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Hill Sphere Question

  1. Nov 15, 2007 #1
    I'm trying to understand the Hill Sphere formula. (Yes, I've already been to Wikipedia.)

    I notice that it doesn't take the mass of the satellite into consideration. Is that because the mass of the satellite has no effect, or is it a simplification based on the assumption that the mass of the satellite is very small with respect to the masses of the other two bodies? How would it apply to twin planets?

    I found this comment which seems to indicate satellite mass does matter. If so, how should the formula be altered to include the effect of a large satellite?

    Also, what about the relative masses of the other two bodies? Is m assumed to be a very small fraction of M? If, for example, you were looking at a planet orbiting one star of a binary system, would the formula need to be altered in any way?

    (I'm looking for a way to calculate reasonable outer limits for stable orbits in a variety of hypothetical situations. I don't need a high degree of accuracy. For example, although it would be nice to include eccentricity, the cases I'm interested in are sufficiently near to circular for that to be a reasonable approximation.)

    Thank you for your assistance.
     
  2. jcsd
  3. Nov 15, 2007 #2

    tony873004

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    The Hill Sphere formula is an approximation for the distance to the L1 and L2 points. It's close, but not perfect.

    Imagine an object in Earth's L1 point. The Sun strongly tugs it one way and the Earth weakly tugs it the other way. When you sum up these tugs, you get a tug towards the Sun that is a little weaker than the tug from the Sun alone. This is really no different than if Earth didn't exist, and the L1 object simply orbited a slightly less-massive star. If it did orbit a slightly less massive star, its circular orbital velocity would be a little less, which would give it a longer period. The L1 point is simply the point where this longer period matches the period of the Earth. If you do all the algebra, it reduces to the Hill Sphere formula.

    In the case where the satellite in the L1 point had appreciable mass compared to the Earth, it would pull the Earth in the same direction that the Sun pulls the Earth. When you add these pulls together, you get a pull on the Earth that is slightly stronger than the Sun alone. It is as if the Earth orbited a more massive star, which would increase its orbital velocity and its period.

    So when the period of the inner object, being tugged by the Sun in one direction and the Earth in the opposite direction, and the period of the Earth, being pulled towards the Sun by both the Sun and the satellite, are equal, you'd have your new Hill Sphere. I imagine the algebra is ugly to prove this. It can be computed with numerical methods, just like the actual distance to the L1 point for a non-massive body can be computed more accurately with numerical methods than with the Hill Sphere approximation.
     
  4. Nov 15, 2007 #3

    pervect

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    The hill sphere formula was derived for a satellite of negligible mass, i.e. for the Sun-Earth-moon system, only the sun/earth mass ratio is relevant, it is assumed that the moon's mass is "small". Calculating the effects of a heavier "moon" would be hard, you'd probably be best off doing simulations if you were really interested in the effects of satellite mass. It probably wouldn't be worth the effort if you don't demand high accuracy.

    The hill sphere formula does NOT directly give the radius of a stable circular orbit. Mathematically, it basically says that the satellite does not have sufficient energy to reach L1 or L2, which is what it needs to escape. The Hill sphere is not actually spherical, it's a zero velocity surface in a coordinate system that corotates with the two primary bodies: in our SunEarthMoon (SEM) example, the appropriate coordinate system would be fixed at the SEM center of mass, and it would corotate with the orbit of the Earth so that the Earth and the Sun appeared to have fixed coordinates. The so-called "radius" of the Hill sphere is the distance from, in our SEM example, the Earth to the Earth-Sun Lagrange points L1 and L2.

    Note that an object can't escape through a closed zero velocity surface, because it'd have to move in order to pass through this surface, and energy constraints say that it doesn't have enough energy to be able to move when it reaches the surface. Hence, it can't escape.

    This zero velocity is closed as long as it doesn't reach L1 or L2, that's why they are critical escape points. You might find the plots at http://www.geocities.com/syzygy303/ helpful.

    Part of the Hill sphere formula is an approximation for calculating the location of L1 and L2. This approximation requires a high mass ratio (for Sun/Earth in our SEM example). To generalize it for when the S/E mass is not large, you could calculate the location of L1 and L2 exactly yourself.

    The detailed formula of the Jacobi integral was already mentioned, i.e. http://scienceworld.wolfram.com/physics/JacobiIntegral.html for starters and I give an observation in https://www.physicsforums.com/showpost.php?p=383703&postcount=24 (which has to be double checked) that simplifies the formula - are you interested in that level of detail?

    Since you aren't interested in zero velocity surfaces, but in circular orbits, you need to make some additional assumptions.

    The very short and very rough answer to your question is that for a circular orbit, somewhere around 1/2 to 1/3 the Hill sphere radius is what you want to use.

    Note that as you approach some fraction of the Hill sphere limit you'll reach a situation where circular orbits will tend to evolve naturally into highly elliptical ones. (You can see this if you do simulations, which is how I observed it and why I don't recall the exact value of the fraction). This may put a kink into your assumption that the orbit was circular in the first place
     
    Last edited: Nov 15, 2007
  5. Nov 15, 2007 #4
    So the moon's mass does matter, but no one has already done the work of deriving it? Thanks, that answers a big part of my question.

    You can't nail down that fraction more specifically?

    Oh well, I guess I see now why I wasn't finding the answer I wanted.

    Do you happen to have a feel (from simulation) which direction the effect of a massive satellite would be? Would it increase or decrease the maximum stable radius? If it increases it, then everything is cool, the calculated maximum will just be a little conservative. If it decreases it, that's more problematic for me.

