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Gravitational Attractions of Small Bodies

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Calculating the gravitational attraction of an arbitrary body Given an elementary body with mass m i at position r i, the associated acceleration g i at any point r is Forces add linearly, so the total acceleration due to a group of n bodies is Where m i = (r i )v i.

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For a continuous mass, let v i go to zero and the sum becomes an integral: Thus, given the density distribution (r’), you can calculate the gravitational acceleration. Two practical comments: 1.Usually we try to take advantage of any symmetry, to make the vector addition easier. 2.Potential is often easier to work with because it is a scalar. Specifically:

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In practice, we measure |g| as opposed to the vector quantity itself. Suppose we are measuring the gravitational anomaly caused by a small body within the Earth. Let g e be the acceleration of the Earth without the body, and g be the perturbation due to the anomalous mass ( g << g e ). Then, the total field is found from Expanding (as (1+x) 1/2 ~ 1 + x/2) we get:

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Or, since gcos is the z component of g: The residual R is therefore Therefore, in calculating the anomaly due to a body, we only worry about the z component. We will now demonstrate how to calculate the gravity anomaly due to bodies with simple shapes.

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Calcuation of g for some simple bodies 1.Sphere of constant density . An example of using the potential U. Recall that In this example, (r’) = constant = o. For a sphere, dv(r’) = r’ 2 sin d d dr’ Using the law of cosines, |r - r’| = [r 2 + r’ 2 - 2rr’cos ] 1/2. Thus The last integral is just 2 . Note that

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Substitute x = cos and dx = -sin d in the above and we get Now consider the case where r > r’ (outside the sphere): Which is the same as the potential for a point mass. Thus

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Now consider the case where r < r’ (inside the sphere): Thus Meaning that only the mass within the radius r contributes to the acceleration.

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A Buried Vertical Cylinder. An Example of Using Symmetry. Let’s try computing the anomaly due to a vertical cylinder within the earth at a point directly above the center of the cylinder. Consider a cylinder of radius a buried in the earth at a depth h 1. The cylinder extends to depth h 2. It has a density 2, while the surrounding earth has a density 1 (so = 2 - 1 ). Horizontal (sin ) attractions cancel because of symmetry. The remaining vertical (cos ) gravitational anomaly caused by a volume element is For a cylinder, dv = rdrdzd . Also, cos = z/(r 2 + z 2 ) 1/2. Therefore

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If we let a become very large, the the equation above reduces to This will turn out to be a very useful equation later on, so remember it!

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Anomalies from Two-Dimensional Structures Very often it is useful to treat structures in the Earth as two- dimensional. It turns out there are clever ways to analyze gravity anomalies from such structures. Here is how you can do it: For a 2-D structure, r = r(x’, z’) (i.e., no y dependence). In what follows, we use primed coordinates (x’, y’, z’) for the anomalous mass, and unprimed coordinates (x, y, z) for the observing station. All measurements are made along the y = 0 axis. The distance R in the x-z plane is R 2 = (x-x’) 2 + (z-z’) 2 And the total distance r is r 2 = R 2 + y’ 2

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The gravitational potential is The integral over y’ is now finite. Later we will let it get big. For now Here comes cute trick #1. Because we are really interested in g and not U, and because g = U, we can add anything we want to U that is not a function of (x,y,z) without affecting g. So, let’s define a constant:

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Then If we let L >> R, then. The argument in the logarithm becomes: So And

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The integral for g is often easy to integrate. In many applications we can assume (x’,z’) = constant and we set z = 0 at the surface. Now comes cute trick #2: let’s use cylindrical coordinates instead of Cartesian. Then Substituting these expressions into the formula for g z gives Noting that we have For a given value of , the limits of integration are from R( ) near to the observer to R( ) from from the observer. We perform this kind of integration from at the top of the body ( top ) to at the bottom ( bottom ).

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Noting that sin dR = dz, we have These two integrals are line integrals that together encompass the body (the minus sign on the second integral reverses the direction. Therefore Which is a line integral around the body. This is a great way to compute anomalies for complicated structures. Note that the line integration must always be done in a clockwise sense! Let’s use this method to calculate the attractions of some simple bodies.

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The Infinite Sheet (Revisited) In this case Which is what we obtained before (but with much more work!)

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The Semi-Infinite Sheet of Thickness t As above

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Now So Note that as x -> ∞ -> g -> 2 G t (i.e., infinite slab) x -> 0 -> g -> G t x -> -∞ -> g -> Also Thus, the slope lessens as the depth of the sheet increases. In fact, the slope can give us some information about the depth of the sheet.

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Hitchhiker’s Guide to Gravity First, since many situations use the infinite slab formula, it’s useful to remember that if the density difference is given in gr/cm 3 and the thickness of the slab h is given in kilometers, then g in mgals is g = 2 G h = 2 x x 6.67 h = 41.9 h ~ 42 h So, if you can remember the answer to Life, the Universe, and Everything, you can remember to how to calculate the infinite slab attraction!

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Practical example #1: A basin filled with sediments Let’s suppose we are standing on top of a basin that is 2 km thick and filled the sediments with a density that is 0.5 gr/cm 3 less than the basement below, we would expect that in the middle of the basin: g mid = 42 x 2 x 0.5 = 42 mgals At the edge of the basin (x=0), g x=0 = g mid /2 = 21 mgals and dg/dx = (2G h/z o )=13.34 mgals/km (note z o = 1 km)

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Practical example #2: The continental margin In this case, we can think of a margin as two semi-infinite slabs on either side of an infinite slab that has a thickness equal to the ocean side of the margin. The upper semi-∞ slab make a contribution, while the lower semi-∞ slab make a contribution. BUT, the contributions don’t cancel because they are at different depths.

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Example of Equivalence and Non-uniqueness Convince yourself that the 3 geometries shown to the right give identical relative gravity anomalies:

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Identical Response from different Mass Distributions

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