A very narrow hole is drilled all the way through the Earth (through its center)

Question

A very narrow hole is drilled all the way through the Earth (through its center) and evacuated (i.e. there is neither air resistance nor friction). A small object of mass m is dropped into the hole at the surface.

a) Describe its motion qualitatively. The Earth has radius 6375 km. Assume that the Earth has uniform density, and state any other assumptions you choose to make. (Hint: Make sure your answer makes physical sense).

b) Does the force an object experiences near the surface of the Earth need to be continuous as it passes from above to beneath the surface? YES NO

c) Does the gravitational potential energy an object experiences near the surface of the Earth need to be continuous as it passes from above to beneath the surface? YES NO

d) What is the force on the object for points within the tube (r < R E )? (Express your answer with a constant, C, and a power, n, of the radial distance from the center, r, i.e. F = Cr n )

e) How long does it take to return to its starting point?

f) What is the gravitational potential energy for points within the Earth (r < R E )?

g) What is the gravitational potential energy at the center of the Earth?

h) What is the speed of the mass at the center of the Earth?

Explanation / Answer

The force of gravity is given by

F = -G*m_b*m_E/r^2 (Eq. 1)

where
F is the force on the ball due to gravity,
m_b is the mass of the ball,
m_E is the mass of the earth,
r is the radius from the centre of earth's mass to the centre of the ball's mass, and
G is the gravitational constant.

Dropping a ball from the surface of the Earth, the "restoring force" is always pulling the ball toward the centre of the earth. The first thought is that this is may work out to be a simple harmonic oscillator; however, let's look closer.

First, let's assume that the Earth is a uniform density (I know it's not, but I doubt your teacher cares).

The gravity inside a spherical shell is zero. (Shell theorem)

So, for any given r (position of the ball relative to the centre of the earth), the mass of the non-enclosed sphere will be

m_E =d*4/3*pi*r^3 (eq 2)

Where d is the density of the earth; this is simply the volume of the sphere the size of r times the density of what's inside the sphere.

Substitute the equation for m_E from (eq 2) into (eq 1)
F = -G*m_b*[d*4/3*pi*r^3]/r^2
simplifying and grouping the constant terms, we get:
F = -[G*d*4/3*pi*m_b] * r (eq 3)

Equation 3 has a form of a simple harmonic oscillator; which is:
F = -kx

in this case, k = G*d*4/3*pi*m_b

Any introductory physics class or ODE math class will teach you how to solve this equation (fairly simple, so I will omit); the characteristic frequency given by:
w = sqrt( k / m_b )
w = sqrt( G*d*4/3*pi*m_b / m_b)
simplifying, we arrive at:

w = sqrt( G*d*4/3*pi )
where
G is the gravitational constant
d is the (assumed constant) density of the earth

Note that this frequency is given in radians per second (angular frequency) and not Hz ("frequency"); divide by 2pi to get the latter.

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