Influence of external gravitational forces on a moving body

[Sriman Dutta]

Suppose that there's a body of mass m moving with uniform velocity of v in the horizontal direction of XX' ( see diagram). There are two massive bodies m1 and m2 which lie at a distance of r from centre O. So, as the body moves, the two bodies will influence its course of motion due to gravity.

My question is how the motion of the body will be affected?

From analysis, I found that if m1=m2, there will be a net force in the XX' direction, causing the body to accelerate upto O. At O, it's velocity is maximum. But, as soon as the body crosses O, there will be a net negative force, causing it to decelerate. If the velocity of the body at O is below the escape velocity, then the body will divide into two equal parts, each part will start describing a circular motion about the massive bodies.

If m2>m1, then the body will bend towards m2 and will move in an arc. If the velocity of the body at any point reaches below the escape velocity, the body will fall into m2. Else, it shall follow a circular course of motion, accelerating and decelerating.

[imatfaal]

Things to bear in mind in a nice investigation:

1. M1 and M2 will feel an equal and opposite force. If they are not fixed what will happen?

2. If M1 and M2 are fixed in space - this is very unphysical . Is there a more realistic example which you could look at?

3. What happens if M1=M2 and the test object cannot split?

4. What shapes can orbits take? How do you engineer (by changing relative masses/velocities) each of them? Is this a simple question - ie can you assume what will happen or do you need to calculate, and is it possible to calculate?

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You might also wish to look at cathode ray tubes, or mass spectrometry. This is a very similar question - and it is something we do all the time

[Sriman Dutta]

I have analysed the case when m1=m2.

The velocity-time graph of the body, assuming the velocity of body doesn't fall below the escape velocity, is here-

When m2>m1, the net force acts in the direction of m2 (vector addition yields this result). So I presume the body will follow a circular or may be an elliptical path.

Edited March 25, 2017 by Sriman Dutta

[swansont]

Flat and then linearly increasing? That can't be right. The body always feels a force, so v will not be flat. And it's accelerating, and the acceleration increases as it approaches the masses.

[Sriman Dutta]

Right swansont. But here I have assumed that the gravitational force will influence only upto a certain limit. Beyond that the moving body won't feel any attraction. It's purely hypothetical though.

And please consider the scenario according to Newtonian gravity.

Edited March 25, 2017 by Sriman Dutta

[studiot]

Right swansont. But here I have assumed that the gravitational force will influence only upto a certain limit. Beyond that the moving body won't feel any attraction. It's purely hypothetical though.

And please consider the scenario according to Newtonian gravity.

 

In which case surely your X and X' and the flatlines beyond them should be on the zero line?

 

Initially said said that the acceleration was initially positive up to O and then negative afterwards, so I would expect to see your triangle part above and part below the line.

 

A word about English and terminology.

 

In ordinary English massive means having very large mass, which is as you seem to have used it for M1 and M2.

 

In scientific English massive means possessing any mass at all, however small, so a neutron or a spaceship is massive, a photon is not.

 

In circumstances like these the neutron or spaceship are called the 'test mass' in scientific English.

 

Finally you have (correctly) posted in the classical section and refer to Newtonian mechanics.

But even in Newtonian mechanics you need to consider the motion of M1 and M2 as well.

Why would m pass along line XX?

By the time m has reached the midpoint between M1 and M2, they will have, in general, moved somewhere else unless all three objects are tied to some other (larger) one.

 

This is rather like the Earth, the Moon and a spaceship all possess a common motion about the Sun so we can make the motion assumption you have made as the spaceship moves across between Earth and the Moon.

[swansont]

Right swansont. But here I have assumed that the gravitational force will influence only upto a certain limit. Beyond that the moving body won't feel any attraction. It's purely hypothetical though.

And please consider the scenario according to Newtonian gravity.

Those two statements conflict with each other; Newtonian gravity extends to infinity so there is no limit. Even with a finite range, speed will not increase linearly with time. With a constant acceleration, speed increases as t^2. If the acceleration is rising, as it will here, it increases faster than that.

[Sriman Dutta]

I knew that: acceleration = velocity /time, or, velocity = acceleration x time

 

If a gravitational force acts on a body it will attract it towards itself with a constant acceleration.

 

Here, the vertical components of acceleration cancel out and their horizontal components align along XX' causing it to accelerate upto O and then decelerate after O. I got this result from vector addition.

[swansont]

I knew that: acceleration = velocity /time, or, velocity = acceleration x time

 

If a gravitational force acts on a body it will attract it towards itself with a constant acceleration.

 

Here, the vertical components of acceleration cancel out and their horizontal components align along XX' causing it to accelerate upto O and then decelerate after O. I got this result from vector addition.

No. v=at is only true for constant acceleration.

 

Since gravity is a function of distance, it is not constant. It can be approximated as one over relatively short distances (e.g. We do this near the surface of the earth)

[Sriman Dutta]

You mean there will be jerk.

This means that two bodies moving towards each other due to gravitational force of attraction will cause a variable acceleration. Or am I getting it wrong?

