1
$\begingroup$

It's clear to me that when an object is in free fall in a uniform gravitational field ("Einstein's Elevator") it feels no tug of gravity. However if it is in free fall in a nonuniform gravitational field does the object (or more specifically an accelerometer) feel a tug as gravity pulls in one direction and then another? Asked another way, is an object in free fall in a nonuniform gravitational field considered to be in an inertial reference system?

$\endgroup$
2
  • $\begingroup$ Ignoring the fact that such an object would need be extremely long, if the object is in a non uniform gravitational field will one part of the object experience a different acceleration than another part? $\endgroup$
    – Bob D
    Commented Jul 8, 2023 at 15:29
  • $\begingroup$ I am assuming that tidal stress can be ignored. In other words I have in mind that the object is sufficiently small that gravitational force may be considered as uniform over its volume, but this force will change over time along the path of free fall. $\endgroup$ Commented Jul 8, 2023 at 17:20

1 Answer 1

4
$\begingroup$

OK, you have confirmed that you are asking about the case where the object is sufficiently small to make tidal effect below detection level.


Another way of asking the same question, it would appear, is to ask whether the following two cases are distinguishable with purely onboard accelerometry:

  1. A satellite is in circular orbit around a primary
  2. A satellite is in significantly eccentric orbit around a primary.

In case 1. the magnitude of the velocity is constant. Also, the magnitude of the orbital acceleration is constant. As we know: if we grant equivalence of inertial and gravitational mass then we expect that an onboard accelerometer will read zero acceleration all the time.

What you are interested in, I gather, is case 2., eccentric orbit, because then the magnitude of the orbiting velocity is not constant, and the magnitude of the orbital acceleration is not constant either.

For eccentric orbit, just as in the case of circular orbit: an onboard accelerometer will read zero acceleration all the time.


Discussion:

All mechanical accelerometers are in one way or another an implementation of the following design: an enclosure around a test mass, and the test mass is suspended with springs.

When the accelerometer is pushed around by a non-gravitational force then the motion of the test mass inside the enclosure will lag behind; the suspending springs first have to transfer the accelerating force to the test mass. That lag is measured: that is what gives the acceleration reading.

But gravity acts on all the constituent parts of any object the same. For gravity all mass is totally transparent. There is no such thing as a test mass inside an enclosure being shielded from gravitational influence. The orbital acceleration is the same for all constituent parts: the enclosure, the suspending springs and the test mass; no lag.

So it makes no difference: the orbit can be circular or significantly eccentric, in both cases an onboard accelerometer will read zero acceleration.


I describe it as 'will read zero acceleration' because that is the literal measurement outcome.

The qualification 'is in an inertial reference system', well, that's already towards the realm of interpretation. My preference is to describe expected measurement outcome.

$\endgroup$
2
  • $\begingroup$ I think Cleonis is saying (his case 2) that a physically small accelerometer will not be able to detect the gradient of the non-uniform gravitational field across such a small distance and, therefore, will indicate no acceleration. But is there some fundamental reason that a skilled technician could not design an accelerometer with "springs" of sufficient sensitivity? $\endgroup$ Commented Jul 9, 2023 at 14:13
  • $\begingroup$ @RobertHarper Erm... no. Unfortunately there is an error in your thinking. Here's the thing: with the type of thinking error you are making: if it would be easy for you to see the error you would have seen it by now. But you haven't. A thinking error of that type cannot be addressed from outside. $\endgroup$
    – Cleonis
    Commented Jul 9, 2023 at 15:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.