Something I don't understand about Lagrange points

Hello everybody. There's something about the Lagrange points I don't
understand. Let's consider a body in L4 or L5 in the orbit of Jupiter
(for example, a Trojan or a Greek asteroid). The forces that affect
that body a

1) gravitacional attraction to the Sun
2) centripetal force
3) gravitational attraction to Jupiter

Forces 1 and 2 are the same strength and opposite direction, and are
balanced. But, as far as I can see, force 3 is not balanced by any
other force. This should make the asteroid start moving towards Jupiter
and finally crash into the planet, or maybe become a satellite of
Jupiter.

There must be a mistake in my reasoning (as Trojan and Greek asteroids
are a fact), but I can't see it. Can somebody help?

Thank you.

Vicent Tur

The center of mass isn't in the center of the sun, it's shifted
somewhat towards Jupiter. The Trojan is orbiting the center of mass.
The Trojan a little further from that point than Jupiter is, because
the Trojan and Jupiter are equidistant from the sun. Don't know if
that helps any.

In article .com,
writes:
Hello everybody. There's something about the Lagrange points I don't
understand. Let's consider a body in L4 or L5 in the orbit of Jupiter
(for example, a Trojan or a Greek asteroid). The forces that affect
that body a

1) gravitacional attraction to the Sun
2) centripetal force
3) gravitational attraction to Jupiter

Forces 1 and 2 are the same strength and opposite direction, and are
balanced.

No. Force 2 is the vector sum of forces 1 and 3. The interesting
property of all five Lagrange points is that (for circular orbits)
the centripetal force is exactly perpendicular to the orbit velocity.
This means there is an equilibrium orbit at all five Lagrange points
(and only at those points). _Stability_ of these equilibria is more
complicated.

--
Steve Willner Phone 617-495-7123
Cambridge, MA 02138 USA
(Please email your reply if you want to be sure I see it; include a
valid Reply-To address to receive an acknowledgement. Commercial
email may be sent to your ISP.)

"Steve Willner" wrote in message
...
In article .com,
writes:
Hello everybody. There's something about the Lagrange points I don't
understand. Let's consider a body in L4 or L5 in the orbit of Jupiter
(for example, a Trojan or a Greek asteroid). The forces that affect
that body a

1) gravitacional attraction to the Sun
2) centripetal force
3) gravitational attraction to Jupiter

Forces 1 and 2 are the same strength and opposite direction, and are
balanced.

No. Force 2 is the vector sum of forces 1 and 3. The interesting
property of all five Lagrange points is that (for circular orbits)
the centripetal force is exactly perpendicular to the orbit velocity.
This means there is an equilibrium orbit at all five Lagrange points
(and only at those points). _Stability_ of these equilibria is more
complicated.

--

The original post worried me, because I studied Lagrange points as part of
perturbation theory 30 years ago, and thought I understood why they worked.
The OP has asked a very simple question, and I still don't know the answer.

Lets assume Jupiter and the asteroids are in a circular orbit, and the mass
of the asteroid mass of Jupiter mass of Sun (as is obviously the
case).

Then the gravitational force of the Sun and the "centripedal" force on the
asteroid point in oppsoite directions - both vectors lie on the radial line
which joins the asteroid and the Sun, but point in opposite directions. The
gravitational pull of Jupiter on the asteroid is a vector which clearly
doesn't lie entirely on that radial line. It must therefore have a component
which is perpendicular to that line, which means the asteroid must have some
nett acceleration in the direction of its travel (tangentially) - presumably
falling towards Jupiter. Yet this doesn't happen. Why not?

Here's a variation of that same argument. Lets imagine an asteroid at a
Lagrange point. Poof, Jupiter dissappears. The asteroid is till in the same
stable orbit. Poof, Jupiter re-appears. Asteroid still in same stable orbit.
How come this bloody great hulking gas giant has no apparent gravitational
effect on the asteroid - the orbit is the same whether its there or not?

Peter Webb skrev:

Here's a variation of that same argument. Lets imagine an asteroid at a
Lagrange point. Poof, Jupiter dissappears. The asteroid is till in the
same stable orbit. Poof, Jupiter re-appears. Asteroid still in same stable
orbit. How come this bloody great hulking gas giant has no apparent
gravitational effect on the asteroid - the orbit is the same whether its
there or not?


The asteroid, Jupiter and Sun form an equilateral triangle and are in a
circular orbit around the center of mass of Jupiter and the Sun. Jupiter
disappears and the center of mass the Sun. The velocity vector of the
asteroid remains the same, but the radius vector has changed both length
and direction. In other words the velocity vector and the radius vector are
no longer perpendicular and the asteroid is on an elliptical orbit instead
of a circular orbit.

--
Oeystein Olsen, oystein.olsen_at_astro.uio.no, http://folk.uio.no/oeysteio
Institute of Theoretical Astrophysics, http://www.astro.uio.no
University of Oslo, Norway

Oystein Olsen wrote:
Peter Webb skrev:

Here's a variation of that same argument. Lets imagine an asteroid at a
Lagrange point. Poof, Jupiter dissappears. The asteroid is till in the
same stable orbit. Poof, Jupiter re-appears. Asteroid still in same stable
orbit. How come this bloody great hulking gas giant has no apparent
gravitational effect on the asteroid - the orbit is the same whether its
there or not?


