Surfacing Profile

What's your weight at the center of the earth???

If you fell through (an imaginary tunnel) from one pole to the other or from one side to the other,

how would your weight profile be as you fall from the surface towards the center of the earth?

The weight of an object increases or decreases from the center of gravity. The effect of gravity is inversely proportional to the distance from the centre. so if you were to fall to the center of the earth, how much would you weigh? the radius of the earth at the center is zero so the force on you is infinite??? will you be crushed by gravity?? any clarifications?

Your weight would be proportional to the distant from the center of the earth. So it would it act like a spring (Hookes law) and oscillate, where you would fall through and pass the center of the earth and assuming the tunnel went through to the other side you would end up on the other side minus the air drag. So with air drag consideration, you would be a damped oscillator, falling through with increasing speed that would "whip" you through the center and almost exactly the same distance on the other side, and then you would fall back through again from the other side. Because of air drag you would not make it quite as far each time you passed the center, so you would eventually end up wiggling around the center of the earth at a weight of zero.

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Mass and Density Profile Question?

How do I find find the mass of a planet with the following density profile?

p(r)= [((ps-pc)/Rp)]r + pc

ps= surface density
pc= center density
Rp= radius of planet

Integrate the density over the volume of the planet. This is most easily done in spherical coordinates. Let ρ(r) be the density as a function of radius. Then the mass is equal to:

M = triple INTEGRAL of {ρ(r) dV}

M = triple INTEGRAL of {ρ(r) * r^2 * sin(φ) dr dφ dθ}

where r is the radius, and the integration runs from r = 0 to r = Rp

θ is the equatorial coordinate (longitude), and runs from 0 to 2*π

φ is the azmiuth (latitude) and runs from 0 to π.

Because the density depends only on r, and not on θ or φ, we can immediately do the integrations involving these coordinates, leaving the single integral:

M = 4*pi INTEGRAL of {ρ(r) dr }

Substituting in the function for the density, we have:

M = 4*π*INTEGRAL of {((r^3)*(ρs-ρc)/R + ρc*r^2) dr }

M = 4*π*((R^4)*(ρs-ρc)/(4*R) + ρc*(R^3)/3)

M = 4*π*(R^3)*(3*ρs + ρc)/12

M = π*(R^3)*(3*ρs + ρc)/3