create a 'bottomless pit of derivatives'? It does NOT. The case you describe has been treated on the pages of this site, many, many times (search and yee shall find). The accelerations are uniform ones, no jerks or higher derivatives involved. – Gert Jan 04 '21 at 18:13
  • @Gert: If that were the case, then all trajectories could be described componentwise by polynomials in time (if the derivatives of a scalar function eventually all vanish, then the function is a polynomial), which is clearly false. – Vercassivelaunos Jan 04 '21 at 18:16
  • @Vercassivelaunos Big sigh. I'm not talking about ALL TRAJECTORIES, I'm talking about the situation described by the OP and correctly derived by JEB (the answer below) – Gert Jan 04 '21 at 18:18
  • Why would one even need to keep considering higher order derivatives? I don't understand your question. For example, in an idealized cased of an object falling with air resistance, the derivatives involve exponentials, so you can keep taking derivatives, but... why? If you know the position and velocity of the object as functions of time then what more do you need? I am also confused on your link of "higher order derivatives" to "neglecting smaller effects"; these do not necessarily go hand-in-hand. – BioPhysicist Jan 04 '21 at 18:19
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    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. – G. Smith Jan 04 '21 at 18:34
  • Possible duplicates: https://physics.stackexchange.com/q/3534/2451 , https://physics.stackexchange.com/q/28519/2451 and links therein. – Qmechanic Jan 04 '21 at 18:36
  • When they get close enough, they can no longer be assumed to be a dimensionless point wrt gravity. If they are spherically symmetric, they behave as if they were points until they touch. This is not obvious but again is a standard result in classical mechanics. – G. Smith Jan 04 '21 at 18:36
  • The solution for two-body radial freefall is here. – G. Smith Jan 04 '21 at 18:38
  • @Gert: Still, then the trajectory in this situation could be described as a polynomial. But the solution of the below differential equation is not a polynomial. Though it's possible that we're talking about different things, like G. Smith said. – Vercassivelaunos Jan 04 '21 at 18:39
  • @Vercassivelaunos The OP is talking about two bodies approaching each other due to gravitational attraction only, as am I. – Gert Jan 04 '21 at 18:43
  • @Gert: Yes, so am I: The position (or even relative position) of two bodies interacting via gravity is not polynomial, which it would be, if the derivatives of high enough order all vanish. I'm reading the OP as asking wether the higher order derivatives of position in said situation vanish at some point. Which they don't. Do you read their question differently? – Vercassivelaunos Jan 04 '21 at 18:47

1 Answers1

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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:

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

and

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

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:

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

JEB
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  • It had not occurred to me that the center-of-mass also moves (in the direction of falling). I had assumed that was impossible. – chasly - supports Monica Jan 04 '21 at 18:48
  • 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. – Michael Seifert Jan 04 '21 at 20:24
  • P.S. I meant the centre of mass of the system not of B. – chasly - supports Monica Jan 04 '21 at 22:41