Dropping a mass in a hole through the Center of Earth

  • Thread starter Pruddy
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  • #1
Pruddy
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Homework Statement


Assume that the Earth is a uniform density spherical mass. (This assumption is not correct but we will use it for simplicity in working the problem.) The deepest hole drilled into the Earth's surface went to a depth of 40,230 ft (Wikipedia.org). Imagine that this hole was straight toward the center of the Earth and that a small 50 kg spherical mass was dropped into the hole and ended up at the very bottom of it. What would be the weight of the spherical mass at that location?


Homework Equations



w=mg


The Attempt at a Solution


This is a two shell theorem problem. Since we all know that w=mg, F=GMm/r2. I think we have to find the gravitational force of the object that is at bottom of the Earth and then multiply it by its mass. This will give us the weight of the object.
 

Answers and Replies

  • #2
tiny-tim
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Welcome to PF!

Hi Pruddy! Welcome to PF! :smile:
I think we have to find the gravitational force of the object that is at bottom of the Earth and then multiply it by its mass. This will give us the weight of the object.

Yes, that's correct. :smile:

What is worrying you about that? :confused:

(btw, don't worry about the duplicated posts, that sometimes happens :wink:)
 
  • #3
Pruddy
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Thanks for your quick reply. I do not know the right equation to use to get the gravitational force of of the mass at the bottom of the earth:confused:.
 
  • #4
tiny-tim
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it's the formula you've already given … F=GMm/r2

(and of course GM/R2 = g where R is the radius of the Earth)

(btw, you mean the bottom of the hole, not the bottom of the Earth :wink:)
 
  • #5
Pruddy
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Thanks again for your quick reply. Yeah you are right, I meant the bottom of the hole. But what happends to the depth of the hole which is 40,230ft. Is it irrelevant to the question? and if it is not, will is be right to add it to the radius of the Earth before squaring it(radius)...
 
  • #6
tiny-tim
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… will is be right to add it to the radius of the Earth before squaring it(radius)...

uhh? :confused:

it's a hole

subtract it! :smile:
 
  • #7
Pruddy
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Thanks Tim, That was very helpful...
 

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