Why Does the Moon Cause Tides despite Sun’s Greater Force?

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    Moon Tides
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Discussion Overview

The discussion revolves around the question of why the moon causes tides on Earth despite the sun having a greater gravitational force. Participants explore the mechanics of tidal forces, the relative distances and masses of the moon and sun, and the implications of gravitational pull differences.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants note that while the sun has a greater gravitational effect due to its mass, the moon's proximity results in a greater tidal influence.
  • One participant mentions that the tidal effect is proportional to the inverse cube of the distance to an object, suggesting that the sun's effect is diminished by its greater distance.
  • Another participant explains that tides are caused by the difference in gravitational pull across the Earth, emphasizing the importance of calculating the gravitational pull on different sides of the Earth.
  • There is a discussion about the calculations involving distance squared versus distance cubed, with some participants expressing confusion over the correct approach to understanding tidal forces.
  • One participant highlights that the variation in gravitational pull, rather than the absolute gravitational force, is what causes the tidal "stretching" effect.
  • Another participant provides specific distances from the Earth to the moon and sun to illustrate why the moon has a larger tidal influence.
  • A later reply introduces the idea that the sun also exerts a significant influence on the moon's motion, adding another layer to the discussion.

Areas of Agreement / Disagreement

Participants generally agree that the moon has a greater tidal effect due to its proximity, but there are competing views regarding the calculations and the relative contributions of the sun and moon to tidal forces. The discussion remains unresolved on some technical aspects, particularly regarding the calculations of gravitational pull.

Contextual Notes

Participants reference various calculations and assumptions about distances and gravitational effects, but there are unresolved mathematical steps and potential misunderstandings regarding the application of distance squared versus cubed in the context of tidal forces.

Parth Dave
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I know the moon causes tides, but I am unsure why. I did a quick calculation and found that the sun has a greater gravitational effect on the Earth than the moon does. My question is, how come the sun does not affect tides despite the fact that the gravitational force is stronger than the moon?
 
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Parth Dave said:
My question is, how come the sun does not affect tides despite the fact that the gravitational force is stronger than the moon?
It does. This is why the tides are strongest when the moon and sun are aligned.
 
The tidal effect is proportional to the inverse cube of the distance to an object. The sun is about 30 million times the mass of the moon and about 400 times the distance. That means that the effect of the sun is 30000000/64000000 or about half of the effect of the moon
 
Yes, choron is correct.

To make it simple.

Tides are caused by the difference of gravitation pull from one point to the next. So to calculate it you would figure the gravitation pull on the far side of the Earth due to the moon and then calculate the pull on the near side of the Earth due to the moon. Then you can do the same for the sun, and you will clearly see the difference.

F= G m1m2/r2

It is just a plug and chug from there. After that you can clearly see high tides low tides ect... (although it is rather easy to visualize).

Nautica
 
Oh so basically the moon causes a greater difference in gravitational pull than the sun does.
 
Parth Dave said:
Oh so basically the moon causes a greater difference in gravitational pull than the sun does.

Agreed, the moon is closer. Do the distance squared thing and you will find it has much more tidal effects than the sun.
 
here are some nice illustrations:

http://oceanlink.island.net/oinfo/tides/tides.html

http://maritime.haifa.ac.il/departm/lessons/ocean/lect19.htm
 
Last edited by a moderator:
Chronos said:
Agreed, the moon is closer. Do the distance squared thing and you will find it has much more tidal effects than the sun.
I did the distance *squared* thing, and the sun's pull came out much larger.

But chronon showed the calculation for distance *cubed*, and then the moon's pull is about half that of the sun... So which one is it? And if the latter, where is the extra distance multiplication coming from? :confused:
 
It's cubed.

If the position of the Moon, or Sun relative to the Earth centre is given by [tex]\vec{s}[/tex] and the position of an ocean point is given by [tex]\vec{r}[/tex] then the tidal acceleration is given by:

[tex]\vec{a} = GM \left( \frac{\vec{s} - \vec{r}}{| \vec{s} - \vec{r} |^3} - \frac{\vec{s}}{| \vec{s} |} \right)[/tex]

expanding of the first term between brackets, filling in, and neglecting small terms gives something that is proportional to ][tex]\frac{GM \vec{r}}{s^3}[/tex]

more details later..maybe
 
  • #10
alpha_wolf said:
I did the distance *squared* thing, and the sun's pull came out much larger.

But chronon showed the calculation for distance *cubed*, and then the moon's pull is about half that of the sun... So which one is it? And if the latter, where is the extra distance multiplication coming from? :confused:
What matters in causing the tides is not the gravitational pull (which goes as distance squared) but the change in gravitational pull across the planet. It's that change that goes as distance cubed.

While it's true that the sun's pull is much stronger, the moon, being closer, exerts a greater tidal effect since its pull (think "force per unit mass") varies more from one side of the Earth to the other. Again, it's the variation in gravitational pull that causing the "stretching" effect that is the tides.
 
  • #11
alpha_wolf said:
I did the distance *squared* thing, and the sun's pull came out much larger.

But chronon showed the calculation for distance *cubed*, and then the moon's pull is about half that of the sun... So which one is it? And if the latter, where is the extra distance multiplication coming from? :confused:
What's the Sun's gravitational pull at the point on the Earth closest to the Sun? ditto, furthest from the Sun? What's the difference?

Now do the same calculations, but use 'Moon' instead of 'Sun'.

For the nitpickers, assume mid-solar (or mid-lunar) eclipse and a point on the Earth's equator.
 
  • #12
Read this one again:
nautica said:
To make it simple.

Tides are caused by the difference of gravitation pull from one point to the next. So to calculate it you would figure the gravitation pull on the far side of the Earth due to the moon and then calculate the pull on the near side of the Earth due to the moon. Then you can do the same for the sun, and you will clearly see the difference.

F= G m1m2/r2

It is just a plug and chug from there. After that you can clearly see high tides low tides ect... (although it is rather easy to visualize).
The diameter of the Earth is 13,000km. The distance from the Earth to the moon is 385,000km, and the distance from the Earth to the sun is 150,000,000km.

So the Earth's diameter is .0087% of the distance to the sun, but 3.4% of the distance to the moon. That's why the moon has a bigger tidal influence.
 
  • #13
Thanks everyone.
 
  • #14
on the same note... the Sun accelerates the Moon more than the Earth does :) really!
 

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