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November 11th, 2014, 06:50 AM  #1 
Newbie Joined: Nov 2014 From: uk Posts: 1 Thanks: 0  Effects of Gravity Falling Through Earth's Centre
I drop a mass into a tunnel bored through the earth's centre and out to the opposite side. Assuming no air resistance, constant density, earth diameter 12,742,000m what velocity is the mass travelling at as it passes the centre? I assume this is the max velocity? Is the reduction in acceleration linear or a more complex function? 
November 11th, 2014, 07:31 AM  #2  
Senior Member Joined: Jun 2013 From: London, England Posts: 1,312 Thanks: 115  Quote:
Hole Through the Earth Example  
November 11th, 2014, 07:57 AM  #3 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 937 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
By universal gravitation the force between the object and the earth is $$ F=G\frac{m_1m_2}{r^2} $$ where $r$ is the distance between the object and the center of the Earth and $m_2$ is the mass of the object. Thus the acceleration experienced by the object is $$ a=G\frac{m_1}{r^2} $$ But you still need to account for the mass of the Earth. Fortunately it's not hard: split the Earth into two parts, a sphere at the center with radius r, and the shell outside that sphere. The shell's gravity perfectly cancels out, so you can treat $m_1$ as just the mass of the inner sphere. So $$ a=G\frac{m_E(r^3/r_E^3)}{r^2}=\frac{Gm_E}{r_E^3}r $$ where $m_E$ is the mass of the Earth and $r_E=6,371,000\text{m}$ is the radius of the Earth. You can see that the acceleration is proportional to the distance from the center. Now you need to integrate. I've run out of time; maybe someone else can help you from here? 
November 11th, 2014, 07:59 AM  #4  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 937 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Quote:
 

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centre, earth, effects, falling, gravity 
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