Something I don't understand with 2 moving masses

[lazykid]

Hello.

This message is not homework. I came up with the problem on my own.

Trying to think about the value measured by an accelerometer in different scenarios, I ended up with a strange conclusion.
Hopefully with your help I will be able to figure out if I'm doing something wrong.

Consider 2 ponctual masses, m₁ and m₂.

By gravitational attraction, m₂ exerts on m₁ a force:
\vec{F_2}

By gravitational attraction, m₁ exerts on m₂ a force:
\vec{F_1}=-\vec{F_2}

Additionally, another force is applied on m₂:
\vec{F_m}=\left(1+\frac{m_2}{m_1}\right)\vec{F_2}
(for example using some kind of motor)

The acceleration of m₁ is given by:
\vec{a_1}=\frac{\vec{F_2}}{m_1}

The acceleration of m₂ is given by:
\vec{a_2}=\frac{\vec{F_1}+\vec{F_m}}{m_2}=\frac{\left(-1+\left(1+\frac{m_2}{m_1}\right)\right)\vec{F_2}}{m_2}=\frac{\vec{F_2}}{m_1}=\vec{a_1}
(I tried to define \vec{F_m} the way it is just for that).

To convince myself that this result makes sense, if I consider the system m₁ + m₂, I can verify that its acceleration is:
\frac{\vec{F_m}}{m_1+m_2}=\frac{\vec{F_2}}{m_1} = \vec{a_1} = \vec{a_2}

But here comes the odd part.
Initially, both masses are at rest and at a distance d > 0 from one another.
If I let the system evolve for a while, it accelerates with a constant acceleration and the distance between m₁ and m₂ stays constant, with the value d.
But then, the only force that is applied to m₁ is the gravitational force of m₂.
And by staying at a constant distance from m₂, this gravitational force is constant and m₁ is always in free fall, moving with the same constant acceleration (as m₂).
So, an accelerometer placed on m₁ would always indicate 0.

If m₂ were the Earth, it would be like maintaining a falling object in mid-air by moving the Earth away from it as it falls.

One example I can think of that almost matches my problem is the ISS in orbit around the Earth : its distance from Earth doesn't change a lot, and astronauts are constantly accelerated by gravity without feeling it.

But m₁ isn't in orbit (?) so it does seem a little paradoxical and I'd like to get your opinions on that: am I making a mistake? and if so, what is it?


Thanks,
Dan

 

[lalbatros]

Do you know if ownwork can be answered? ;)

 

[lazykid]

Do you know if ownwork can be answered? ;)

When I posted my message, there was a red warning saying that I should post homework (as in "work given by teachers" as I understand it) in a dedicated forum.

But no teacher asked me to work on that problem, and it doesn't even come from a book (that I know of).

Still, if I'm in the wrong forum, then I'm sorry; could you please tell me where I should post?

 

[lalbatros]

The additional force exerted on m2 depends on the force exerted on m1 (by m2).
In addition, the dependence is very specific.
No surprise that with such a kind of (unphysical) unnatural feedback force you would get (unphysical) unnatural results.

You should analyze a similar problem with a more general "additional force".
You would then see why a specific choice leads to a specific result.

You should also try to imagine a real physical example of this exercice.

I don't see how the additional force could be implemented except, maybe, by a kind of electronic feedback. In this case, a measurement of the force acting on m1 could be use to create an additional force on m2, and "compensate". This could be done with a rocket propeller for example.

Once you have found out a practical implementation of this system, you will realize that it is totally unrelated to any natural planetary system. Then you will discard any irrelevant intuition.

Ooh, well, maybe there is a real physical example?

 

[lalbatros]

When I posted my message, there was a red warning saying that I should post homework (as in "work given by teachers" as I understand it) in a dedicated forum.

But no teacher asked me to work on that problem, and it doesn't even come from a book (that I know of).

Still, if I'm in the wrong forum, then I'm sorry; could you please tell me where I should post?

I was just joking.
It occurred quite often that my thread was mistaken for an homework.
I am a few years away from retirement!

I don't care for rules, since no rule forbids that!

 

[lazykid]

The additional force exerted on m2 depends on the force exerted on m1 (by m2).
In addition, the dependence is very specific.

Yes, F_m can be computed knowing m₁, m₂ and the distance d between them.

You should analyze a similar problem with a more general "additional force".

Maybe m₂ could be tracted with a light cable.
But I can't use gravity because it might also affect m₁ and then the problem would be different.
Also, I can't use a stick to maintain both masses at the same distance because that would add reaction forces (propagated by contact through the stick) to the problem and the outcome would be completely different (no net acceleration, and even immobility because nothing was moving from the start).

You would then see why a specific choice leads to a specific result.

I think I sort of see that now...
But it doesn't really explain to me what's happening in my (specific) example.
Or are you saying that by applying a punctual force on m₂ then a reaction force "appears" (without contact) on m₁?
I didn't know it could do that, but I guess it would change the problem, like adding a stick.
But does that means that if I use a cable then the system has a uniform motion (I can't imagine it staying still because m₁ isn't actually attached to anything)?

You should also try to imagine a real physical example of this exercice.

I don't see how the additional force could be implemented except, maybe, by a kind of electronic feedback. In this case, a measurement of the force acting on m1 could be use to create an additional force on m2, and "compensate". This could be done with a rocket propeller for example.

Ok.
Let's say a space shuttle is moving closer to take a look at an asteroid (but without touching it).
When he's done, the pilot turns around and tries to leave by slowly accelerating (low but constant force).
The shuttle is designed so that any eventual exhaust wouldn't affect the asteroid.
At first the pilot will notice that the distance is still decreasing.
(the asteroid and the shuttle are falling toward each other because of their masses)
Knowing the constant mass m₁ of the asteroid, the (assumed) constant mass m₂ of the shuttle, and the currently varying distance d between them (using a laser telemeter for example), he is able to compute the minimum force his engines must generate to prevent the asteroid from coming any closer:
1) |F_m| = (1 + m₂ / m₁) G m₁ m₂ / d² (if he assumes there is no reaction force in he asteroid, like in my initial problem)
or
2) |F_m| = G m₁ m₂ / d² (if he assumes there is in the asteroid a contactless reaction force to his engine)
At this point, I'm still not sure how to save the astronaut...

Anyway, the choice made, F_m can be constant, and there is no more need for "feedback" because the distance will not change anymore.
Now, the pilot measures his acceleration to be non-null, and the asteroid doesn't move in relation to the shuttle.
The asteroid doesn't have propellers and is not physically attached to the shuttle.

If the asteroid measures a null acceleration, then I have my previous strange conclusion.

If the asteroid measures a non-null acceleration, then I expect it to result from the balance state that would exist if the asteroid and the shuttle were attached.
(like in the example above with the stick/reaction force).
In that case, the system asteroid + shuttle would be either at rest or in uniform motion in space, I can't tell (it seems to me that it might depend on whether the shuttle uses rocket propellers or is tracted by a cable).

The two scenarios don't seem equivalent to me, so I guess one of them is false.
I can't really test it so I'll have to trust you on that one (unless you had something different in mind) and choose number 2, meaning my problem is false because there is an additional contactless reaction force on m₁.

What do you think?

Once you have found out a practical implementation of this system, you will realize that it is totally unrelated to any natural planetary system. Then you will discard any irrelevant intuition.

I know that it is impossible to move the Earth like I described.
But what would happen (theoretically) to the falling object, would it "feel" pulled down?

 

[willem2]

Ooh, well, maybe there is a real physical example?

There's the idea of a Gravity tractor: An ion engine strapped to a large weight, that will hover for years near an asteroid to deflect it away from the earth.