Rate of Attraction

[StringJunky]

Consider 2 separate systems of masses; A and B, C and D. B,C and D are all the same mass. A is some mass a lot smaller than the others. The only gravitational influence is from the masses within in each system. Assuming the edge of each mass is lined up the same as in the image, will the two systems make contact at the same time?

 

Edited June 6, 2015 by StringJunky

[Robittybob1]

If the difference between the outside edges are the same (something you don't say) they will touch at different times the smaller one will get there first as the center of mass is closer in the situation with the smaller one.

[md65536]

If the difference between the outside edges are the same (something you don't say) they will touch at different times the smaller one will get there first as the center of mass is closer in the situation with the smaller one.

Closer to what? If the A mass is negligible, A must traverse the entire distance to B. Yet, C and D will touch halfway between their original distance. Masses C and D are closer to their touch point.

 

 

 

You could definitely rig the system with wording and unsaid assumptions. The masses might not have uniform density, mass A might be small only in size but just as massive, etc. But I'd assume the most straightforward case.

Edited June 7, 2015 by md65536

[StringJunky]

Closer to what? If the A mass is negligible, A must traverse the entire distance to B. Yet, C and D will touch halfway between their original distance. Masses C and D are closer to their touch point.

 

 

 

You could definitely rig the system with wording and unsaid assumptions. The masses might not have uniform density, mass A might be small only in size but just as massive, etc. But I'd assume the most straightforward case.

Yes, the three large masses are identical and the small sphere has similar density. I imagined that they would touch at the same time but not on the same line horizontally. A will move quicker towards B than the other way round.

Edited June 7, 2015 by StringJunky

[Robittybob1]

Closer to what? If the A mass is negligible, A must traverse the entire distance to B. Yet, C and D will touch halfway between their original distance. Masses C and D are closer to their touch point.

 

 

 

You could definitely rig the system with wording and unsaid assumptions. The masses might not have uniform density, mass A might be small only in size but just as massive, etc. But I'd assume the most straightforward case.

They touch via their outside edges not their center of masses (COM). So since gravity acts from the COM A-B is closer than C-D, therefore the force/unit mass will be stronger between A-B than C-D. As A accelerates it will cover more distance/time than the larger masses

Edited June 7, 2015 by Robittybob1

[swansont]

Consider 2 separate systems of masses; A and B, C and D. B,C and D are all the same mass. A is some mass a lot smaller than the others. The only gravitational influence is from the masses within in each system. Assuming the edge of each mass is lined up the same as in the image, will the two systems make contact at the same time?

 

4 Spheres.png

 

I don't think you have provided enough information to solve the problem.

[StringJunky]

 

I don't think you have provided enough information to solve the problem.

What else do you need? They are in open space in an otherwise flat spacetime apart from that curvature generated by each object.

Edited June 7, 2015 by StringJunky

[swansont]

What else do you need? They are in open space in an otherwise flat spacetime apart from that curvature generated by each object.

 

Sorry, I misread it. The large masses hit first. You have to look at the accelerations of each body. A, C and D will all have equal accelerations. The acceleration of B will be smaller (edit: for equal center separations).

 

edit: not universally correct. see below

[md65536]

The closing distance is the same. Assuming they all start at rest, the rate of acceleration determines who touches first. The closing acceleration is proportional to total mass M1+M2 and inversely proportional to 1/r^2, where r is the distance between the centers of the masses. The question suggests that the mass is the more important factor, but there should be values where the r is more important.

[StringJunky]

 

Sorry, I misread it. The large masses hit first. You have to look at the accelerations of each body. A, C and D will all have equal accelerations. The acceleration of B will be smaller.

Thanks. I, mistakenly, thought that the combined accelerations of A and B would match up with the accelerations of C and D.

[swansont]

Thanks. I, mistakenly, thought that the combined accelerations of A and B would match up with the accelerations of C and D.

 

No. The mass of A is smaller, so the acceleration it can cause is smaller.

[Robittybob1]

 

Sorry, I misread it. The large masses hit first. You have to look at the accelerations of each body. A, C and D will all have equal accelerations. The acceleration of B will be smaller.

