Velocity for an object at the center of the earth?

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The discussion focuses on calculating the velocity of a 20kg object dropped through a hypothetical hole that passes through the center of the Earth, assuming ideal conditions such as uniform density and spherical shape. The gravitational force acting on the object can be modeled as a linear function, akin to a spring, due to the properties of a hollow spherical shell. By integrating the gravitational force from the Earth's surface to the center, one can determine the work done on the object, which translates to its kinetic energy at the center. The formula E=(mv²)/2 is then used to derive the object's velocity at that point.

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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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