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) Does the force an object an object experiences near the surface of the Earth need to be continuous as it passes from above to beneath the surface?

b) 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?

c) What is the force on the object for points within the tube (r < RE)? (Express the answer with a constant, C, and a power, n, of the radial distance from the center, r, i.e. F = C* r^n).

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

e) What is the gravitational potential energy for points within the Earth (r < RE)?

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

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

People did elaborate lots on its motion, but I would like to hear explanations on all parts, not part a... Thanks in advance.

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). Next, with this little tidbit: http://hyperphysics.phy-astr.gsu.edu/hba… The gravity inside a spherical shell is zero. 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 a 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.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.