Rate of freefall of Earth-sized object towards Earth

[StringJunky]

Assumption: The acceleration of objects in freefall in Earth’s field is 9.8m/s/s in a vacuum with grams, kilos or tonnes difference between them the rate is measured the same.

Would the rate of an Earth-sized object within the gravitational field of this Earth ‘fall’ towards it at the same rate - or towards each other - to give a 9.8m/s/s figure or is there a difference at this larger scale because with much smaller objects the differences in mass are measurably negligible relative to the mass of the Earth , for computing purposes, and are thus considered to have the same rate of freefall?

I put this in Relativity but a Newtonian explanation will do if it's easier.

Edited July 18, 2016 by StringJunky

[Markus Hanke]

The motion of ( uncharged, non-spinning ) test bodies is independent from any measurable property other than its initial position and momentum. Therefore, it does not make any difference how massive they are, when it comes to relative linear acceleration between them. Do take note though that the 9.8m/s^2 figure is valid only on the surface of the Earth, and when both bodies are of comparable mass, then to obtain the figure one has to take into account the motion of both of them - i.e., we have a full 2-body problem, instead of just an isolated test particle in free fall. The numerical value itself will not be affected though.

[StringJunky]

The motion of ( uncharged, non-spinning ) test bodies is independent from any measurable property other than its initial position and momentum. Therefore, it does not make any difference how massive they are, when it comes to relative linear acceleration between them. Do take note though that the 9.8m/s^2 figure is valid only on the surface of the Earth, and when both bodies are of comparable mass, then to obtain the figure one has to take into account the motion of both of them - i.e., we have a full 2-body problem, instead of just an isolated test particle in free fall. The numerical value itself will not be affected though.

Thanks. In my scenario I'm assuming the two objects initially start from a stationary position relative to each other. Is your answer still the same?

[Markus Hanke]

It would still be the same, but note that 9.8m/s^2 is coordinate acceleration, not proper acceleration, so its numerical value depends on how you set up the coordinate system. It is also only valid on the surface of the Earth, not anywhere else.

[StringJunky]

It would still be the same, but note that 9.8m/s^2 is coordinate acceleration, not proper acceleration, so its numerical value depends on how you set up the coordinate system. It is also only valid on the surface of the Earth, not anywhere else.

Right OK. Thanks Markus.

 

From Wiki:

 

In relativity theory, proper acceleration[1] is the physical acceleration (i.e., measurable acceleration as by an accelerometer) experienced by an object. It is thus acceleration relative to a free-fall, or inertial, observer who is momentarily at rest relative to the object being measured. Gravitation therefore does not cause proper acceleration, since gravity acts upon the inertial observer that any proper acceleration must depart from (accelerate from). A corollary is that all inertial observers always have a proper acceleration of zero.

 

Proper acceleration contrasts with coordinate acceleration, which is dependent on choice of coordinate systems and thus upon choice of observers.

Edited July 18, 2016 by StringJunky

[Enthalpy]

The Newtonian answer is that the Earth-mass object will accelerate as much as a sand grain does.

 

But then, Earth too accelerates, and more so if the other object is more massive, so while at identical distance the object's acceleration is the same, the speed curve over time differs.

 

The observer has a non-accelerated motion of course, or if you forgot to switch off the Sun's attraction, then at least the observer shall orbit the Sun like Earth does, and near enough and for short enough that tidal effects remain negligible, but far enough to not feel Earth's field, and so on. Or better, he compensates his observation from his own acceleration.

[StringJunky]

The Newtonian answer is that the Earth-mass object will accelerate as much as a sand grain does.

 

But then, Earth too accelerates, and more so if the other object is more massive, so while at identical distance the object's acceleration is the same, the speed curve over time differs.

 

The observer has a non-accelerated motion of course, or if you forgot to switch off the Sun's attraction, then at least the observer shall orbit the Sun like Earth does, and near enough and for short enough that tidal effects remain negligible, but far enough to not feel Earth's field, and so on. Or better, he compensates his observation from his own acceleration.

I was considering no other influences. A linear path between them was assumed. I just thought of them as two test masses in flat spacetime, or otherwise, zero gravity.. Thanks.

Edited July 19, 2016 by StringJunky

[Sensei]

"Earth-sized object" is term which you rather should not use, it's too vague and confusing,

because "size" is related to radius and volume of object, but does not say anything about mass of object.

Newtonians equation takes mass of object as input, not its volume (although you could rearrange formula m=V*p, when volume/radius of sphere is known and average density is known).

 

Earth-sized object, made of water, will have the same radius, the same volume, as our Earth, but its mass will be 5.5 times smaller than our Earth.

Earth-sized object, made of iron, will have mass 43% higher mass.

 

So suppose so you by "Earth-sized object", you meant the same mass as Earth.

 

[math]F=\frac{GMm}{r^2}[/math]

where G=6.67*10^-11 N*m^2/kg^2

M = mass of Earth,

m = mass of small object, significantly smaller than Earth's mass m<<M

r = variable, at different distances from object will give different force/acceleration

Center of mass in such system is located near center of massive body, in our case Earth.

 

But if m=M=Earth mass,

center of mass will be outside of each body, somewhere in the middle distance between them.

And either "Earth" will be orbiting around this center of mass location.

 

Acceleration toward one object, is cancelled by acceleration toward other object.

You can also find location were both accelerations cancels nearly perfectly and average F is close to 0.

 

Situation in which one-Earth-sized object is so close to other-Earth-sized object smaller distance ( r ) than 2*6371 km = ~ 12742 km is not possible (draw two circles with one common point, min distance between centers of circles will be 2r).

Any closer distance would be collision.

Edited July 19, 2016 by Sensei

[StringJunky]

"Earth-sized object" is term which you rather should not use, it's too vague and confusing,

I thought it was quite clear that I meant two Earths,.or an equivalent mass, at that scale Markus and Enthalpy didn't seem confused but thanks for making me aware anyway