- Thank you received: 0
tidal effects
- tvanflandern
- Topic Author
- Offline
- Platinum Member
Less
More
20 years 3 months ago #10334
by tvanflandern
Reply from Tom Van Flandern was created by tvanflandern
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Jim</i>
<br />According to Newton about 1/5 of the force that causes tides is supplied by the sun.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">That would be 2/5, not 1/5.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">How does that much force develop if it is the diameter of Earth causing the effect?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Earth's diameter doesn't cause the effect. But it does modulate it. If Earth were a point mass, the forces on it would just be the inverse-square pull of gravity. But because Earth is much larger than a point, the inverse-square force on its center is a bit different from the inverse-square force on some surface point. If Earth had no cohesion to hold it together, the slightly different pulls on different parts of the Earth would pull the Earth apart, with each of the bits going into a slightly different orbit.
But because Earth has enough cohesion to stay in one piece, the different tugs on different parts just cause the Earth's shape to change slightly. This is called a "tide". The bigger Earth's diameter is in proportion to the distance to the source mass doing the pulling, the greater is the difference in gravity on Earth's parts, making for a greater tide. If r is Earth's radius and R is the distance to the tide-raising body, then the strength of the tides is the force of gravity multiplied by r/R. That extra power of R is where the inverse-cube-of-distance law comes from.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">I know it is true the third power law points this out very clearly, but since the tide is a second power effect what does invoking a bogus third power mean?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">If you think the third power law is bogus, you will have a very difficult time explaining why the Moon orbits Earth at all. Do a quick calculation of mass over distance squared in any consistent units, and you will verify that the Sun's inverse-square gravitational pull on the Moon is twice as strong as Earth's gravitational pull on the Moon. So why should the Moon orbit Earth instead of having an independent orbit around the Sun, given that the Sun's gravity dominates?
The answer lies in understanding tidal forces and that extra power of R. Chapter six of "Dark Matter, Missing Planets and New Comets" is devoted to explaining the basics of orbits and the solution to this paradox. -|Tom|-
<br />According to Newton about 1/5 of the force that causes tides is supplied by the sun.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">That would be 2/5, not 1/5.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">How does that much force develop if it is the diameter of Earth causing the effect?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Earth's diameter doesn't cause the effect. But it does modulate it. If Earth were a point mass, the forces on it would just be the inverse-square pull of gravity. But because Earth is much larger than a point, the inverse-square force on its center is a bit different from the inverse-square force on some surface point. If Earth had no cohesion to hold it together, the slightly different pulls on different parts of the Earth would pull the Earth apart, with each of the bits going into a slightly different orbit.
But because Earth has enough cohesion to stay in one piece, the different tugs on different parts just cause the Earth's shape to change slightly. This is called a "tide". The bigger Earth's diameter is in proportion to the distance to the source mass doing the pulling, the greater is the difference in gravity on Earth's parts, making for a greater tide. If r is Earth's radius and R is the distance to the tide-raising body, then the strength of the tides is the force of gravity multiplied by r/R. That extra power of R is where the inverse-cube-of-distance law comes from.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">I know it is true the third power law points this out very clearly, but since the tide is a second power effect what does invoking a bogus third power mean?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">If you think the third power law is bogus, you will have a very difficult time explaining why the Moon orbits Earth at all. Do a quick calculation of mass over distance squared in any consistent units, and you will verify that the Sun's inverse-square gravitational pull on the Moon is twice as strong as Earth's gravitational pull on the Moon. So why should the Moon orbit Earth instead of having an independent orbit around the Sun, given that the Sun's gravity dominates?
The answer lies in understanding tidal forces and that extra power of R. Chapter six of "Dark Matter, Missing Planets and New Comets" is devoted to explaining the basics of orbits and the solution to this paradox. -|Tom|-
Please Log in or Create an account to join the conversation.
20 years 3 months ago #10230
by Jim
Replied by Jim on topic Reply from
The Earth and moon are in the same orbit around the sun so why is it odd to have them in orbit about each other while they orbit as a pair unit? Why would the moon be in a different orbit anyway? The moon can orbit the sun at one AU even if the Earth is not factored into it's orbit.
Please Log in or Create an account to join the conversation.
20 years 3 months ago #10919
by Jim
Replied by Jim on topic Reply from
Tom, I notice you are using a factor(r/R) to create a third power effect which will predict a correct tidal effect. Newton used a different factor to get a good predictor. He observed the tidal effect was proportional to the third power of the distance to the moon and the sun. I am not real sure but I think that method is used in modern tidal models that get very near the actual tidal effect any where on Earth. Can you tell me when the r/R factor was first used? thanks
Please Log in or Create an account to join the conversation.
