Particle dropped in hole through the earth

In summary: You are given two periods of orbital revolution (in days), and you are asked when two high tide points happen on the earth. The solution is actually quite simple--you just have to solve for the two high tide points in terms of the two periods of orbital revolution.
  • #1
Barley
17
0
I'm having trouble with 2 problems in my mechanics textbook.

First: A particle dropped into a hole drilled straight through the center of the earth. Neglecting rotational effects, show that the particles motion is simple harmonic if you assume Earth has uniform density. Show that the particle of oscillation is about 84 minutes.

Problem does not say if the hole is through the equator, or from pole to pole, does this matter. Now, I had the idea to show that the particle has the acceleration of a(t) = -omega^2x(t). Meaning that somehow I've got to show that the acceleration of the particle is proportional to the displacement of the particle but opposite in sign, and the quantities are related by the square of the angular frequency.

In the case described by the problem, is the acceleration through the center of the Earth the same as the accelration near the surface of the Earth ? What happens to g force in the middle--it must be zero right? And how do I figure the angular frequency the particle ?



2nd problem: The orbital revolution about the Earth is about 23.7 days and is in the same direction as the as the Earth's rotation every 24 hours. Use this information to show high tides occur everwhere on Earth every 12 h and 26 minutes.

Looked in an astronomy book and found that the tidal force has a 1/r^3 dependence. My text goes a little further and derives the Tidal force as a vector with Fx and Fy allowing you to determine tidal force at an point on the earth. But, I don't see how this information helps with the problem stated above. I'm given 2 periods of orbital revolution.


Clues, I need clues.

Thanks
 
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  • #2
For the first problem:

It is not necessary that the hole pass through the center of the Earth but the resistive forces are to be absent or negligible. The acceleration due to gravity at distance r from the center of Earth (r < R) is given by g' = g(r/R) ; R is the radius of Earth and g is at surface. Find the component of this force along the hole as a function of distance x, from the midpoint of the hole and you will get the required relation.
 
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  • #3
mukundpa said:
The acceleration due to gravity at distance r from the center of Earth (r < R) is given by g' = g(r/R) R is the radius of Earth and g is at surface.

This might seem mysterious at first but it is key to solving the problem. To understand how you get this formula, you need to use a theorem proven by Newton that says in a spherically symmetric mass distribution, the gravitational force is determined solely by the matter contained inside a sphere centered on the origin of the distribution, with radius out to where you are calculating the field. (i.e. think of it as being the sphere that you would be sitting on the surface of.) It is quite an amazing theorem which essentially proves that all the extra matter outside your radius completely cancels out! (One result of this is that inside a hollow sphere sitting in a uniform mass distribution, there is no gravitational field.)

Once you apply this theorem, you'll find the gravitational field strength as a function of radius from the center of the earth, and then you just have to show that the field strength (and hence the acceleration) is proportional to the radius to get sumple harmonic motion.
 
  • #4
Barley said:
2nd problem: The orbital revolution about the Earth is about 23.7 days and is in the same direction as the as the Earth's rotation every 24 hours. Use this information to show high tides occur everwhere on Earth every 12 h and 26 minutes.

Here is just a hint that might or might not get you started. Have you ever solved a problem that asks you when the minute hand and second hand on a clock are perfectly aligned? This problem is equivalent to that.
 

1. What would happen if a particle is dropped in a hole through the Earth?

If a particle is dropped in a hole through the Earth, it would experience a gravitational force towards the center of the Earth. As it falls, it would gain kinetic energy and its speed would increase until it reaches the center. At the center, the particle would stop briefly and then start accelerating towards the other side of the Earth due to the pull of gravity.

2. How long would it take for the particle to reach the other side of the Earth?

The time it takes for the particle to reach the other side of the Earth would depend on the depth of the hole and the density of the Earth. On average, it would take about 42 minutes and 12 seconds for the particle to reach the other side. This is assuming the Earth is a perfect sphere with uniform density.

3. What would happen to the particle's speed as it reaches the other side of the Earth?

As the particle reaches the other side of the Earth, it would start to slow down due to the gravity pulling it towards the center of the Earth. It would lose kinetic energy and its speed would decrease until it reaches the surface on the other side.

4. Would the particle be able to escape the Earth's gravitational pull?

No, the particle would not be able to escape the Earth's gravitational pull. The escape velocity for an object on the surface of the Earth is about 11.2 km/s, which is much greater than the speed the particle would reach when falling through the hole.

5. What would happen if the Earth's density was not uniform?

If the Earth's density was not uniform, the time it takes for the particle to reach the other side would vary. In some cases, the particle may not even reach the other side and would instead oscillate back and forth within the Earth. This is known as a "gravity train" scenario and was first proposed by physicist Isaac Newton.

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