We have just completed digging a tunnel through the center of the Earth to New H
ID: 1481384 • Letter: W
Question
We have just completed digging a tunnel through the center of the Earth to New Haven’s antipodal point, and now want to construct a series of physics experiments. You are tasked with doing some of the calculations for this.
(a) A kilogram mass is dropped into the tunnel. How long does it take to return?
(b) You repeat the experiment with a 2 kg mass. How does your answer to the previous problem change?
Background information: this is a continuation of a previous exercise we did (for which I therefore already have the answer), which asked:
We have just completed digging a tunnel through the center of the Earth to New Haven’s antipodal point, and now want to construct a series of physics experiments. You are tasked with doing some of the calculations for this.
(a) What is the force on a mass m when it is a distance r from the center of the Earth? You may assume r < RE , the radius of the Earth. What direction is the force in? answer: F = (G*M*m*r/Re^3) rˆ where M is mass of earth, Re is radius of earth
(b) Show, by direct comparison, that the force equation is that of a spring. What is the spring constant k? Answer: k=(G*M*m/Re^3)
(c) Compute, by integration, the potential energy of a mass m, a distance r from the center of the Earth. Unlike the usual gravitational energy calculation, choose your reference point (U = 0) to be the center of the Earth.
Answer= 0.5 (G*M*m/Re^3) r^2
(d) Compare the answer above with the potential energy of a spring. Do you get the same spring constant k as before? yes
(e) You drop a mass m from the surface. What is its velocity as it passes the center of the Earth? Answer: sqrt (G*M/Re)
(f) Describe the motion of the mass. Answer: Since the force is that of a spring, the mass acts the same as if it were attached to a spring. It oscillates around the center of the Earth, moving up and down in the tunnel.
Explanation / Answer
(a) Since this is oscillating about the centre in the tunnel and moving up and down,
Therefore the natural frequency will be
w = (K/M)1/2 where K is spring constant and M is mass
We have the value of K and M
Now the time period is
w = 2Pi/T
T = 2Pi/w
T = 2Pi*(M/K)1/2
We will use the value of K calculated in second equation.
(b) And the time period becomes (2)1/2 times on increasing the mass to 2 kg.
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