Is that a guess or did you do the calculations? What is the ratio of the distances between their center of masses?

[swansont]

Is that a guess or did you do the calculations? What is the ratio of the distances between their center of masses?

 

You're right. That was my initial thought when I said there wasn't enough info. Then I confused myself. I guess I didn't misread the OP after all.

 

A, C and D have the same acceleration if the center separations are the same. For edge separation, there should be a set of solutions where A and B hit first, and where they strike at the same time.

[Robittybob1]

 

You're right. That was my initial thought when I said there wasn't enough info. Then I confused myself. I guess I didn't misread the OP after all.

 

A, C and D have the same acceleration if the center separations are the same. For edge separation, there should be a set of solutions where A and B hit first, and where they strike at the same time.

Has anyone got a method to check it out mathematically?

[swansont]

Has anyone got a method to check it out mathematically?

Yes. Solve the kinematics equations, if the accelerations are known

[Robittybob1]

Yes. Solve the kinematics equations, if the accelerations are known

It must be like holding the masses apart and then dropping them. You can't actually get two masses attracted to each other without them approaching each other prior to measurements being taken.

Like they can be orbiting each other, in which case they won't be approaching each other.

Or they were approaching each other from infinity?

So one would wonder how the experiment could be run. OK maybe it could be done on the ISS. There might be a spot in the central space where it is gravitationally neutral.

So we place in space two masses of equal mass and another two masses of unequal mass with the same distance apart (measurement of the separation, the "gap" not the centers of mass) on the ISS and see which pair touches first.

 

What I was thinking was to look at the extreme of a very small mass that has no significant radius, so the distance apart for the purposes of gravitational attraction would be the radius of the larger mass and the distance apart (gap). The two larger masses would have a gravitational distance of two radius "r" plus gap "g".

Newtonian gravitation force:

F = GmM/(r^2)

 

Force 1 = GmM/(r+g)^2

 

Force 2 = GMM/(2r+g)^2

It seems that if radius and gap were equal the acceleration of the smaller mass is greater than twice the acceleration of the two larger masses.

 

When the gap is more than 1.5 radius then the two larger masses accelerate at a combined rate greater than the vastly different sized masses.

 

That would mean if they were started off far apart the larger masses would contact first but if they were close to start off with the smaller mass would be first.

So where they are separated by twice the radius (g=2*r), it would be close. I haven't figured that out yet.

It would start off giving the advantage to the two larger masses and then as they got closer the advantage would change over to the disparate masses.

Edited June 10, 2015 by Robittybob1

[Robittybob1]

If they are made from the same stuff there is a relationship between mass and radius for all the objects so I'll try and give the small mass some nominal radius and compare the mass difference between the large and the small spheres.

[md65536]

If they are made from the same stuff there is a relationship between mass and radius for all the objects so I'll try and give the small mass some nominal radius and compare the mass difference between the large and the small spheres.

It should be only the total mass that matters.

 

The force is proportional to the product of the two masses, but the acceleration of a mass acted on by a given force, is inversely proportional to its own mass. So only the mass of the gravitational object affects how fast another mass falls toward it.

 

One mass falls toward the other, and the other falls toward the first. The closing acceleration will be the sum of the two's accelerations. It should be the same, with the same r (distance between the centers of the two masses) and same total mass, regardless of how the mass is distributed between the two.

 

 

You should be able to manipulate the initial acceleration to favor either AB or CD, by balancing only two variables: r and total mass. I'm not sure if it's possible to have one begin accelerating faster, and have the other catch up and touch first, by also manipulating the gap.

Edited June 11, 2015 by md65536

[imatfaal]

snip

 

The force is proportional to the product of the two masses, but the acceleration of a falling mass is inversely proportional to its own mass. So only the mass of the gravitational mass affects how fast another mass falls toward it....snip

 

F₁₂=-GM₁M₂/r^2

a₁₂=-GM₁/r^2 - it is directly proportional to the gravitating mass of the attracting object and independent of the falling objects mass

snip..

 

One mass falls toward the other, and the other falls toward the first. The closing acceleration will be the sum of the two's accelerations. It should be the same, with the same r (distance between the centers of the two masses) and same total mass, regardless of how the mass is distributed between the two.