- tvanflandern
- Topic Author
- Offline
- Platinum Member
Less
More
- Thank you received: 0
20 years 3 months ago #11285
by tvanflandern
Replied by tvanflandern on topic Reply from Tom Van Flandern
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Jim</i>
<br />I notice you are using a factor(r/R) to create a third power effect which will predict a correct tidal effect. Newton used a different factor to get a good predictor. He observed the tidal effect was proportional to the third power of the distance to the moon and the sun.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">So both of us arrived at a third power law. Then I don't understand your question. What is different about Newton's approach? -|Tom|-
<br />I notice you are using a factor(r/R) to create a third power effect which will predict a correct tidal effect. Newton used a different factor to get a good predictor. He observed the tidal effect was proportional to the third power of the distance to the moon and the sun.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">So both of us arrived at a third power law. Then I don't understand your question. What is different about Newton's approach? -|Tom|-
Please Log in or Create an account to join the conversation.
20 years 3 months ago #11367
by Jim
Replied by Jim on topic Reply from
Well, we know Newton invented the 3rd power law, but he was observing something different than you are saying. I'm not going to get this right so please correct any errors as you see fit. He observed the 3rd power of the distance to the moon and sun divided by the mass of each was about proportional to the difference of spring and neap tides and so being a very practcal man used that to construct a model of the tides. He also was quite clear in saying the moon perplexed him and gave up on any more attempt to understand it. You are saying the effect the sun has on tides is a result of the radius of the Earth divided by the distance to the sun and that factor also gets the spring and neap tides. To be clear on my view here,I say the sun has no effect on the spring/neap tide cycle and that the cause is the moon only. So, I am at odds with both Newton and you and everyone else on this detail as well as several other details. And also I know I may be wrong in some or all these matters as I have been on other matters in the past.
Please Log in or Create an account to join the conversation.
- tvanflandern
- Topic Author
- Offline
- Platinum Member
Less
More
- Thank you received: 0
20 years 3 months ago #10284
by tvanflandern
Replied by tvanflandern on topic Reply from Tom Van Flandern
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Jim</i>
<br />You are saying the effect the sun has on tides is a result of the radius of the Earth divided by the distance to the sun...<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">That r/R factor multiplies the normal force of gravity, GM/R^2, making the tidal force GMr/R^3. So I am saying the same thing as Newton and everyone else.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">...and that factor also gets the spring and neap tides.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Water tides are a whole different animal. Up to now, the discussion has been about solid body tides. The latter lifts land and water by the same amount, producing no water tides.
The reason for ocean tides is that water is free to flow, whereas land is not. So when the Sun or Moon pulls on the surface of Earth by a different amount than the pull on the center, water responds to the horizontal part of those pulls by flowing. And when an ocean of flowing water runs into a coastline, it pushes water onto the land. Newton didn't have a clue about the actual mechanics of this, which only began to be understood with the work of Sir Harold Jefferys about 1931.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">To be clear on my view here, I say the sun has no effect on the spring/neap tide cycle and that the cause is the moon only.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Do a quick calculation. The gravitational force of the Sun on the Earth is 160 times stronger than that of the Moon. Why would water respond to the Moon's gravity but not the Sun's stronger gravity? Please explain your thinking. -|Tom|-
<br />You are saying the effect the sun has on tides is a result of the radius of the Earth divided by the distance to the sun...<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">That r/R factor multiplies the normal force of gravity, GM/R^2, making the tidal force GMr/R^3. So I am saying the same thing as Newton and everyone else.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">...and that factor also gets the spring and neap tides.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Water tides are a whole different animal. Up to now, the discussion has been about solid body tides. The latter lifts land and water by the same amount, producing no water tides.
The reason for ocean tides is that water is free to flow, whereas land is not. So when the Sun or Moon pulls on the surface of Earth by a different amount than the pull on the center, water responds to the horizontal part of those pulls by flowing. And when an ocean of flowing water runs into a coastline, it pushes water onto the land. Newton didn't have a clue about the actual mechanics of this, which only began to be understood with the work of Sir Harold Jefferys about 1931.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">To be clear on my view here, I say the sun has no effect on the spring/neap tide cycle and that the cause is the moon only.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Do a quick calculation. The gravitational force of the Sun on the Earth is 160 times stronger than that of the Moon. Why would water respond to the Moon's gravity but not the Sun's stronger gravity? Please explain your thinking. -|Tom|-
Please Log in or Create an account to join the conversation.
Time to create page: 0.902 seconds