The asteroid, Jupiter and Sun form an equilateral triangle and are in a
circular orbit around the center of mass of Jupiter and the Sun. Jupiter
disappears and the center of mass the Sun. The velocity vector of the
asteroid remains the same, but the radius vector has changed both length
and direction. In other words the velocity vector and the radius vector are
no longer perpendicular and the asteroid is on an elliptical orbit instead
of a circular orbit.

What you are saying above sounds correct, but I find it rather hard to
follow (and, if you don't mind me saying so, I'm afraid Vicent may also find
it rather hard to follow).
So here's my explanation. First of all, let's make the simplifying
assumption that Jupiter's orbit is exactly circular. It's pretty close to
circular, so that's a good approximation. Also, of course, let's ignore the
effect of the other planets, and let's ignore the gravitational force exerted
by the asteroid---it can feel gravity, but its gravitational force on
other objects is negligible.
Next, let's look at everything in a reference frame that co-rotates with
Jupiter. So it's as if you are somewhere in space looking down on the Solar
System, and you slowly twirl your camera so that the camera's rotation matches
Jupiter's. From your viewpoint, Jupiter itself will then remain stationary.
The picture will be something like the below.


S

CM


A



J



The symbol S above is the Sun, and J is Jupiter. CM is the center of mass
of the Sun-Jupiter System. We take CM as the center of our frame of reference.
Both the Sun and Jupiter will revolve around the CM---but since our frame of
reference co-rotates, both the Sun and Jupiter will appear stationary.
Since the Sun is about 1000 times more massive than Jupiter, the Sun is
actually 1000 times closer to the CM than Jupiter is. So the relative distances
as shown above are greatly exaggerated. But the relatively small distance
between the Sun and the CM turns out to be significant.
Now consider the asteroid A at a stable Lagrange point. It will experience
gravitational forces from the Sun and from Jupiter. It will also experience
centrifugal force (really a pseudo-force that can be treated as if it were
a genuine force when working in a rotating reference frame).
Naturally enough, the gravitational force from the Sun will point straight
at the Sun; and the gravitational force from Jupiter will point straight towards
Jupiter. But the centrifugal force will point directly away from the CM. So
the gravity from the Sun will NOT be exactly cancelled out by the centrifugal
force. In order for that cancellation to occur, the centrifugal force would
have to be directed straight out from the Sun---and it isn't.
Those two forces will come reasonably close to cancelling each other
out, however. A relatively small "residue" of force will remain. And what
cancels out that residual force? The gravity of Jupiter! Since Jupiter is so
much smaller than the Sun, the force from Jupiter will indeed be relatively
small, but it will (for an asteroid in just the right position) be the right
force to cancel out the "residual" force.
So the net force on the asteroid is zero, and it remains stationary. If
it is stationary in our rotating reference frame, that really means that it
is revolving around the CM with the same orbital period as Jupiter. So this
is exactly the well-known behavior of Trojan asteroids.



Kevin

OK, I get it.

I made the explicit assumption that the mass of the Sun is mass of
Jupiter, which if the Sun is only a thousand times bigger than Jupiter is
not true (as it is, as I now know).That obviously places the CM way outside
the Sun itself, not that it matters.

Interesting that the orbits are stable at pi/3 irrespective of the mass of
the body - Jupiter, in this case. The movement of the CM caused by earth is
far less - certainly inside the Sun itself, and this makes the stable
Lagrangian zone far smaller.

Thanks




"Kevin" wrote in message
...
Oystein Olsen wrote:
Peter Webb skrev:

Here's a variation of that same argument. Lets imagine an asteroid at a
Lagrange point. Poof, Jupiter dissappears. The asteroid is till in the
same stable orbit. Poof, Jupiter re-appears. Asteroid still in same
stable
orbit. How come this bloody great hulking gas giant has no apparent
gravitational effect on the asteroid - the orbit is the same whether its
there or not?


The asteroid, Jupiter and Sun form an equilateral triangle and are in a
circular orbit around the center of mass of Jupiter and the Sun. Jupiter
disappears and the center of mass the Sun. The velocity vector of the
asteroid remains the same, but the radius vector has changed both length
and direction. In other words the velocity vector and the radius vector
are
no longer perpendicular and the asteroid is on an elliptical orbit
instead
of a circular orbit.

What you are saying above sounds correct, but I find it rather hard to
follow (and, if you don't mind me saying so, I'm afraid Vicent may also
find
it rather hard to follow).
So here's my explanation. First of all, let's make the simplifying
assumption that Jupiter's orbit is exactly circular. It's pretty close to
circular, so that's a good approximation. Also, of course, let's ignore
the
effect of the other planets, and let's ignore the gravitational force
exerted
by the asteroid---it can feel gravity, but its gravitational force on
other objects is negligible.
Next, let's look at everything in a reference frame that co-rotates
with
Jupiter. So it's as if you are somewhere in space looking down on the
Solar
System, and you slowly twirl your camera so that the camera's rotation
matches
Jupiter's. From your viewpoint, Jupiter itself will then remain
stationary.
The picture will be something like the below.


