A woman has a large stone in her yard. She lives at the equator and

[lildrea88]

A woman has a large stone in her yard. She lives at the equator and the date is September 21st. It is midnight and she is thinking of moving the stone. She wants to know, as a fraction of its current weight, how much lighter the stone will be if she waits until noon to move it. Here are some facts you can use to answer her question:
Newton’s law of gravitation says that the force of attraction between two bodies with masses M and m is equal to GMm/r^2 where G is the gravitational constant, and r is the distance between their centers of mass. You may find the following quantities useful and they are even approximately true (at any rate you can treat them as if they are true for this problem).
93000000 miles is the distance from the sun to the Earth (center to center).
The Earth's diameter is 8000 miles.
The mass of the Earth is 1/330000 of the mass of the Sun.
The mass of the stone is whatever it is, if you analyze this correctly, you won’t need to know what it is.


From the diagram that I drew. I know that the mass for the rock is not needed because they will cancel out. But not sure how my formula should look. Can anyone help. **Just a bonus question in which I'm curious about**

 

[khemist]

F = (g1g2/r^2)G

Though with your use of sig figs, the rock will be the same weight. If I had to guess, I would think the difference would be EXTREMELY small, maybe even not enough to measure.

 

[jetwaterluffy]

I think the moon would have more of an impact than the sun in this measurement, and it is 384,403 km away, with a mass of 7.36 × 10²²kg. The effect would be noticeable, as we can see it in the tides.

 

[halli]

I think it will always weight the same, no matter what other gravity comes along.

If the Moon, the Sun or anything else is pulling the rock it is also pulling the Earth,(and the Earth "it") in same direction.

 

[DaveC426913]

I think it will always weight the same, no matter what other gravity comes along.

If the Moon, the Sun or anything else is pulling the rock it is also pulling the Earth,(and the Earth "it") in same direction.

No. We have ocean tides because the moon and sun pull on different parts of the Earth differently.

 

[halli]

You are probably much better at this then I am and I am not going to argue.

But as I see it gravity "pulls" not just what it "sees", it "pulls" the whole thing. Front, center and the behind.
As I see it the tides are because of the Earth is spinning and there for liquid matter as water flows around, because of Earths position regarding to other bodies effecting it. But a 1 ton stone will always be a 1 ton stone.

 

[DaveC426913]

But as I see it gravity "pulls" not just what it "sees", it "pulls" the whole thing. Front, center and the behind.

The near side of Earth is slightly closer to the Moon than the centre of mass of the Earth. This means gravity pulls on the near side more than on the bulk of the Earth. Likewise, the centre of the Earth is slightly closer to the Moon that the far side of the Earth, so gravity pulls slightly more on the centre than on the far side.

The effect of a gravitational gradient (or tide) is that it wants to stretch a body along the axis aligned with the other body.

If Earth were just a loose bunch of pebbles and much less massive, the ones near the Moon would tend to rise toward the Moon and the ones far from the Moon would fall away. (There is a secondary effect of orbital velocity since they are now in different orbits. They will pull away orbitward and antiorbitward. This is how Saturn's ring was formed. A moon was pulled apart by gravitational tides.)The Roche Limit is the radius from a body, nearer than which a body will break up due to this pull:

http://en.wikipedia.org/wiki/Roche_limit

In our case, rather than a big planet tearing up a small moon, it is the moon applying the same forces to Earth. It fails to tear it apart, but the forces are there.

 

[halli]

Well Dave

I do not doubt you. I will have to go back to the drawing board :)

I am a family man and there is not always a free quality time for physics available. But I will use the time I have to "play" the next week or so to "digest" this, and looking forward to.

I might pop up later with another naive comment or opinion on this matter, I hope you will still be there to correct me.

Best regards

halli

 

[DaveC426913]

Well Dave

I do not doubt you. I will have to go back to the drawing board :)

I am a family man and there is not always a free quality time for physics available. But I will use the time I have to "play" the next week or so to "digest" this, and looking forward to.

I might pop up later with another naive comment or opinion on this matter, I hope you will still be there to correct me.

Best regards

halli

I am a family man too. Well, my kids are almost grown now. I'm an armchair physics buff who got everything from reading. And all I know is that I know so little.

 

[mrspeedybob]

Here is a video that explains it pretty clearly...



Look at time-stamp 47:00 through about 52:30.