Author Topic: Problem 4  (Read 3633 times)

Fanxun Zeng

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Problem 4
« 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.

Fanxun Zeng

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Re: Problem 4
« Reply #1 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

Calvin Arnott

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Re: Problem 4
« Reply #2 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 $r$.
« Last Edit: November 28, 2012, 10:14:45 PM by Calvin Arnott »

Victor Ivrii

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Re: Problem 4
« Reply #3 on: November 29, 2012, 06:27:54 AM »
It is very simple:

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

  • If it is concentrated as $\{r \ge R\}$ then inside of the cavity $\{r<R\}$ the pull is $0$
  • If it is concentrated as  $\{r \le R\}$ then in $\{r>R\}$ the pull is as if it was created by the same mass but concentrated in the center

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 $2\pi R$).
« Last Edit: November 29, 2012, 09:12:42 AM by Victor Ivrii »