Thread: Is the far side of the moon older than the earth facing side

Originally Posted by grapes

The strength of the gravity of the sun is two and a half times stronger than the earth's, on the moon.

I did not know that... Thankyou. I respect this point... ( but but but...)

The moon effects Earth's tides more than the Sun does, yes.

So surly the larger body of Earth would have a greater gravity effect at lunar distance than the sun ?

Is my miff with the maths of this reasonable ?

We know the age of the moon and to make suggestion of age differential because of gravity effect... seems less than real.

and might I point out that as the moon DOES rotate about its axis all sides receive equal sunlight.. or shaded times.. So what are you all on ?

( about. )..

Originally Posted by astromark

I did not know that... Thankyou. I respect this point... ( but but but...)

The moon effects Earth's tides more than the Sun does, yes.

So surly the larger body of Earth would have a greater gravity effect at lunar distance than the sun ?

Is my miff with the maths of this reasonable ?

The tidal effect is inversely proportional to the cube of distance, whereas gravity itself is inversely proportional to distance.

The lunar tide is twice the solar tide, on earth, so since earth is 80 times bigger than the moon, the earth tide on the moon is 160 times the solar tide. But the sun is 400 times farther away, so in order to have even that small tidal effect, its gravity must be 400/160 times greater than the earth's, at the moon.

ETA: To make that number more precise, copy the following into wolframalpha.com:

((mass of the sun)/(distance to sun)^2) / ((mass of the earth)/(distance to moon)^2)

My first impulse response...without too much thought is this:

In a gravitational field, if you are standing on the surface of the source, you are accelerating. The only non-accelerated frame is one in free-fall.

Planets and moons are in a state of free-fall...that is, they are in the most energy conservative state within a gravitational field.

The moon is close to a state of equilibrium...within the realm of inertial and gravitational forces.

So I want to say that clocks are going to read identical on both sides. Just my thoughts. Blast away.

It's close to a state of equilibrium, except for the differences in gravity from one side to the other that result in the creation of tides.

Which is what this thread is about, the differences from side to side.

Originally Posted by grapes

The far side of the moon gets deeper into the gravity well of the sun than the near side, though, and that is stronger

Hmmm ... BUT ... doesnt it also get further away ... it is only closer a portion of the time. Whereas the effect from the earth is constant.

Remember the tides are from a difference in gravitational field. The greater the gravitational pull the less of a "difference" there is .....
This is why there is spaghettification near the EH for a small BH and almost no tidal effects for a super massive BH at the EH. It is because the difference in gravitational pull between say the head and the feet of an infaller is almost nothing for the latter.

Originally Posted by astromark

I did not know that... Thankyou. I respect this point... ( but but but...)

The moon effects Earth's tides more than the Sun does, yes.

So surly the larger body of Earth would have a greater gravity effect at lunar distance than the sun ?

Is my miff with the maths of this reasonable ?

We know the age of the moon and to make suggestion of age differential because of gravity effect... seems less than real.

and might I point out that as the moon DOES rotate about its axis all sides receive equal sunlight.. or shaded times.. So what are you all on ?

( about. )..

But the center of the earth is younger than the surface of the earth.

H

Originally Posted by tom

Remember the tides are from a difference in gravitational field. The greater the gravitational pull the less of a "difference" there is .....
This is why there is spaghettification near the EH for a small BH and almost no tidal effects for a super massive BH at the EH.

Not quite that simple. Just because gravity is greater doesn't mean the tidal force is less. Earth gravity is 80 times the gravity of the moon, at the same distance, and its tidal force would also be 80 times.

It is because the difference in gravitational pull between say the head and the feet of an infaller is almost nothing for the latter.

True that, though

I am not understanding some of this... and that bothers me. I stand and watch a near full moon rise ( last night )

and talk with over 70 children of the ecliptic and the plain of the Solar System... as we see it.. and it went well until.

A small voice asked this.. ^ "Does the gravity force change strength much ?" >>> Hmmmm.. "No, not much." I said.

My thinking was general and non specific in that the Earth Moon distance does sort of stay the same..

with a bit of a wobble here and there... What's a few thousand km between friends..

But watching this thread has cast some shadows.. Help...

.. and yes I see that this is about some sort of time slowing due to a gravity well strength..

or not but was dismissive of it as we do have rotations of all parties..

which I would have thought cancelled any change in time rate.. or gravity change..

Originally Posted by astromark

which I would have thought cancelled any change in time rate.. or gravity change..

But the far side of the moon, although it gets closer and further from the sun, always stays further from the earth than the near side.