S

CM


A



J



The symbol S above is the Sun, and J is Jupiter. CM is the center of
mass
of the Sun-Jupiter System. We take CM as the center of our frame of
reference.
Both the Sun and Jupiter will revolve around the CM---but since our frame
of
reference co-rotates, both the Sun and Jupiter will appear stationary.
Since the Sun is about 1000 times more massive than Jupiter, the Sun
is
actually 1000 times closer to the CM than Jupiter is. So the relative
distances
as shown above are greatly exaggerated. But the relatively small distance
between the Sun and the CM turns out to be significant.
Now consider the asteroid A at a stable Lagrange point. It will
experience
gravitational forces from the Sun and from Jupiter. It will also
experience
centrifugal force (really a pseudo-force that can be treated as if it were
a genuine force when working in a rotating reference frame).
Naturally enough, the gravitational force from the Sun will point
straight
at the Sun; and the gravitational force from Jupiter will point straight
towards
Jupiter. But the centrifugal force will point directly away from the CM.
So
the gravity from the Sun will NOT be exactly cancelled out by the
centrifugal
force. In order for that cancellation to occur, the centrifugal force
would
have to be directed straight out from the Sun---and it isn't.
Those two forces will come reasonably close to cancelling each other
out, however. A relatively small "residue" of force will remain. And
what
cancels out that residual force? The gravity of Jupiter! Since Jupiter
is so
much smaller than the Sun, the force from Jupiter will indeed be
relatively
small, but it will (for an asteroid in just the right position) be the
right
force to cancel out the "residual" force.
So the net force on the asteroid is zero, and it remains stationary.
If
it is stationary in our rotating reference frame, that really means that
it
is revolving around the CM with the same orbital period as Jupiter. So
this
is exactly the well-known behavior of Trojan asteroids.



Kevin

In article ,
Kevin writes:
So here's my explanation.
....
Next, let's look at everything in a reference frame that co-rotates with
Jupiter.

The snipped analysis looks correct to me. Some of the other posters
might suffered from confusion over terminology. In case so:

"Centripetal force" is a real force on an orbiting body, specifically
the vector component directed towards the orbit center. In this
case, it comes from the Sun and Jupiter. If working in an inertial
coordinate system, F=ma, and F is the centripetal force plus any
tangential force.

"Centrifugal force" is a fictitious force that one introduces if
working in a rotating coordinate system. It is related to the
rotation velocity of the coordinate system, not to any real object.

Just for completeness, "coriolis force" is another fictitious force
that arises in rotating coordinate systems. It isn't important for
the original question in this thread.

One can perform calculations in any coordinate system that's
convenient, but you have to be consistent about which forces you
include.

--
Steve Willner Phone 617-495-7123
Cambridge, MA 02138 USA
(Please email your reply if you want to be sure I see it; include a
valid Reply-To address to receive an acknowledgement. Commercial
email may be sent to your ISP.)

Peter Webb (or somebody else of the same name) wrote thusly in message
:

OK, I get it.

I made the explicit assumption that the mass of the Sun is mass of
Jupiter, which if the Sun is only a thousand times bigger than Jupiter is
not true (as it is, as I now know).That obviously places the CM way
outside the Sun itself, not that it matters.

Interesting that the orbits are stable at pi/3 irrespective of the mass of
the body - Jupiter, in this case. The movement of the CM caused by earth
is far less - certainly inside the Sun itself, and this makes the stable
Lagrangian zone far smaller.

Are there asteroids or any other natural bodies at Earth's Trojan positions?
(Have any spacecraft actually been there to look?). What about the Trojan
positions of the Earth-Moon system?
--
A couple of questions. How do I stop the wires short-circuiting, and what's
this nylon washer for?

Interchange the alphabetic letter groups to reply

Prai Jei wrote:
Peter Webb (or somebody else of the same name) wrote thusly in message
:


OK, I get it.

I made the explicit assumption that the mass of the Sun is mass of
Jupiter, which if the Sun is only a thousand times bigger than Jupiter is
not true (as it is, as I now know).That obviously places the CM way
outside the Sun itself, not that it matters.

Interesting that the orbits are stable at pi/3 irrespective of the mass of
the body - Jupiter, in this case. The movement of the CM caused by earth
is far less - certainly inside the Sun itself, and this makes the stable
Lagrangian zone far smaller.


Are there asteroids or any other natural bodies at Earth's Trojan positions?
(Have any spacecraft actually been there to look?). What about the Trojan
positions of the Earth-Moon system?

Searches for Earth Trojans have never found anything larger than a few
hundred meters.
There are reports of clouds of dust at the Earth-Moon Lagrangian points
but the only spacecraft that have investigated them (the Japanese Hiten
probe, equipped with dust detector) failed to find anything