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Gravity at a spot between the Earth and the Sun Summing to 0.

  1. May 5, 2009 #1
    I had a friend who told me that there is a distance between the Earth and the Sun which the components of gravity add to zero. Is this true?

    If so, does this mean that something could get stuck there in that spot?
     
  2. jcsd
  3. May 5, 2009 #2

    Born2bwire

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    They are called Lagrangian Points.

    http://en.wikipedia.org/wiki/Lagrangian_point

    Yes, things can get stuck there though I imagine that perturbations from other gravity sources can, after long time periods, shift them a little. A number of satelites, sent up and pending, have been sent to Langrangian points.
     
  4. May 5, 2009 #3
    Hi there,

    From the pratical point of view, this is correct. From the physical point of view, your friend should have formulated the comment differently.

    "There is a point between the Sun and the Earth (or between the Earth and the Moon) where their gravitaional pull compensate each other: [tex]\sum_{\text{body}} \vec{F_g} = 0[/tex]"

    I'm sorry for the little extra comment. I don't know I pinpoint on details like that.

    Cheers
     
  5. May 5, 2009 #4

    D H

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    The components of gravity do not sum to zero at the Lagrange points (aka Lagrangian points, aka libration points).

    The easiest way to envision the Lagrange points is in the reference frame rotating with the planet's orbit. Ignoring orbital eccentricity and perturbations by other planets, the planet's position is constant in such a rotating frame. Fictitious forces are needed to make Newton's laws of motion appear to be valid in non-inertial frames. For rotating frames one needs to invoke the fictitious centrifugal and coriolis forces. In this rotating frame, the Lagrange points are points at which the net gravitational force plus the fictitious centrifugal force sums to zero. The only remaining force is the coriolis force, and this of course depends on the (rotating frame) velocity. A small test object placed exactly at one of these Lagrange poiints with zero velocity (rotating frame) will have a net force (rotating frame) of zero: The object will be stationary.

    So, what do these points look like in an inertial frame? The net true force (gravity only) at the Lagrange points is non-zero. It is instead exactly equal to the centripetal force needed to make an object placed at such a point orbit the central mass (the Sun, in this case) with exactly the same period as the planet. In short, gravity does not cancel at the Lagrange points.

    So, is there a point at which gravitational forces do truly cancel instantaneously? The answer is of course yes. This point (and there is only one, as opposed to five Lagrange points) is located between the two objects. This point, unlike the Lagrange points, has very little (if any) physical significance.
     
  6. May 5, 2009 #5

    Nabeshin

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    It is also worth noting that the equilibrium point between the earth-moon, sun-earth, or any two body system is an unstable equilibrium. If there is even a small perturbation the object will fall towards either of the gravitating bodies, so I wouldn't expect to find anything there.
     
  7. May 5, 2009 #6

    Mapes

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    But also note that L4 and L5, which lie somewhat away from the axis between the two bodies, are stable points for small perturbations.

    The existence of stable points in a two-body gravitational system is probably the most amazing and surprising thing I learned in undergraduate engineering dynamics.
     
  8. May 5, 2009 #7
    since the sun and earth are in constant motion and their relative distance changes minutely by the minute, the Lagrange points should be consistently shifting, shouldn't they?
     
  9. May 5, 2009 #8
    Thanks for the information. I think I understand what Lagrange points are now. :). So from the sound of things, the object would rotate around the sun at the same rate as the earth? And if it is exactly the same rate(correct me if I am wrong), does that mean they rotate instantaneously in sync? Or would they be out of sync by a small amount?
     
  10. May 6, 2009 #9

    D H

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    That's correct. Suppose a planet is exactly in a circular orbit around a star and that this star has no other planets. Suppose a (small) object is placed exactly at one of the libration points and is moving exactly in sync with the motion of the libration point. The combined gravitational forces on the object coupled with the motion of the object will keep the object exactly in sync with the libration point, forever. The key word here is "exactly". What happens if the object is not exactly at the libration point and moving with the libration point? My response below to Nabeshin addresses this question.

    Because the planet is following a curved path, the libration points are also following curved paths. In order for an object to follow a curved path, there must be some force acting on the object. It is precisely because gravitational forces do not cancel at the L1 point that an object exactly at the L1 point and moving exactly with the L1 point will follow the L1 point exactly.

    What about the point where the gravitational forces truly do cancel one another? That point, like the L1 point, follows a curved path. An object placed at this cancelation point and instantaneously moving with the cancelation point would just move in a straight line (initially). Think of it in terms of Newton's first law. Meanwhile this null gravity point will curve off away from the object. The point where the gravitational forces cancel is rather uninteresting and is not very useful.

    The collinear libration points (L1, L2, and L3) are indeed metastable (metastable = unstable equilibrium, such as a ball balanced on top of another ball). What metastable means is that any deviation from the equilibrium state will grow and grow. Compare to a stable equilibrium point, where sufficiently small deviations from the equilibrium state become smaller or remain bounded.

    Objects can be maintained at or near a metastable point with the application of external force. Example: An inverted pendulum. Put a broom with the business end up and the other end in the palm of your hand. You can keep the broom balanced by moving your hand about appropriately.

    An object could similarly be maintained at one of the collinear libration points, but doing so would require a lot of energy. However, there exist pseudo-orbits about these points that, while still unstable, require significantly less expenditure of energy to maintain the orbit than would be needed to stay at the point.

    The world's space agencies take advantage of these orbits because the Earth-Sun L1 and L2 points are very, very useful as locations for spacecraft. Three satellites are presently operating in pseudo-orbits about the Sun-Earth L1 point: SOHO, ACE, and Wind; WMAP operates about the Sun-Earth L2 point.
     
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