Mass-energy in an accelerated system

[Markus Hanke]

Freefall is the state when the only force is gravity

 

 

Better be careful here - when you put an accelerometer into free fall, it will read zero at all times, so where's the "force"

[swansont]

How it could be beside the point?

He asked "And you can verify this by the tidal forces on its surface (and interior) (of Moon) ?"

So, I explained him, how observation of moving away Moon is performed (by precise measurement of distance using Retroreflector in dozen years).

 

 

Actually, Moon does not move in a circle, but ellipse, with perigee 356400-370400 km, and apogee 404000-406700 km.

 

 

We're talking about acceleration of an object in orbit.

[geordief]

He asked "And you can verify this by the tidal forces on its surface (and interior) (of Moon) ?"

So, I explained him, how observation of moving away Moon is performed (by precise measurement of distance using Retroreflector in dozen years).

 

 

My idea was that movements such as moonquakes might be used by a Moon inhabitant as a "natural" accelerometer.

 

If the Moon was simply freefalling into the Earth (so not orbiting) and not accelerating these would not be occurring.

 

That is what I was trying to say.in that post As I seem to put my foot in it every second post I may well be wrong but that is why I said it.

 

I had not considered movement away from the Earth but it does also seem relevant and shows acceleration as well.

[swansont]

My idea was that movements such as moonquakes might be used by a Moon inhabitant as a "natural" accelerometer.

 

If the Moon was simply freefalling into the Earth (so not orbiting) and not accelerating these would not be occurring.

The moon could be stationary and still have tides. The earth is relatively stationary with respect to the moon, and it causes tides here.

 

 

 

I had not considered movement away from the Earth but it does also seem relevant and shows acceleration as well.

That's a recessional speed, which is not an acceleration. Hence its irrelevance to the discussion.

[J.C.MacSwell]

My idea was that movements such as moonquakes might be used by a Moon inhabitant as a "natural" accelerometer.

 

If the Moon was simply freefalling into the Earth (so not orbiting) and not accelerating these would not be occurring.

 

That is what I was trying to say.in that post As I seem to put my foot in it every second post I may well be wrong but that is why I said it.

 

I had not considered movement away from the Earth but it does also seem relevant and shows acceleration as well.

There are still tidal forces while dropping in freefall. There are gradients in the horizontal and the vertical.

[geordief]

There are still tidal forces while dropping in freefall. There are gradients in the horizontal and the vertical.

I wonder how I got that impression.Under what circumstances does an accelerometer read 100% zero.? At infinity?

[swansont]

There are still tidal forces while dropping in freefall. There are gradients in the horizontal and the vertical.

 

 

In a uniform field?

[imatfaal]

[ot]

 

 

 

 

Better be careful here - when you put an accelerometer into free fall, it will read zero at all times, so where's the "force"

 

I did say I was talking Newtonian - and they only measure proper acceleration; which is the acceleration relative to the freefall. The force relative to freefall when in freefall must be zero thus no acceleration.

 

And no one would claim there is no force on something in the vomit comet (gravity doesn't disappear because a plane describes a parabola) but an accelerometer would read zero because its proper acceleration, the acceleration relative to freefall is zero. For an accelerometer to show a reading there must be a force and thus acceleration other than gravity

 

[/ot]

[Markus Hanke]

I wonder how I got that impression.Under what circumstances does an accelerometer read 100% zero.? At infinity?

 

 

Either that, or if your free fall frame is small enough so that tidal forces are negligible.

[J.C.MacSwell]

 

 

In a uniform field?

No. Free falling into the Earth.