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APM346-2012 => APM346 Math => Home Assignment 8 => Topic started by: Fanxun Zeng on November 28, 2012, 09:29:25 PM

Title: Problem 4
Post by: Fanxun Zeng on November 28, 2012, 09:29:25 PM
Just to note, homework 8 PDF states Newton Screening theorem is in L26 in fact it is in L27.
Title: Re: Problem 4
Post by: Fanxun Zeng on November 28, 2012, 10:06:44 PM
Anyone knows full solution of Problem 4? One of my ideas is gravitational potential approaches zero at infinity to set B=0 as:
http://www.math.toronto.edu/courses/apm346h1/20129/L27.html
Title: Re: Problem 4
Post by: Calvin Arnott on November 28, 2012, 10:13:06 PM
Anyone knows full solution of Problem 4? One of my ideas is gravitational potential approaches zero at infinity to set B=0 as:
http://www.math.toronto.edu/courses/apm346h1/20129/L27.html

I'm curious to see how everybody else approached this one as well, I didn't have time this week to write an answer I was happy with. The idea I was going to use was essentially take equations (2)/(3) in lecture 27, and take a rotationally symmetric density function $f(x) = c$ for some constant $c$. Evaluate the integral to be some other constant $k$. Then use that the $r^{2-n}$ in $n = 3$ gives us a term proportional to $r$.
Title: Re: Problem 4
Post by: Victor Ivrii on November 29, 2012, 06:27:54 AM
It is very simple:

NSP says. Let us have a spherically symmetric density. Then


So, outer layers don't pull us and the pull of the layers below is $G\times \frac{4}{3}\mu r^3 \times  r^{-2}= \frac{4G\mu}{3} r$.

In particular travelling to the center of the Earth in the framework of this model we would see decaying gravity. Also if we drill a tunnel and drop something then movement (barring air resistance) will be described by $r'' = - gr/R$ where $R$ is the radius and $g$ a gravity acceleration on the surface. Then period is $\frac{2\pi }{\sqrt{g/R}}=2\pi \sqrt{\frac{R}{g}}\approx 5071 sec \approx 84 min$ which coincides with the period of the  circular orbit with the radius $R$ (it is given by the same formula as the speed  is $\sqrt{gR}$ and the length of the orbit is $2\pi R$).