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Orbital speed of Earth
19 years 11 months ago #11868
by Jim
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I need to edit my last post to read the Earth is nearer the sun at full moon-not the moon is nearer.
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19 years 11 months ago #11869
by Jim
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Are the magnetudes of the tidal bulges caused by the moon and sun proportional to the distance from Earth to them? As you explained in another post the diameter of Earth is the same in either calculation so is the moon making a bigger tidal bulge-how do you calculate that proportion? The diameter of Earth is 1.3x10E7 meters and the sun is 1.5x10E11m from Earth. How does that make a bulge almost as big as 1.3X10E7m and 4x10E8m? You did say the diameter of Earth was a factor in this-right?
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19 years 11 months ago #11870
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 />Are the magnetudes of the tidal bulges caused by the moon and sun proportional to the distance from Earth to them?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">They are inversely proportional to the cube of distance.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">As you explained in another post the diameter of Earth is the same in either calculation so is the moon making a bigger tidal bulge-how do you calculate that proportion?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">The gravitational force on Earth with mass m caused by another mass M at a distance R is GmM/R^2. If Earth's radius is r, then the gravitational force on the front side of Earth is GmM/[R-r]^2, and on the back side of Earth is GmM/[R+r]^2. So the excess force on the front side over the back side is (using some algebra and some reasonable approximations such as R >> r): 4GmMr/R^3.
So that is the size of the small force leading to a tidal bulge. Note that the albebra shows why it must be inversely proportional to R^3. When we calculate this force acting in one direction over several hours (because the solid Earth spins), and allow for Earth's resistance to deformation, we arrive at the result that the tidal bulge from the Moon is typically three feet (relative to the Earth's center); whereas that from the Sun (farther away but much more massive) is 2.2 times smaller, or a bit over one foot.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">The diameter of Earth is 1.3x10E7 meters and the sun is 1.5x10E11m from Earth. How does that make a bulge almost as big as 1.3X10E7m and 4x10E8m?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">You lost me there. Solid-body tidal bulges are tiny -- a meter or less. -|Tom|-
<br />Are the magnetudes of the tidal bulges caused by the moon and sun proportional to the distance from Earth to them?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">They are inversely proportional to the cube of distance.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">As you explained in another post the diameter of Earth is the same in either calculation so is the moon making a bigger tidal bulge-how do you calculate that proportion?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">The gravitational force on Earth with mass m caused by another mass M at a distance R is GmM/R^2. If Earth's radius is r, then the gravitational force on the front side of Earth is GmM/[R-r]^2, and on the back side of Earth is GmM/[R+r]^2. So the excess force on the front side over the back side is (using some algebra and some reasonable approximations such as R >> r): 4GmMr/R^3.
So that is the size of the small force leading to a tidal bulge. Note that the albebra shows why it must be inversely proportional to R^3. When we calculate this force acting in one direction over several hours (because the solid Earth spins), and allow for Earth's resistance to deformation, we arrive at the result that the tidal bulge from the Moon is typically three feet (relative to the Earth's center); whereas that from the Sun (farther away but much more massive) is 2.2 times smaller, or a bit over one foot.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">The diameter of Earth is 1.3x10E7 meters and the sun is 1.5x10E11m from Earth. How does that make a bulge almost as big as 1.3X10E7m and 4x10E8m?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">You lost me there. Solid-body tidal bulges are tiny -- a meter or less. -|Tom|-
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19 years 11 months ago #10998
by Jim
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I see you are using a lot of math to get from point A to point B that is not transparent. The meters in r and the meters in R are not at all as interchangable as the math allows don't you think?
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19 years 11 months ago #11000
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 see you are using a lot of math to get from point A to point B that is not transparent. The meters in r and the meters in R are not at all as interchangable as the math allows don't you think?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">You can't ask "How do you calculate that proportion" if you haven't yet mastered even algebra, which is the math of symbols and equations.
If you just need the intermediate steps spelled out for you, that is one thing. If you understand F = ma (Newton's second law) but not F = GmM/R^2 (Newton's universal law of gravitation), then I would be at a loss for how to answer your "how do you calculate" question.
Would someone else like to have a go at it? (Show the steps in the formula manipulation and/or explain Newton's universal law and/or explain something else, as Jim wishes.) -|Tom|-
<br />I see you are using a lot of math to get from point A to point B that is not transparent. The meters in r and the meters in R are not at all as interchangable as the math allows don't you think?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">You can't ask "How do you calculate that proportion" if you haven't yet mastered even algebra, which is the math of symbols and equations.
If you just need the intermediate steps spelled out for you, that is one thing. If you understand F = ma (Newton's second law) but not F = GmM/R^2 (Newton's universal law of gravitation), then I would be at a loss for how to answer your "how do you calculate" question.
Would someone else like to have a go at it? (Show the steps in the formula manipulation and/or explain Newton's universal law and/or explain something else, as Jim wishes.) -|Tom|-
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19 years 11 months ago #12078
by Jim
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Well, as it usually is the way things go this is getting way off the point. Can we restart from a few laps back? Or lets get bogged down in calling me stupid and lacking in book learning.
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