Edited March 25, 2017 by Sriman Dutta

[KipIngram]

The big question that someone asked above and I didn't notice an answer for is whether or not M1 and M2 are fixed. If you have to consider their motion as well this gets a lot more complicated. If they're fixed then 1) if M1=M2, your test body should stay on the centerline at all times - it will accelerate until it gets to the vertical centerline through M1 and M2 and then decelerate. If it's velocity is high enough it will go on off to infinity, slowing down the whole time. If it's not high enough, it will oscillate back and forth along the horizontal centerline.

 

For non-equal M1 and M2, the test object path will curve toward the heavier one. If it's velocity is high enough it will follow a curved path to infinity. The really interesting case is for unequal M1 and M2 without escape velocity - your test object will enter some kind of orbit, and I suppose it's possible that orbit might hit M1 or M2 (more likely the lower mass one). But you'd have to calculate that.

[zztop]

The big question that someone asked above and I didn't notice an answer for is whether or not M1 and M2 are fixed. If you have to consider their motion as well this gets a lot more complicated. If they're fixed then 1) if M1=M2, your test body should stay on the centerline at all times - it will accelerate until it gets to the vertical centerline through M1 and M2 and then decelerate. If it's velocity is high enough it will go on off to infinity, slowing down the whole time. If it's not high enough, it will oscillate back and forth along the horizontal centerline.

 

For non-equal M1 and M2, the test object path will curve toward the heavier one. If it's velocity is high enough it will follow a curved path to infinity. The really interesting case is for unequal M1 and M2 without escape velocity - your test object will enter some kind of orbit, and I suppose it's possible that orbit might hit M1 or M2 (more likely the lower mass one). But you'd have to calculate that.

It is more complicated than that. The problem has a rigorous mathematical formalism, it is known as the three-body problem

[KipIngram]

Yes, if all the masses are free to move. The three body problem I'm most familiar with is a satellite moving in the Earth / Moon system, where you can neglect the mass of the satellite. I don't think the fully general three-body problem has a closed form solution, does it?

[zztop]

Yes, if all the masses are free to move. The three body problem I'm most familiar with is a satellite moving in the Earth / Moon system, where you can neglect the mass of the satellite. I don't think the fully general three-body problem has a closed form solution, does it?

Yes, it has a closed solution.

[KipIngram]

Hmmm. that link someone posted above says this:

 

In 1887, mathematicians Heinrich Bruns^[4] and Henri Poincaré showed that there is no general analytical solution for the three-body problem given by algebraic expressions and integrals.

 

That's what I meant by "no closed form solution." Seems that a number of specific solutions, for specific cases have been found.

[zztop]

Hmmm. that link someone posted above says this:

 

In 1887, mathematicians Heinrich Bruns^[4] and Henri Poincaré showed that there is no general analytical solution for the three-body problem given by algebraic expressions and integrals.

 

That's what I meant by "no closed form solution." Seems that a number of specific solutions, for specific cases have been found.

You should kept on reading.

[KipIngram]

Ok, fair enough. It doesn't look terribly useful for practical purposes, though. But you're right - there it is.

[Sriman Dutta]

Now, I have another question.

Suppose there are two bodies of mass [math]m_1[/math] and [math]m_2[/math] respectively, separated by a finite distance of [math]S[/math]. They will attract each other according to Newton's law of universal gravitation as -

[math]F=G\frac{m_1m_2}{S^2}[/math]

Since their distance will change at every moment, each of them will experience a variable acceleration. Considering the case of [math]m_1[/math],

[math]a(t)=G\frac{ m_2}{S(t)^2}[/math]

And [math]j=\frac{da(t)}{dt}[/math] , where [math]j[/math] is jerk.

 

Am I on the right track???

Edited March 26, 2017 by Sriman Dutta

[zztop]

Now, I have another question.

Suppose there are two bodies of mass [math]m_1[/math] and [math]m_2[/math] respectively, separated by a finite distance of [math]S[/math]. They will attract each other according to Newton's law of universal gravitation as -

[math]F=G\frac{m_1m_2}{S^2}[/math]

Since their distance will change at every moment, each of them will experience a variable acceleration. Considering the case of [math]m_1[/math],

[math]a(t)=G\frac{ m_2}{S(t)^2}[/math]

And [math]j=\frac{da(t)}{dt}[/math] , where [math]j[/math] is jerk.

 

Am I on the right track???

yes

[Sriman Dutta]

Thanks for clarifying that zztop.

[Sriman Dutta]

Now I do realize that the problem would never be practical because it is scientifically wrong. Long before the small body of mass m arrived, the two bigger masses would have attracted each other and bumped together.

If we consider the theoretical assumption that their field ceases to exist beyond a distance of a, and consider that r=a ( see figure), then also the moving body wouldn't have felt any force. This is on account of the fact that O, the vertical components of gravitational force sums to zero and their horizontal components are zero. So there's no acceleration and the body can continue in its state of uniform motion.