Another strange physical event: This time regarding gravity.

In summary: As for what observers would see, it would depend on the location and orientation of the hole. If it is drilled through the poles, the person would simply appear to fall straight down and back up again. But if it is drilled at an angle or on the equator, observers may see the person appear to float or bounce around in the hole before falling back to the other side. However, this scenario is purely hypothetical and impossible to achieve in reality. In summary, jumping into a perfectly drilled hole through the Earth would result in simple harmonic motion and a back and forth journey between the two sides of the hole. Observers would see the person appear to float or
  • #1
Pshock92
7
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?
 
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  • #2
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
 
  • #3
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.
 
  • #4
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.
 
  • #5
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.
 
  • #6
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.
 
  • #7
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:
  • #8
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.)
 
  • #9
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)
 

FAQ: Another strange physical event: This time regarding gravity.

What is the event that occurred regarding gravity?

The event that occurred is a strange physical phenomenon that affects the force of gravity in a particular area.

How does this event affect objects in the affected area?

This event causes objects to experience a change in their weight and the rate at which they fall due to the altered force of gravity.

What could be the potential cause of this event?

The cause of this event could be a disruption in the space-time continuum, the presence of a large mass nearby, or a change in the fundamental laws of physics.

Is this event a cause for concern?

It depends on the magnitude and duration of the event. If it is a small and temporary change, it may not have a significant impact. However, if it is a large and long-lasting event, it could potentially have serious consequences for our planet and its inhabitants.

What further research is needed to understand this event?

To fully understand this event, further research is needed to gather more data, analyze the data, and develop theories and models to explain the phenomenon. Collaborative efforts from various fields of science, such as physics, astronomy, and geology, may be necessary to gain a comprehensive understanding of this event.

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