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Gravitational Attractions of Small Bodies. Calculating the gravitational attraction of an arbitrary body Given an elementary body with mass m i at position.

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Presentation on theme: "Gravitational Attractions of Small Bodies. Calculating the gravitational attraction of an arbitrary body Given an elementary body with mass m i at position."— Presentation transcript:

1 Gravitational Attractions of Small Bodies

2 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.

3 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:

4 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:

5 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.

6 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

7 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

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

9 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

10 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!

11 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

12 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:

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

14 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 ).

15 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.

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

17 The Semi-Infinite Sheet of Thickness t As above

18 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.

19 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!

20 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)

21 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.

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

23 Identical Response from different Mass Distributions

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