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What has a greater influence on water, the Sun or the Moon?


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Guest mickgraham
Posted

the question has been the topic of debate amongst my friends and I recently...

Posted

The moon has the greater effect on tides. Even though it's smaller, it's closer. for proof, just see how the tide tables line up with the moon. The high and low tides decrease by 50 min. everyday because the earth and moon are both rotating, at a lose of 50 min. per day.

Posted
the question has been the topic of debate amongst my friends and I recently...

 

 

Hehe i think you forgot the Earh. The earth has the greater influence on water :D hehe j/k *It keeps it all here*

Guest mickgraham
Posted

Thanks for the replies guys. Personally i'd vote for the moon, but without the sun would the water freeze? ... also, i suppose the earth does keep it where it is :P .. but are they influences?

Posted
If you are referring to the tides' date=' I think it is the Moon.

 

If you mean the temperature, it is definitely the sun. :D[/quote']

 

I think you missed his post then. It effectively sums up the effect of both celestial bodies.

Posted

i think the sun , i mean leave the tides. without the sun,th water wud freeze. i mean as the other person said

Posted
i think the sun , i mean leave the tides. without the sun,th water wud freeze. i mean as the other person said

 

The question is, which has a greater influence on water: the sun or the moon? I think it is both because they both affect the water in some way.

 

Perhaps the question should be reworded to specify exactly what type of "influence" is being talked about.

Posted

Hmmm....Without the sun the water would freeze....thats a little bit to strong since we would all be dead anyway.

 

In the spirit of the question. The sun heats the water and causes evaportation. But the moon moves it.....

 

Heating the ocean is one thing, but moving it shows real power. If I had to place a bet. I pick the moon as having the greater influence overall.

 

Bettina

Posted

While I considered the question dull when I first read it (my 1st guess was: The moon: The sun doesn´t induce tides) it gets more interesting the more I think about it - but that´s of course also due to the vague expression "influence".

 

If you consider the ammount of energy stored in the water I´d bet it´s the sun. The direct heating with sunlight should easily exceed the ammount of heat that´s stored in the water due to the internal friction that occurs due to the water moving because of the moon.

 

If you consider gravitational effects only, then things perhaps aren´t as trivial as they seem. The gravitational force on the water can of course be easily calculated:

Sun: F_Sun = dm * M_sun/d_sun² = dm * 332000 / (149,597,870 )² ~= 3*10^5 /2.25*10^16 ~= 10^-11 dm

Moon: F_Moon = dm * M_Moon/d_Moon² = dm * 0.01 / (384,400)² ~= 10^-12 dm

(distances in km, masses in multiples of earth mass, numbers from http://www.solarviews.com/eng/)

The numbers show that the gravitational attraction due to the sun is (didn´t bother to look for a pen or to even use a calculator, so they´re only an approx) 10 times greater than the gravitational attraction due to the moon.

 

Now the question arises: Why does the moon have a visible influence on the water in a sence of tides while the sun hasn't? I see two possible reasons for this:

1) Earth´s rotation. I will not consider this here as this would really force me to grap a pencil :cool:. I also do not expect much of it when compared to the factor 200 from 2).

2) If the whole earth was in a constant gravitational potential there wouldn´t be any tides at all since all of earth matter is accelerated by the same ammount. The tides therefore must stem from either 1) or the inhomogenity of the gravitational field which causes objects nearer to the moon/sun to be attracted more than those that are more distant. The inhomogenity can be described (in an approx, of course) by the derivative of the gravitational field with respect to distance. This leads to:

I_Sun = d/dr F_Sun = 1/r_sun * F_Sun = 1.5*10^-8 * F_Sun = 1.5*10^-7 * F_Moon

I_Moon = d/dr F_Moon = 1/R_Moon * F_Moon = 3*10^-5 * F_Moon

This handwaving argumentation therefore assumes the tidal effects of the moon to be 200 times greater than that of the sun even though the force excerted by the sun is 10 times larger than that of the moon.

 

As a note for those who are confused by my math: I´m perfectly aware that I didn´t use SI units or took care of signs here. However, as I´m only talking about ratios here, this doens´t play a role at all. The assumtion that the inhomogenity of the grav field is responsible for the tides is only the first approach to estimating the effects quantitatively that came to my mind so it´s probably far from perfect.

Posted

Because the mechanical energy has to go somewhere and thermal energy was my first guess where it might go. You shouldn´t take my post as a very qualified one. It was rather an attempt to bring this discussion on a scientific (=quantitative, in this case) level rather than a "I guess it´s the ..."-one.

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