Time Dilation and space missions

[Doc Al]

Here is another puzzle to me so it may be an opportunity for someone to explain it to me. GR says clocks will tick faster at lesser gravities (ie. higher up). Yet pendulum clocks do the opposite and tick slower at lesser gravities. Don't pendulum clocks count as clocks?

Take away the Earth and the pendulum clock wouldn't work at all. You must consider the Earth itself as a key part of a pendulum or pendulum clock.

 

[Janus]

Here is another puzzle to me so it may be an opportunity for someone to explain it to me. GR says clocks will tick faster at lesser gravities (ie. higher up). Yet pendulum clocks do the opposite and tick slower at lesser gravities. Don't pendulum clocks count as clocks?

Well for one thing, the swing of a pendulum clock depends on the gravitational "force" and the Gravitational time dilation depends on the gravitational "potential". Gravitational force varies by the inverse of the square[i/] of the distance from the center of the gravity field and gravitational potential varies by the inverse of the distance.

Thus it is possible create a situation where the gravitational force felt by two pendulum clocks are equal, but they are at different gravitational potentials. For example, one clock sitting on the surface of the Earth and the other sitting on the surface of a body with 4 times the mass of the Earth and twice the radius. The two clocks would feel the same gravitational force but would be at different gravitatonal potentials, and the second clock would run faster (as seen by a distant observer) than the first.

 

[gonegahgah]

Thanks Doc. Yep. At zero g time ceases to exist for a pendulum clock - if not for us - because there is nothing to pull the pendulum down.

My prophecy is that "aging time" has something to do with sub-atomic spin (whereas pendulums rock back and forth) but that is personal conjecture and neither here nor there.
In that respect if something is at 2 from the centre then it would be 1/2 for potential and 1/4 for force, and at 3 from the centre it would be 1/3 for potential and 1/9 for force regardless of the size or density of the planet.

Is that incorrect? Otherwise can you help me with the difference?

It is also my understanding, though you were probably just providing a simplified model, that the closer you get to a large gravitational body, the less attraction is experienced down (although small by proportion) and more is experienced sideways and cancels out. So that, close to the surface of the Earth the mass directly below is a direct attraction down but mass off to the sides below you has an attraction with a component sideways and proportionally less down. This is in the same way that once you drop below the surface then you become attracted upwards towards the mass above you cancelling out some of your attraction to mass below. Is this correct?
Hi Janus. Surely gravitational potential varies "by the inverse [not the square acknowledged] of the distance" "from the centre of the gravity field" as well.

 

[gonegahgah]

Thanks Doc. Yep. At zero g time ceases to exist for a pendulum clock - if not for us - because there is nothing to pull the pendulum down.

My prophecy is that "aging time" has something to do with sub-atomic spin (whereas pendulums rock back and forth) but that is personal conjecture and neither here nor there.

Hi Janus. (Rewritten; accidentally deleted) Doesn't gravitational potential vary "by the inverse [not squared acknowledged] of the distance" "from the center of the gravity" also.

In that respect if something is at 2 from the centre then it would be 1/2 for potential and 1/4 for force, and at 3 from the centre it would be 1/3 for potential and 1/9 for force regardless of the size or density of the planet.

Is that incorrect? Otherwise can you help me with the difference?

It is also my understanding, though you were probably just providing a simplified model, that the closer you get to a large gravitational body, the less attraction is experienced down (although small by proportion) and more is experienced sideways and cancels out. So that, close to the surface of the Earth the mass directly below is a direct attraction down but mass off to the sides below you has an attraction with a component sideways and vector-wise less down. This is in the same way that once you drop below the surface then you become attracted upwards towards the mass above you cancelling out some of your attraction to mass below. Is this correct?