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Particle dropped in hole through the earth

  • Thread starter Barley
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  • #1
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I'm having trouble with 2 problems in my mechanics text book.

First: A particle dropped in to 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
 

Answers and Replies

  • #2
mukundpa
Homework Helper
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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.
 
Last edited:
  • #3
David
Science Advisor
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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
David
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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.
 

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