# Tides due to the moon vs. the sun?

1. Oct 31, 2003

### nautica Tides due to the moon vs. the sun???

The acceleration of gravity on the earth due to the sun is 177 greater than gravity on earth due to the moon.

Why are the tides predominantly due to the moon and not the sun, in spite of this number???

Nautica

2. 3. Oct 31, 2003

### mathman I can't give you any numbers, but the moon is so much closer that its effect on the oceans is much larger than that of the sun.

4. Oct 31, 2003

### russ_watters ### Staff: Mentor

To expand, tides are due to the DIFFERENCE in gravitational force between the sun/moon and the near and far sides of the earth. Since the sun is so far away, the near and far sides of the earth are very close to the same distance from the sun - so not much tidal force.

5. Oct 31, 2003

### Janus Staff Emeritus
To expand further, tidal force falls of by the cube of the distance while g-force falls off by the square of the distance.

Thus, the sun is 26730000 times more massive than the sun and 400 times the distance.

therefore its g-force on the Earth is 26730000/400² = 167 times that of the moon's, but its tidal effect is 26730000/400³ = 0.4 times, or a little less than 1/2 that of the moon's/

6. Oct 31, 2003

### nautica The sun's effect is only 1/2 that of the moon.

But the 177 x's is based on how close the moon is to the earth relative to the distance of the sun to the earth - distance has already been taken into consideration. So, I do not see how distance from the sun to the earth vs the moon to the earth makes a difference in the effect of the tides.

Thanks
Nautica

7. Oct 31, 2003

### Janus Staff Emeritus
The near side of the Earth is 12756 km closer to both the sun and the moon than its far side.

The sun is 149,000,000 million miles away from the Earth so the distance difference is about .00856 %. This means that the difference in force accross the Earth varies by about .017%

The moon is 384000 km away, so the % distance difference is 3.3%, and the force difference is about 6%

If the acceleration due to gravity of the sun is 167 time that of the moon, then its tidal effect (the difference in force between far and near side of the Earth) is 167 * .017% = 2.839%, when compared to the moon's, or again, about 1/2.

8. Oct 31, 2003

### Hurkyl Staff Emeritus
Tidal forces aren't caused by the strength of gravitational attraction; they're caused by the differences in gravitational attraction at different points on an object.

Here's an exercise for you.

We know the mass of the sun is 1.99 * 10^30 kg.
The mass of the moon is 7.35 * 10^22 kg.
The earth's radius is 6.38 * 10^6 m.

(assume all of these values are exact)

At a particular point in earth's orbit, its center is 1.50 * 10^11 m from the center of the sun. Compute the acceleration due to the sun's gravity at the point of the earth nearest to the sun, and the acceleration at the point of the earth farthest from the sun.

The tidal stress on the earth due to the sun is (proportional to) the difference of these two accelerations.

At a particular point in the moon's orbit, its center is 3.84 * 10^8 m from the center of the earth. Do the same calculation as you did for the sun. You'll find the tidal stress caused by the moon is greater!

9. Oct 31, 2003

### nautica Is that not what I did, when I determined that the pull of the sun was 177 times greater than that of the moon???

Thanks
Nautica

10. Oct 31, 2003

### Janus Staff Emeritus
No, all you did was calculate the pull of each body on the center of the Earth.

The near side to the sun is 149,000,000 - 6378 km = 148993622 from the sun, and the far side is 149,000,000 + 6378 km form the sun. You need to calculate the acceleration due to gravity at each of these distances and the sun's mass, and then take the difference between the two to get the net force acting across the Earth which causes solar tides.

For the moon, the distances are 384000 - 6378 = 377622 km and 384000+6378 = 390378 km.

If you calculate the net difference for these distances, using the moon's mass, you will find that it is twice as great as that of the Sun.

11. Nov 1, 2003

### Loren Booda Tidal forces are important in relativity's spacetime, where gravitation acts not only radially but angularly between finite masses.

12. Nov 1, 2003

### nautica Nice, I am a little slow, but have now figured out what you are saying.

Thanks
Nautica

13. Nov 1, 2003

### nautica I did what you said and it appears that the difference of gravitional pull on each side due to the moon is 1.068678428 x's stronger on the near side than the far side.

The sun is 1.000170496 greater on the near side than the far side. So how does this account for the fact that the sun only has half of the effect when the numbers show it should have 93% of the affect.

Thanks
Nautica

14. Nov 1, 2003

### russ_watters ### Staff: Mentor

You didn't account for the difference in gravitational force - the number you posted in the beginning which got you confused in the first place. That number is how you relate the actual tidal force caused by each.

So convert those two numbers to percent (chop off the 1 then multiply by 100%), then multiply the sun's by the difference in gravitational force.

15. Nov 1, 2003

### Hurkyl Staff Emeritus
Remember I said compute the difference, not the ratio.

16. Nov 1, 2003

### nautica The difference in the pull of the moon is 2.20 x 10^-6

The difference in the pull of the sun is 1.01 x 10^-6

Which is .457 or roughly 1/2.

You guys are unbelievable.

Thanks
Nautica

17. Feb 24, 2004

### Anne-ke Thanks people!
This information was really useful to me!!

BUT: WHY is it the DIFFERENCE between forces on the far and the near sides of the earth that determines the tides??

Would there not be tides when these forces were (theoretically) equal?
I thought the tides are determined by the difference between the gravitational force and the centrifugal force. And that the differences between the far and the near sides of the earth arise because of the difference in direction of the centrifugal force on both sides...

Can somebody explain me why??

18. Feb 24, 2004

### Kalimaa23 Last edited by a moderator: Apr 20, 2017
19. Feb 24, 2004

### russ_watters ### Staff: Mentor

"Why" gets to be kinda a fuzzy question: there was a force that was useful and it needed a name.

Adding the forces does give you a useful number, its the total force on the object as a whole. But the difference is important too because the difference gives you the internal forces on the object. It may be hard to visualize how two forces pulling in the same direction can result in a force pulling an object apart, but the reason is the internal forces on an object must be consistent. So if you apply two forces to an object, there must be a resultant internal force.
Correct.
Since centrifugal force is equal on all sides (along equal lattitude lines), there is no force imbalance with which to create a tidal force. Centrifugal force does create the bulge at the equator, similar but not quite the same as a tidal bulge.

20. Feb 24, 2004

### Nereid Staff Emeritus
Welcome to Physics Forums Anne-ke!

Assuming you're now clear on the magnitude, and cause, of tides, and are ready to dive into some more fascinating aspects of the Earth-Moon system ...

Consider:
- the Earth's rotational axis is tilted ~23.5o to the plane of its orbital motion about the Sun
- the Moon's orbit in inclined ~5o to the plane of the Earth's equator
- the Moon completes one revolution about the Earth in a time which is much greater than the time the Earth takes to spin on its axis (the Moon's 'year' is longer than the Earth's 'day')
- neither the Earth's orbit around the Sun (actually the Earth-Moon's centre of mass orbit) nor the Moon's orbit around the Earth are circles

... and you've got a lot of extra forces that the 'point masses in circular orbits' ideal case doesn't account for!

It all leads to some really good-fun maths ... and profound long-term effects on the Earth's climate (among other things).

21. Feb 24, 2004

### Loren Booda Nereid,

Do these take into account precession or nonlinearity?