# Hometest help

1. Dec 7, 2004

### Gramps

Hometest help (fixed)

If i drill a hole through the center of the earth that cross the entire planet and then i drop a ball into the hole:

1. What is going to be the speed of the ball at the other end of the planet?

2. How much time is going to take the ball of going from one end to another?

When the ball reach the core there will be a balance in forces and the kinetic energy will transform in potential energy again, thats why the ball will move to the other end of the planet and not stop in the core.

I think the speed at the end of the planet is going to be 0 m/s because it would had reached a max altitude and the gravity will pull the ball down the hole again.

I'm still having problems finding the time, i think it has to do with gravitational attractions.

Any help will be welcome.

Last edited: Dec 7, 2004
2. Dec 8, 2004

### Tide

If the mass of the Earth were distributed uniformly throughout its volume then the local acceleration due to gravity varies linearly with distance from the center so that

$$g(r) = g(r = R) \frac {r}{R}$$

where R is the radius of the Earth and r is the distance from the center. In other words, the motion of the object you drop will be like a harmonic oscillator whose frequency is

$$\omega = \sqrt {\frac {g(r = R)}{R}}$$

Last edited: Dec 8, 2004
3. Dec 8, 2004

### Gramps

we haven't talked about frequency yet.

Can i use thos forumla to find time:

t^2 = 2h/g ?

4. Dec 8, 2004

### HallsofIvy

Staff Emeritus
The first problem should be trivial: what is the difference in potential energy at the two ends of the hole? What does that tell you about the kinetic energy of the ball?

"Can i use thos forumla to find time:

t^2 = 2h/g ?"

Since you didn't mention "h" in your original post, I have no idea!

Assuming that the earth is uniformly dense (which isn't true!) with mass M and radius R, its density is 3M/(4 pi R3). Taking r to be the distance from the center of the earth at each point on the hole, the mass below the ball is M(r/R)3 so the acceleration due to gravity is GM(r/Rsup3). Taking g to be the acceleration at the surface of the earth, that means that the acceleration of the ball at distance r from the center of the earth is gr. You will need to solve the differential equation r"= -gr for r(t) with r(0)= R, r'(0)= 0 and then set that equal to R.

5. Dec 8, 2004

### Gramps

Man, i'm having real trouble understanding this.
Ok, I have:

Mass of Earth= 5.974 * 10 ^24 Kg
R of Earth = 6. 374 * 10 ^ 6 m
G = 6.67 * 10 ^ -11
diameter of Earth = 1.27 *10^7 m

I need to know the speed the ball use to go from one end of the hole to the other end. I thought the speed was 0 m/s but i think i'm wrong.

I think that with: $$V= \sqrt {2gd}$$

Now i have to look for the time it takes the ball to travel from one point to the other. So the ball is going to pass Earth's core. I don't think there is any height in this problem, i don't know why i gave you that formula.
One point i want to clear is that r = R.
The proffesor told me that we need to use earth's diameter to find time.

If there is something wrong with something i said, please correct me.

Last edited: Dec 8, 2004
6. Dec 8, 2004

### Tide

Let's try it this way. Energy conservation gives you

$$v^2 = \frac {g_0}{R}(R^2 - r^2)$$

so that

$$\frac {dr}{dt} = - \sqrt { \frac {g_0}{R} (R^2 - r^2)}}$$

where $g_0$ is the acceleration due to gravity at the Earth's surface. Can you determine (integrate!) from that the time it takes for r to go from R to 0? If so, then just double it to find the time to cross from one side to the other.

7. Dec 8, 2004

### Gramps

The acceleration due to gravity is 9.8 m/s?

R is the radius of Earth, and r is the radius between the object and the Earth's core, right? So r = R.

Am i right?

8. Dec 8, 2004

### dextercioby

Leave any assumptions/speculations aside and do what Tide asked u to do.It' s the only way to find the answer.Solving that simple 1-st order ODE.
You should thank him for showing about 90% of the problem,not question his logics.That's bacause HE IS RIGHT.

Daniel.

9. Dec 8, 2004

### Gramps

i know he is right, i'm just confused in the values i need to use. Because if R = r then (R^2 - r ^ 2) = 0, so why adding it in the formula?
I just want to be 100% sure that what i'm doing is right.

But i know Tide knows exactly what he is talking about, but i'm a "n00b" in physics and it's hard for me to understand all this things. <_>

10. Dec 8, 2004

### Tide

Gramps,

Have you had any calculus at all?

11. Dec 9, 2004

### Shockwave

hard

Physics is the hardest subject probably to major in. Thank god I majored in Math. I don't think I could handle all of the problems I have seen via this site.

12. Dec 9, 2004

### Gramps

yes, Calculus I. But math is not my strength, so i have to survive. In other hand, Chemistry I and II was very easy for me.

Thankfully, Physics I and II are the last math classes i have to take