How Does Gravity Change Inside the Earth's Surface?

[blackwing1]

Dang, some good pressure. I'm trying to make some synthetic diamond so I can get marry with less cost. I'm encountering problem with heat source. Where can I get 4000K ? no metal would hold that temp even ceramic. I'm thinking concentrate sunlight intersecting beams +conventional fire. Crap, it seems easier buy that stone.

doesn't a diamond evaporate somewhere 2000- 3000 degrees fahrenheit? or was it somewhere in the 4000s, i forgot...

 

[Feldoh]

Not so misleading if you drop a mass from the surface into a hole through the planet. It will oscillate under ideal conditions between the opposit surfaces, which would correspond to a zero height orbit. Real low Earth orbits are 60 to 90 miles, which is a small fraction of a percent of the radius. So it really does model a low Earth orbit in some sense.

Now here's an interesting realization (to me anyway): according to Newton, the gravitational force felt beneath the surface of a homogenous planet is due to the mass contained in a spherical radius between you and the center. That makes the force proportional to x^3, and inversely proportional to x^2 ... therefore, so ma=-kx, and we actually do get harmonic motion.

So how does it change if the amplitude exceeds the radius? Once the mass leaves the surface the rule reverts to inverse square. The nice harmonic oscillation breaks, and the nifty model of a circular motion projected onto the center line fails too. The projection of the orbit will always be sinusoidal, but the oscillating mass won't remain sinusoidal once the amplitude exceeds the radius. Or did I miss something?

There's a conservation of energy, so unless you started with an velocity >= escape velocity you would continue to oscillate, I believe.