# The gravitational tug of the Sun and the Moon on Earth - a question

#### shirvanshah

Hi! When it comes to physics, I'm a total layperson (so, please, be kind ;-)), I actually study English, but recently my interest in physics has been growing more and more. So, yesterday I started watching the first of Feynman's Cornell University lectures. And that led me to the question: why there are two high tides each day? What's especially striking to me is that there are two high tides simultaneously, I mean, on both "sides" of the Earth at the same time. So, I tried to get some insight on that, and I've found the following paper: http://www.jal.cc.il.us/~mikolajsawicki/tides_new2.pdf [Broken]
And what is quite perplexing here is the following passage:
Consider the point C on Earth closest to the Sun and the point F on a far side of Earth. The
Sun pulls harder on a unit mass at the point C, not as hard on a unit mass at Earth center O,
and weaker yet on a unit mass at point F. The acceleration as of Earth as a whole in free fall
towards the Sun is determined by the gravitational pull of the Sun on Earth’s center. Hence
the unit mass at C has a tendency to accelerate towards the Sun with acceleration as + as,
i.e. more than the center of Earth, while a mass at the far side F has a tendency to accelerate towards the Sun with acceleration as - as, i.e. to lag behind the center of Earth.
This difference in Solar gravitational pulls would have lead to a disintegration of Earth, had
Earth’s own gravity been to weak to hold Earth together. To an observer on Earth it would have looked like rocks at point C and F were lifted away from the surface of Earth.

And I believe it's a very similar case with the Moon's gravitational pull and this is why there are two high tides. So, my question is this - why the rocks at point F would be lifted away from the Sun and not pulled in the direction of Sun, that is, towards Earth's center? Why the direction of that force is not in the same direction as the gravitational pull of the sun (towards the Sun's center) but, as it seems, away from it?
I probably should have payed more attention to science in high school, but I would really appreciate a kind and an easy-to-grasp answer ;-).

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#### shirvanshah

Yes, but still I don't quite understand why the force of the Moon's gravitational pull on the side farthest from the moon is directed, as it seems, away from the moon, and not towards the moon and towards the Earth's center. It seems to me that on the side farthest from the Moon the force of the Moon's gravity and the force of the Earth's gravity have the same direction, so why they are subtracted rather than added?
I realize that there must be some profound misconception in what I wrote above, and I would be really glad to know what the misconception is exactly about.

#### DaveC426913

Gold Member
It's not that rock F is pulled away from the Sun, it's that Earth is pulled away from rock F.

Think of it this way:

Three rocks, a, B and c are resting together, 93 million miles from the sun:

(SUN) . . . . . . . . . . aBc

a is 1kg
B is 10^23 kg
c is 1kg

a is pulled most strongly toward the Sun;
B is pulled toward the sun but less strongly than a;
finally, c is pulled toward the sun but less strongly than B.

The net effect is that
aBc separate neatly and evenly to become a B c.

See?

#### shirvanshah

Hm... Is it really that simple? Well, it seems I get it now, thank you very much :D

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