Another strange physical event: This time regarding gravity.

Click For Summary
SUMMARY

This discussion explores the hypothetical scenario of a person jumping into a perfectly drilled hole through the Earth. It concludes that, neglecting temperature and air resistance, the individual would experience simple harmonic motion due to the gravitational forces acting on them. As they fall, gravitational force decreases linearly to zero at the center, allowing them to reach the opposite side in approximately 42 minutes before falling back. The net gravitational force is defined by the formula F = R M g / 6378,000, where R is the distance from the center, M is the mass of the person, and g is the acceleration due to gravity (9.81 m/s²).

PREREQUISITES
  • Understanding of gravitational forces and their behavior in a spherical mass.
  • Familiarity with simple harmonic motion principles.
  • Knowledge of basic physics formulas, particularly those involving force and mass.
  • Concept of uniform density and its implications on gravitational acceleration.
NEXT STEPS
  • Study the principles of simple harmonic motion in physics.
  • Research gravitational force variations within spherical bodies, focusing on Earth.
  • Examine the effects of air resistance on free-fall motion.
  • Learn about the implications of drilling through the Earth, including geological and physical challenges.
USEFUL FOR

Physics enthusiasts, educators, students studying mechanics, and anyone interested in gravitational phenomena and theoretical physics scenarios.

Pshock92
Messages
7
Reaction score
0
Imagine,
You have dug a hole clear through the Earth. From one side to the other. Don't ask me how, you just did.
Let's pretend that the hole is totally perfect: Same diameter thoughtout, no cave-ins, and it somehow doesn't get hot or filled with magma in the center. It's just a clear shot of a hole.

Now, picture someone jumping into it. I have two discussion topics:

1. Would the person fall through the entire hole and come out on the other side?
- I mean, wouldn't gravity stop them in the center, as it comes at the person from all sides? Would this cause them to sort of float in the center of the Earth (remember, we're pretending that there's no magma in the center, so our guy is safe).
- Or, would the great pressure in the center of the Earth crush our jumper to death?

2. What if he made it to the other side?
- What would the people watching him leave the hole see? Would he come out the ground, followed by gravity forcing him down back to Earth?
- Or, would his velocity keep him falling (or in this case, flying) out the hole, into the atomsphere? (This is obviously not possible, but whatever)

What do you think?
 
Physics news on Phys.org
Pshock92 said:
Imagine,
You have dug a hole clear through the Earth. From one side to the other. Don't ask me how, you just did.
Let's pretend that the hole is totally perfect: Same diameter thoughtout, no cave-ins, and it somehow doesn't get hot or filled with magma in the center. It's just a clear shot of a hole.

Now, picture someone jumping into it. I have two discussion topics:

1. Would the person fall through the entire hole and come out on the other side?
- I mean, wouldn't gravity stop them in the center, as it comes at the person from all sides? Would this cause them to sort of float in the center of the Earth (remember, we're pretending that there's no magma in the center, so our guy is safe).
- Or, would the great pressure in the center of the Earth crush our jumper to death?

2. What if he made it to the other side?
- What would the people watching him leave the hole see? Would he come out the ground, followed by gravity forcing him down back to Earth?
- Or, would his velocity keep him falling (or in this case, flying) out the hole, into the atomsphere? (This is obviously not possible, but whatever)

What do you think?

Ignoring issues with temperature and so on, just worrying about gravity, the force downwards will be proportional to your distance from the center. A particle in a hole moving under gravity alone will therefore trace out simple harmonic motion, from one side of the Earth to the other.

The net gravitational at a point inside the Earth is proportional to all the mass that is closer to the center than you are. The forces from everything further from the center than you are cancels... it corresponds to a spherical shell of mass, for which there is no net gravitational force inside the shell.

The mass close to the center than you is proportional to volume, or R3. The force also varies as the inverse square of your distance from the center, or R-2. Hence gravitational force is proportional R, and you have simple harmonic motion.

Cheers -- sylas
 
As you fall through the hole, at distance R from the center, only mass inside this radius will attract you, but you will be a distance R from the center, so the net gravitational force F =R M g/6378,000 (where 6378,000 is radius of Earth in meters, g = 9.81 m/sec^2, and M is your body mass).
If the hole is drilled from the North Pole to the South pole, you will fall straight through, and just reach the other pole in about 42 minutes. If somebody doesn't catch you, you will fall back through in another 42 minutes. If the hole is drilled on the equator (like Brazil to India maybe), you will have to drill a curved hole to account for the fact that the Earth is rotating, and the rotational velocity is 1000 miles per hour at the equator. If you drill a straight hole on the equator, you will bounce around and get stuck at the center of the Earth.
 
