Velocity for an object at the center of the earth?

AI Thread Summary
In an ideal scenario where a hole passes through the center of the Earth and the density is uniform, the gravitational force on a 20kg object can be calculated as a function of its distance from the center. The gravitational field inside a hollow spherical shell is zero, allowing for simplification in calculations. By integrating the force from the Earth's surface to the center, one can determine the work done on the object. This work translates into kinetic energy, which can be used to calculate the object's velocity at the center using the equation E=(mv²)/2. Thus, the velocity at the center can be derived through these steps.
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Let's assume there is a hole that goes through the center of the Earth with ideal conditions (earth is spherical and gravity is the only force). If you dropped an object with a mass of, say 20kg, what is its velocity at the center of the earth? I know the force is much like that of a spring and thus is linear but have no clue how to calculate this. Please show work.
 
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ukballer1012 said:
Let's assume there is a hole that goes through the center of the Earth with ideal conditions (earth is spherical and gravity is the only force). If you dropped an object with a mass of, say 20kg, what is its velocity at the center of the earth? I know the force is much like that of a spring and thus is linear but have no clue how to calculate this. Please show work.

You'll need to make another ideal-conditions assumption as well: that the density of the Earth is uniform.

If you make this assumption, you'll be able to the calculate the gravitational force on your mass when it's at a distance r<RE (where RE is the radius of the earth) from the center of the earth. Because the gravitational field of a hollow spherical shell is always zero in the interior of the shell, you can ignore everything outside of r when calculating the force at r.

Once you have the force as a function of r, you can get the work done by dropping the object into the hole from the surface of the Earth by integrating between RE and 0. That will give you the kinetic energy of the dropped object at r=0, and E=(mv2)/2 will get the velocity from there.
 
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