    Thanks for your help.
     
  6. Nov 15, 2007 #5

    tony873004

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    You'd have to define how round is round. The Earth's Hill Sphere is about 1.5 million kilometers. The Moon orbits the Earth about 1/4 of the way to the edge of the Hill Sphere, and its orbit is pulled slightly out of round. The further out you go, the more your orbit is pulled out of round.

    Here is a screen shot from a simulation:

    http://orbitsimulator.com/gravity/images/ehs.GIF

    The Moon's orbit is green. For simplicity, it is set to round, and it is massless so it won't perturb the other objects. 10 test particles are added at various distances from just inside the Moon's orbit out to just over 1/2 of the Hill Sphere. They were all started out in round orbits, but pertubations from the Sun cause them to be immediately pulled out of round. You can see that the deeper they sit inside the Hill Sphere, the rounder they are. As pervect points out, they stay pretty round out to about 1/2 of the Hill Sphere.
     
  7. Nov 15, 2007 #6

    tony873004

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    Here's a snipet of Visual Basic code that computes the distance to any object's L1 point. It begins by approximating it with the Hill Sphere formula, then numerically zeros in on the answer accurate to 1 meter. There's nothing tricky in this coding, and it should easily translate to any language.
    Code (Text):

    Private Sub btnComputeL1_Click()
        Dim Pi As Double, G As Double, Ms As Double, Mt As Double, a1 As Double, a2 As Double
        Dim accs As Double, acct As Double, accx As Double, acctot As Double, accnet As Double
        Dim hs As Double, v As Double
       
        Pi = Atn(1) * 4 'define Pi
        G = 6.6727004301751E-11 'define gravitational constant
        Ms = Val(txtM1.Text) ' Get mass of primary body from the GUI
        Mt = Val(txtM2.Text) ' Get mass of secondary body from the GUI
        a1 = Val(txtSMA.Text) ' Get semi-major axis of secondary body from the GUI
        p1 = 2 * Pi * Sqr(a1 ^ 3 / (G * (Ms + Mt))) 'period of Earth
        hs = a1 * (Mt / (3 * Ms)) ^ (1 / 3) 'compute the hill sphere as a starting point
        lb = hs * 0.98: ub = hs * 1.02 'as an upper bound and a lower bound, choose 98% and 102% of the hill sphere
       
        While Abs(ub - lb) > 1 'While |ub - lb| is greater than 1 gives a precision of 1 meter
            a2 = (ub + lb) / 2 'semi-major axis of the L1 particle with respect to the secondary body is the average of the upper and lower bound
            accs = G * Ms / (a1 - a2) ^ 2 'acceleration by the primary body on the L1 satellite
            acct = G * Mt / (a2) ^ 2 'acceleration by the secondary body on the L1 satellite
            acctot = accs - acct 'compute net acceleration
            accnet = acctot / accs 'multiplier to convert primary mass into simulated primary mass
            p2 = 2 * Pi * Sqr((a1 - a2) ^ 3 / (G * Ms * accnet)) ' compute period of satellite around the simulated primary mass
            If (p1 < p2) Then 'set new upper or lower bound
                lb = a2
            Else
                ub = a2
            End If
        Wend
        txtL1.Text = a2 'output the answer to the GUI
        v = 2 * Pi * (a1 - a2) / p1 'compute the orbital velocity of the L1 particle
        txtVelocity.Text = v 'output the orbital velocity to the GUI
    End Sub
     
     
  8. Nov 15, 2007 #7

    D H

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    The Hill sphere involves several simplifying assumptions, foremost among them that the satellite's mass can be ignored. The Hill sphere is itself an approximation. (The real boundary, based on Roche lobes, is not a sphere). Hill derived the equation for the Hill sphere from Hill's equations, aka the Clohessy-Wiltshire equations, or simply the C-W equations, see http://ccar.colorado.edu/asen5050/lecture12.pdf.

    Like the Jacobi integral pervect mentioned, the C-W equations are derived in a reference frame rotating with the satellite's orbit. In the case of the C-W equations, the frame origin is the center of the satellite. In the case of the Hill sphere, the origin is the location of the satellite in the unperturbed orbit.

    Note that the C-W equations themselves are an approximation: they are a linearization of the non-linear equations of motion in this rotating frame.
     
  9. Nov 18, 2007 #8

    pervect

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    If you start to consider a massive moon, you have to start considering how it distorts the shape of what it is orbiting (i.e. the Earth). There isn't any direct effect due to the mass of the moon (this is a consequence of the fact that all masses fall at the same rate, hence they orbit at the same rate, this is one way of expressing Einstein's equivalence princple). However, there are such indirect tidal effects of the mass of the moon.

    In fact, Earth tides cause "tidal locking", and such tidal locking is what is propelling our moon's orbit outwards (towards a region that is less stable). But this is really a somewhat different topic, though you might also need to know about it, I suppose. It's important to this tidal locking effect that the Earth is rotating.

    A usenet poster whom I personally trust (Brian Davis) has mentioned the figure of 1/2 to 1/3 a_hill several times. If you do a google, you probably find his posts on the topic, and his references to a U of Arizona book on satellites (which I've never seen, unfortunately, though it sounds very interesting). You might also want to look up "worldbuilding", there are several online references, including (but not limited to)

    http://curriculum.calstatela.edu/courses/builders/
    http://www.geocities.com/Area51/Chamber/2838/wbcookbook2.pdf


    The main effects are going to be due to tides, and won't be modelled by most simple simulators :-(.
     
    Last edited: Nov 18, 2007
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