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Typical treatments of free-fall experiments treat the the Earth as fixed. Using this model, object being dropped moves toward Earth, but the Earth doesn't move toward the object.

A more detailed analysis would have both the Earth and object accelerating toward each other due to the mutual third-law interaction.

Is it really safe to ignore the acceleration—and hence movement—of the Earth during such an interval?

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  • $\begingroup$ The general procedure for answering this kind of question is: compute the size of the neglected effect to first order and compare to your uncertainty in actually performing such an experiment. If the neglected effect is comparable your uncertainty, then you should probably insist on it. If it is much smaller than your uncertainty then you can safely forget it. If you improve the precision of your measurement you may have reconsider such questions. $\endgroup$ Jul 23, 2014 at 1:20

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Newtonian mechanics should be accurate enough.

There are general procedures that can help answer the question "is it safe to ignore this effect." See, for example, dmckee's comment. This answer is more of a question-specific approach.

The force on the "falling" object by Earth is equal in magnitude to the force on the Earth by the falling object. This can be shown using Newton's third law, which is also inherent in universal gravitation. Since the forces are equal, the accelerations $a=F/m$ experienced by each body is inversely proportional to its mass. Thus, the much more massive Earth will experience a significantly smaller acceleration as compared to the falling object.

To put a number to it, consider an object of $m\approx 1000\text{ kg}$. The ratios of the accelerations will be $\frac{1000\text{ kg}}{M_\text{Earth}}=1.7\times 10^{-22}$. Quite small indeed!

That alone might be used as an argument that the Earth can be treated as stationary over typical short time intervals of free fall. But it might also be interested to approximate the distance that the Earth moves toward the falling object as well.

The tallest buildings in the world are roughly 800 meters high, and it will take close to 13 seconds for an object to hit the ground once released from rest. Air drag would increase this free-fall time, but here we seek only order of magnitude estimates.

The forces on Earth will be approximated as constant with magnitude $F=GM_Em/R_\text{Earth}^2$. The distance that the Earth travels during free-fall is then $$d_E=\frac{1}{2}at^2 = \frac{1}{2}\frac{GM_Em/R_E^2}{M_E}t^2 = \frac{Gmt^2}{2R_E^2}.$$

Using a $1000\text{ kg}$ mass, I calculate that the Earth moves a distance $d_E \approx 10^{-19}\text{ m}$. For comparison, the size of a nucleus is on the order of femtometers ($10^{-15}\text{ m}$). That's nucleus, not atom.

I would say this is small enough to ignore.


$^1$ This estimation of 13 seconds was calculated assuming a stationary Earth. Plenty accurate for these purposes.

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Skipping the (high-school level) math for a moment, and apply a layer of common sense gives us this:

Objects attract each other, inertia depends on mass. Given the mass of the earth is large one can reasonably assume it to be stationary for human-scale purposes. If we are dropping a moon-sized object we may want to consider the earth's motion, but it won't be an issue for very long.

If you really, really (really) need to consider the earth's acceleration (maybe the LHC people?) then you must consider the whole system. If you drop 1000kg in Concordia, Argentina at the same time 20 baggage handlers drop suitcases at Shanghai Airport the accelerations are going to cancel out.

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