You're going to want to ignore air friction too, or your intrepid explorer will find his ride wimpering to a stop in the middle.
 
Bob S said:
As you fall through the hole, at distance R from the center, only mass inside this radius will attract you, but you will be a distance R from the center, so the net gravitational force F =R M g/6378,000 (where 6378,000 is radius of Earth in meters, g = 9.81 m/sec^2, and M is your body mass).
If the hole is drilled from the North Pole to the South pole, you will fall straight through, and just reach the other pole in about 42 minutes. If somebody doesn't catch you, you will fall back through in another 42 minutes. If the hole is drilled on the equator (like Brazil to India maybe), you will have to drill a curved hole to account for the fact that the Earth is rotating, and the rotational velocity is 1000 miles per hour at the equator. If you drill a straight hole on the equator, you will bounce around and get stuck at the center of the Earth.

You are neglecting though that the force of gravity will decrease as you approach the center. So you will continue to accelerate to a very high speed as you fall towards the center, but the acceleration will decrease to zero at the center, and then gradually increase in the opposite direction, slowing you down. If there are no resistances then you would stop once you reach the other end of the hole, and then fall back through.
 
QuantumPion said:
You are neglecting though that the force of gravity will decrease as you approach the center.
He doesn't neglect that fact, he states it in the part that you quoted.
 
Doc Al said:
He doesn't neglect that fact, he states it in the part that you quoted.

But he said he assumed a constant acceleration of 9.8 m/s^2? Wouldn't the acceleration due to gravity would decrease linearly to zero at the center (assuming uniform density for simplicity)?

eta: Here's a nice picture from wiki showing the variation by density: http://en.wikipedia.org/wiki/File:Earth-G-force.png"
 
Last edited by a moderator:
QuantumPion said:
But he said he assumed a constant acceleration of 9.8 m/s^2?
No, he said g = 9.8 m/s^2. (g is just a constant.)
Wouldn't the acceleration due to gravity would decrease linearly to zero at the center (assuming uniform density for simplicity)?
Yes, exactly as he described. See his formula for net gravitational force, which is linearly proportional to the distance from the center.

(I think you just misread what he meant.)
 
QuantumPion said:
But he said he assumed a constant acceleration of 9.8 m/s^2? Wouldn't the acceleration due to gravity would decrease linearly to zero at the center (assuming uniform density for simplicity)?
Please re-read the following from my previus post:
"As you fall through the hole, at distance R from the center, only mass inside this radius will attract you, but you will be a distance R from the center, so the net gravitational force F =R M g/6378,000 (where 6378,000 is radius of Earth in meters, g = 9.81 m/sec^2, and M is your body mass)."

Force at radius R is = R M g /6378,000, where 6378,000 is radius of Earth, so it is linearly proportional to radius.
 
  • #10
Bob S said:
Please re-read the following from my previus post:
"As you fall through the hole, at distance R from the center, only mass inside this radius will attract you, but you will be a distance R from the center, so the net gravitational force F =R M g/6378,000 (where 6378,000 is radius of Earth in meters, g = 9.81 m/sec^2, and M is your body mass)."

Force at radius R is = R M g /6378,000, where 6378,000 is radius of Earth, so it is linearly proportional to radius.

Oh I see what you mean now. Nevermind :p
 
  • #11
Pshock92 said:
Imagine,
You have dug a hole clear through the Earth. From one side to the other. Don't ask me how, you just did.
Let's pretend that the hole is totally perfect: Same diameter thoughtout, no cave-ins, and it somehow doesn't get hot or filled with magma in the center. It's just a clear shot of a hole.

Now, picture someone jumping into it. I have two discussion topics:

1. Would the person fall through the entire hole and come out on the other side?
- I mean, wouldn't gravity stop them in the center, as it comes at the person from all sides? Would this cause them to sort of float in the center of the Earth (remember, we're pretending that there's no magma in the center, so our guy is safe).
- Or, would the great pressure in the center of the Earth crush our jumper to death?

2. What if he made it to the other side?
- What would the people watching him leave the hole see? Would he come out the ground, followed by gravity forcing him down back to Earth?
- Or, would his velocity keep him falling (or in this case, flying) out the hole, into the atomsphere? (This is obviously not possible, but whatever)

What do you think?

I have actually been wondering this question as long as I can remember. Thank you for posting it! (sorry I didn't contribute to the topic)
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 22 ·
Replies
22
Views
983
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 51 ·
2
Replies
51
Views
5K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 69 ·
3
Replies
69
Views
8K
  • · Replies 23 ·
Replies
23
Views
3K