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Phoebe Floats !

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Phoebe Floats!

Presented by Bernard Moore

Author: Ezra Brown

Virginia Polytechnic University

The College Mathematics Journal Vol.36 No.2 (March 2005)

- Ezra’s inspiration for this problem came from his students at Virginia Tech and was named “Phoebe Floats!” for his granddaughter Phoebe Rose.
- Unlike most articles, Ezra incorporates a storyline in the article. Ezra plays the role of the teacher and his students are the intrigued students that continue to surprise Ezra with their ideas.

- Suppose we have an object that is less dense than water, it wont sink fully, but how can we determine how far the object will submerge in water?
- What are the characteristics of Newton’s Method under iteration and why is it applicable to the problem of floating bodies?
- What are attracting and repelling points…and what do they have to do with floating bodies?

- Does Phoebe Float? Phoebe is a moon of Saturn, it is less dense than water, so if we dropped it in the ocean it would float.
- Phoebe’s diameter is 220 km and its mass is 4 x 1018 kilograms, compared to earth which has a diameter of about 12,000 km and mass of 6 x 1024 kilograms.
- What is the density of Phoebe?

- Phoebe’s density is about 7/10 kilograms per cubic meter. Compared to water which has a density of 1000 kilograms per cubic meter.
- To standardize the problem, we work with a ball of radius 1 and density of 7/10.
- Archimedes’ Law of Floating Bodies states that a body immersed in a fluid displaces an amount of fluid equal in weight to the weight of the body, provided the body is less dense than the fluid.

- We place the center of the circle of radius 1 at the origin. We will denote its distance underwater by r. In the case that the density is ½, half the ball is submerged and the depth r=1. Since Phoebe’s density is greater than ½, it’s clear that more than half of Phoebe is submerged, and so we expect its distance underwater to be between 1 and 2.
- Our mission is to find the value r so that the segment of the ball between y=-1 and -1+r has mass of (4π/3)(7/10) – since that is how much is displaced.

- We now set up an integral to find the volume of the segment
- Since the equation of the circle is given by x2+y2 =1. A segment of thickness Δy has radius (1-y2)1/2.
- Then the volume V can be approximated by Σπ(1-y2)Δy.
- So we want to solve for r, where

- So we have
- This gives us

- Thus r satisfies the equation
Do we have a method to solve for r in this cubic polynomial?

Yes. The Cardano-TartagliaFormula.

- For a cubic polynomial of the form ax3 +bx2+cx+d=0, the formula states that the three solutions x1, x2, and x3 can be found provided b2 -3ac ≠0.
- Nature of Solutions: We define
If Δ>0 the equation has 3 distinct real roots, if Δ=0 the equation has real but multiple roots, and if Δ<0 then we have one real and two complex conjugate pairs.

- So we can easily see that Δ>0 we expect 3 real roots.
- We can now use the formulas to find the three roots… get ready…

- From seeing the graph below, we can expect that there is a root in the intervals (-1,0), (1,2), and (2,3).

- So we compute x1, x2, and x3 as -0.852523, 1.273485, 2.579038. These answers makes sense, we have one for each of the intervals that we listed.
- So what does this mean? Remember that r gives us the depth of the unit ball and only one of the solutions makes sense here and that is the solution x2=1.273458. Why?

- But this r value was with a ball of radius 1, we want to know how deep Phoebe sinks.
- Saturn’s satellite Phoebe has a radius of 110km, so how far would it sink? 140.083km.

- How else can we find the roots of the equation
- Newton’s idea was to start with some close approximation to a root of a function and then define a sequence that, under favorable conditions, converges to the actual root.

- Start with some approximation x = x1 as a root of the function, then the favorable condition is having
be closer to the actual root than is x1.

- We can see that the distance between x1 and x2 is given by .
- Using a recurrence formula Newton saw that he could get a closer and closer approximation to the actual root.

- Newton defined the sequence of numbers {xn} by setting
- So we now will use Newton’s method to approximate the roots of the equation

- We begin by finding
- Then we define our “Newton function” by

- We start with some approximation for each root name these x0. Then we define N(x0) as x1, N(x1) as x2, and so on. Our plan is to keep iterating. We will pick our starting numbers for x0 so that they will be close to each root, how about -3, 1, and 5.

- We iterate using the recurrence relation to obtain the following

- The chart showed us that at we keep iterating the function N(x) we actually did get closer and closer to the actual roots that we found earlier using the Cardano-Tortaglia Formula.
- If all goes well the iterates converge to a root. It doesn’t always go well. For f(x) = x3 -3x2 +14/5 it appears N(x) behaves itself and everything works fine.

- Newton’s Method is very sensitive to it’s initial approximation values for xo.
- Instead, take the starting values 1.86693, 1.86695, and 1.86697. What will happen if we iterate with these as our starting values instead of -3,1, and 5?

- See iterate and we have the following table of iterates, again approaching the actual roots

- There is sensitive dependence on the initial choice of x0 that causes this odd behavior.
- There are answers to why this is…
- We look to dynamical systems theory.

- The following theorem explains the odd behavior that we seen with Newton’s Method. We see that the points x = 1.86693 and .32488 are both repelling points which explains their escaping behavior that we seen in the chart of iterates. We may replace g by N below, where the fixed points of N are the roots of f.

- Remember that the starting values for our first table of iterates were -3, 1, and 5 which aren’t near the repelling points at .325 and 1.87. When choosing starting points we will avoid repelling points for better approximations.

Did you know Saturn floats too?!