# Homework Help: Gravity, acceleration and velocity question

1. Dec 25, 2015

### sona

1. The problem statement, all variables and given/known data

Assume that there is a hole dug from one side of the earth's surface to another. A ball is dropped through the surface. Explain what happens to acceleration and velocity as the ball passes through the hole.

2. Relevant equations

3. The attempt at a solution

I am not sure if I am correct but this what I have attempted.

As the ball is moving to the middle, the acceleration should be increasing because the force of gravity is measured from the center and since the distance from the center to the ball is decreasing, acceleration should be increasing and therefore, velocity should also be increasing.

At the center, r = 0 which means a = 0. Not sure about velocity ?

As the ball is moving away from the center of the earth to the opposite side, a should be decreasing and so velocity should also be decreasing.

2. Dec 25, 2015

### Staff: Mentor

What is the formula you used to reach that conclusion? What are the variables you plug in, and what are the assumptions you need to use that formula?
What do you expect to see 1 meter away from the center of earth and right at the center, and what does your approach predict?
Does the second part really follow from the first one? It is right that velocity increases, but does the acceleration have to decrease for that?

Concerning velocity at the central point: it increases before and decreases afterwards, so what do you expect?

3. Dec 25, 2015

### TSny

Hello, sona. Welcome to PF.

In order to be able to answer the question regarding the acceleration, you need to be familiar with the gravitational force exerted on a particle by a spherical shell of mass when the particle is outside the shell and when the particle is inside the shell.

See https://en.wikipedia.org/wiki/Shell_theorem

4. Dec 25, 2015

### sona

Fg = Gm1m2/r^2 but I am not sure if this is the right equation to use becasue this equation is for rotating objects.

On Earth's surface, Fg = mg

5. Dec 25, 2015

### Brian T

This is not the right equation since you are inside the sphere. If you look at the equation you're trying to use, as r goes to 0, Fg goes to infinity. But if you think about it, the force at the center should be 0 (why?). If you are familiar with Gauss law, you can apply it to this problem to find the correct equation.

6. Dec 25, 2015

### FactChecker

I wonder what level of physics class and answer is required. The question just asks you to "explain" what happens, so it may not be necessary to use any equations. They would be complicated. As you said, you know that the velocity starts at 0 at the surface. At the center, all gravitational forces are balanced out, but there has been acceleration up to then. Then it moves to the other side where the forces are in exactly the opposite direction. You can fill in the details in terms of velocities and accelerations increasing and decreasing.

7. Dec 25, 2015

### Staff: Mentor

The equation has nothing to do with rotations.
It is an equation for point-masses, or objects where you can approximate them by a point-mass. If you are inside Earth, approximating Earth as a point-mass in the center doesn't work. The cross-checks I suggested would have highlighted that problem.

8. Dec 25, 2015

### sona

The question didn't ask to use or show how to work this out using any equations. The answer just requires a conceptual explanation.

9. Dec 25, 2015

### Fullmphysicist

In addition to only thinking about the forces it would be helpful to think about the exchange between kinetic and potential energy. By only using energy conservation you can for example make a statement about the velocity you'll have when you emerge on the other side of the earth.

10. Dec 26, 2015

### 2nafish117

the falling object actually performs simple harmonic motion https://en.wikipedia.org/wiki/Simple_harmonic_motion
knowing this makes the explanation simpler
also when you are under the surface of the earth you can prove that only the mass underneath you pulls you in .
so as you go down the hole ,the amount of mass that pulls you in ∝r^3 (to the cube of the distance between you and the centre of the earth) .
i am assuming the earth is uniform in density and is perfectly spherical.
and since the force of attraction is ∝m/r^2⇒ force∝r (r^2 cancels in the numerator and denominator) ⇒acceleration of the body ∝r
which is the condition for simple harmonic motion(not exactly F∝-x is the condition)
thus acceleration decreases as you go to the centre but is maximum at the surface
now about the velocity think in terms of energy conservation principles asfullmphysicist said.