 

 

You should be able to manipulate the initial acceleration to favor either AB or CD, by balancing only two variables: r and total mass. I'm not sure if it's possible to have one begin accelerating faster, and have the other catch up and touch first, by also manipulating the gap.

 

yeah - maybe I musunderstood your phrasing of first section as this bit I agre with

 

[latex]|a_{12}|+|a_{21}|=\frac{G}{gap+radius_1+radius_2} \cdot (M_1+M_2)[/latex]

Edited June 11, 2015 by imatfaal
missing minus sign

[md65536]

F₁₂=-GM₁M₂/r^2

a₁₂=-GM₁/r^2 - it is directly proportional to the gravitating mass of the attracting object and independent of the falling objects mass

Yes, I've edited my wording to be closer to this.

 

The mass M_2 of the object being acted on by a gravitational force F_12 cancels out in the acceleration equation above, since

F₁₂ = M₂ * a₁₂

Edited June 11, 2015 by md65536

[Robittybob1]

Do we still look at the addition of both accelerations i.e. how 1 accelerates towards 2 and how 2 accelerates towards 1?

[md65536]

Do we still look at the addition of both accelerations i.e. how 1 accelerates towards 2 and how 2 accelerates towards 1?

Post #19 does that.

 

 

Another thing to consider: The total mass stays constant during the experiment, while r decreases. So the final acceleration relative to initial acceleration should be greater in the smaller mass with smaller r setup AB. If you begin with M_total/r^2 being just slightly greater in the CD setup, it should start with a higher acceleration and be overtaken by AB.

 

My intuition says units don't matter??? G will be different with different units and so will the time the experiment takes, but not the outcome?

 

So for total mass M_AB, if M_AB / r_AB^2 is greater than or equal to M_CD / r_CD^2, the smaller mass will easily touch first.

 

If you want to ensure that the larger masses touch first, an easy way (with overkill) is to make it so the end (maximum) acceleration is greater in that case than in the AB case. Ie. let r_final equal the distance between the centers of the two masses when touching, rather than the initial distance, and make the CD side have a greater M/r_final^2.

Edited June 11, 2015 by md65536

[Robittybob1]

Post #19 does that.

 

 

Another thing to consider: The total mass stays constant during the experiment, while r decreases. So the final acceleration relative to initial acceleration should be greater in the smaller mass with smaller r setup AB. If you begin with M_total/r^2 being just slightly greater in the CD setup, it should start with a higher acceleration and be overtaken by AB.

 

My intuition says units don't matter??? G will be different with different units and so will the time the experiment takes, but not the outcome?

 

So for total mass M_AB, if M_AB / r_AB^2 is greater than or equal to M_CD / r_CD^2, the smaller mass will easily touch first.

 

If you want to ensure that the larger masses touch first, an easy way (with overkill) is to make it so the end (maximum) acceleration is greater in that case than in the AB case. Ie. let r_final equal the distance between the centers of the two masses when touching, rather than the initial distance, and make the CD side have a greater M/r_final^2.

I think this is similar to what I found the first time, but when I gave the mass some physical dimensions and mass according to those dimensions I have become less certain of the outcome.

I'll have to look at the calculations again to see where I was going astray.

[DimaMazin]

I think this is similar to what I found the first time, but when I gave the mass some physical dimensions and mass according to those dimensions I have become less certain of the outcome.

I'll have to look at the calculations again to see where I was going astray.

Where are correct calculations here? Reducing distance increases gravitational force and accelerations. We can't solve the question without integral equations. Sorry, I don't know them.

[Robittybob1]

Where are correct calculations here? Reducing distance increases gravitational force and accelerations. We can't solve the question without integral equations. Sorry, I don't know them.

I do it via a spreadsheet and vary one parameter at a time and I get the feeling what happens in different situations.

So on the condition that the material has a consistent density you can select r1 and r2 and the spacing (gap) g therefore you can formulate the amount of force, and later divide by the mass based on the chosen radius and gap values to get accelerations.Starting from zero, acceleration determines speed and speed determines distance in time. So at this stage I'm just considering the effects on acceleration.