Gravitational Potential inside a mass

AI Thread Summary
Gravitational potential exists inside a mass, such as 500 km deep within the Earth. While the gravitational field intensity can be calculated using the mass within a specific radius, the same method cannot be applied directly to calculate gravitational potential. For a uniform density sphere, the potential formula leads to incorrect force calculations due to the potential's non-uniqueness, as constants can be added without affecting the physics. The correct approach involves calculating the force first and then deriving the potential from that force. Understanding these concepts requires a solid grasp of the underlying principles of gravitational fields and potentials.
Knightycloud
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I got an exam tomorrow and while i was studying, this popped in.
Is there a gravitational potential inside a mass (eg : 500 km deep from the Earth surface)?

We can calculate the Gravitational field intensity inside a mass by taking the distance from the center to that inner-point as x and calculating the mass within that radius; then taking that calculated mass into the G.I = -GM/x2

Can we do the same thing to calculate the potential? I googled and those graphs showed a similar behavior, we get as the potential inside a conductive sphere. Why can't us calculate it like that above method? I only understand GCE A/L by the way.
 
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Knightycloud said:
Is there a gravitational potential inside a mass (eg : 500 km deep from the Earth surface)?
Of course.

Can we do the same thing to calculate the potential?
No.

Consider a uniform density sphere. Use \Phi(r) = -GM(r)/r and take the gradient. This should yield the additive inverse of the force since \vec F(\vec r) = -\nabla \Phi(\vec r). However, since M(r)=4/3\pi \rho r^3 for a uniform density sphere, this Φ(r) yields F=+8/3\pi G \rho r \hat r. The sign is wrong and the magnitude is twice as large as it should be. The potential must yield F=-\partial \Phi(r)/\partial r = -4/3\pi G \rho r \hat r.
 
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Potentials are not uniquely defined - you can always add a constant. That is fine if you just consider a single potential, but it ruins the approach to combine the potential for each depth in that way.

You can calculate the force, and based on this force you can calculate the potential - that approach works.
 
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mfb said:
You can calculate the force, and based on this force you can calculate the potential - that approach works.
That makes sense. Because my level of knowledge is limited to the basic principles and i haven't used δf(x)/δx applications any where and that describes how few I know about this topic. :/

Thanks a lot for the support people.
 
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