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# the Earth Have you ever speculated as to what would happen to an object dropped into a tunnel bored through the center of the earth to the other side?...

How do i do questions 1 and 3

MA 113 220 1 Summer 2013 Assignment 2 Journey through the Center of the Earth Have you ever speculated as to what would happen to an object dropped into a tunnel bored through the center of the earth to the other side? In this exercise, we will explore this problem. The equations which describe the motion of an object which is dropped from rest into a tunnel bored through a sphere of uniform density can be derived from the geometry of the problem and from Newton's laws. Let us consider a ball of radius R of uniform density and total mass M . This sphere has a tunnel bored through its center, connecting opposite poles of the sphere. We will assume that the mass removed from the sphere in creating the tunnel is negligible compared to M, the mass of the ball. This is equivalent to presuming that the radius of the tunnel is much less than the radius of the sphere. Now that you have dug a hole all the way through the Earth, imagine that a ball of mass m is dropped from rest into the tunnel. In order to find the equation of motion for the object, we need to calculate the force on the object at an arbitrary distance, r , from the center, and then apply Newton's Second Law of Motion. We will use the following fact from basic physics: the force on the object at a distance r from the center is due only to that portion of the ball which is a distance r or closer to the center, and not at all on the mass of the ball which is farther away than r. This says that if the Earth were a hollow shell and you were inside, you would be weightless. Now we need to apply Newton's Law of Gravitation to calculate the force F on the object when at a distance r from the center: 33 44 33 2 (/ ) mM r R FG r   . The term in parentheses represents the ratio of the volumes of the spheres with radius r and R ; when multiplied by M this gives the effective mass which is attracting the object. The coefficient G is the universal gravitational constant. If g represents the acceleration due to gravity at the sphere’s surface, then 2 mMG mg R , so that we find R r m M
MA 113 220 2 Summer 2013 mgr F R  . Finally, Newton’s Second Law of Motion tells us that: 2 2 dr g r dt R  . So, here g is the constant acceleration due to gravity at the surface of the sphere, which is the end of the tunnel. Since the object starts from rest at the surface of the sphere, we must have the initial conditions (0) , initial distance from the center (0) 0, initial velocity rR r Your mission, whether or not you decide to take it, is the following. 1. Show by substitution that the function   () cos / rt R t g R satisfies both the defining equation and the initial conditions set out above. Thus, this expression for r ( t ) gives the radial distance from the center as a function of time. 2. Since the sphere is to be the Earth, then R = 3960 miles, approximately, and g =32 ft/sec 2 . Graph r ( t ), ( ) rt , and ( ) rt  for 0 170 t minutes. Notice that the graphs of the different functions will generally require different scales for the vertical axes. From your graphs determine a. Where is the velocity of the object greatest, and where is it least? b. Where is the acceleration of the object greatest, and where is it least? Recall that ( ) rt  is the acceleration and () rt is the velocity. c. From your graph estimate the maximum velocity of the object, in miles per hour. 3. Using your student intuition, GUESS how long YOU think it will take the object to fall from one side of the earth to the other. Then do the following. a. Estimate this time from your graphs, i . e ., estimate the first value of t for which r ( t )=– R =–3960 miles. b. Show that the transit time is exactly / R g by solving this equation for t . Evaluate using the parameters for the Earth. How good was your original guess?

Zman! I completed your... View the full answer

3a) in the previous result, we have shown that
We can interpret this as the time taken to travel the distance r is dependent of the radius of the sphere
R, constant gravitational pull g and the...

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