Planet Earth as a Spring. Assume that the Earth is perfect sphere of mass M=5.98x10^24kg, radius 6,370km, and uniform density. Imagine a long, straight tunnel through the center of the Earth, connecting the North and South poles . A capsule of mass m is dropped into this tunnel. It falls through the center of the Earth, stops when it reaches the opposite pole, and then returns. This is an example of simple harmonic motion. A.Calculate the force constant for simple harmonic motion of this capsule.(Hint:Show that F=-kr, and find k) B.Find the time it takes for the capsule, dropped into the tunnel at the North pole, to return to the North pole C. Explain in 2-3 sentences what would happen if we instead drilled a tunnel through the center of the Earth from the equator, and dropped the capsule down there. Would the resulting motion be different from the pole-to-pole case? PLEASE SHOW ALL ANSWERS AND UNITS. AND HOW YOU CAME UP WITH THE EQUATION FOR THE ANSWERS>

Physics

A) To calculate the force constant for the simple harmonic motion of the capsule, we need to utilize Newton's law of universal gravitation. The formula for the force of gravity between two objects is F = G * (m1 * m2) / r^2, where G is the gravitational constant, m1 and m2 are the masses of the objects, and r is the distance between their centers.

In this case, since we are considering the capsule falling through the tunnel, we assume the mass of the Earth M is much greater than the mass of the capsule m. Thus, the force exerted by the Earth on the capsule can be calculated as F = G * (M * m) / r^2.

Now, since we want to calculate the force constant k for simple harmonic motion, we can write F = -k * x, where x is the displacement of the capsule from its equilibrium position. Since the displacement in this case is proportional to the distance traveled by the capsule from one pole to the other, we have x = 2 * r.

Equating the gravitational force and the restoring force, we have G * (M * m) / r^2 = -k * (2 * r). Simplifying, we can solve for k:

k = -G * (M * m) / (2 * r^3)

Substituting the given values, we have k = -6.67430 x 10^-11 N(m/kg)^2 * (5.98 x 10^24 kg * m) / (2 * (6.37 x 10^6 m)^3).

Evaluating this expression, we find k ≈ -1.95 N/m.

B) To find the time it takes for the capsule to return to the North pole, we can use the equation for the period of a simple harmonic oscillator: T = 2π * √(m / k).

Substituting the given values, we have T = 2π * √(m / (-1.95 N/m)).

Notice that the mass of the capsule cancels out when calculating the period, so it is not important for finding the time. Therefore, the time it takes to return to the North pole is solely dependent on the force constant k. The calculation does not require the given mass of the capsule.

C) If we instead drill a tunnel through the center of the Earth from the equator and drop the capsule down there, the resulting motion would be different from the pole-to-pole case. In this scenario, the symmetry of forces would be broken since the gravitational force would not act purely in the radial direction. This would introduce forces that alter the motion, deviating from simple harmonic motion.