hstarbuck wrote:the Sun's pull on the Earth is ~18 times that as the Moon's pull on the Earth yet the Moon has a greater effect on the Earth's tides. I am thinking that the difference in force from the front side to the back side of the Earth is greater for the Moon's pull but I have not done this calculation. Basically if you think of an inverse square graph and go out further into the field than the difference in force is less for same distance (Earth's diameter).
The Sun's pull on the Earth is a lot more than ~18 times that as the Moon's pull on the Earth. (Maybe more like 180 times.)
But the difference in force from the front side to the back side of the Earth is indeed greater for the Moon's pull as you state.
Tidal forces drop off with the cube of the distance not the square. Since volume also increases with the cube of the distance and the moon just covers the sun during a solar eclipse it turns out that the difference between lunar tides & solar tides is just due to their different densities.
The moon is 2.4 times denser than the sun on average; hence, it's tidal effect is 2.4 times larger.
hstarbuck wrote:If the moon and Earth were removed from the Solar system and placed in empty space with the same velocities and distance a month would still be a month right? I calculate 27.4 some odd days with formula from my physics book. So what does any of this mean? What effect does it have?
Under that situation the moon does take a few minutes shorter time to orbit the earth than it currently does (due to the orbiting around the sun).
If such a 'free range' moon were out at a distance of 1.5 million km then it would take about 211 days to orbit the earth.
However, artificial satellites out at 1.5 million km in the
L1 & L2 Lagrange points actually take 365 days to orbit the earth due to the influence of the sun.
[quote="hstarbuck"]the Sun's pull on the Earth is ~18 times that as the Moon's pull on the Earth yet the Moon has a greater effect on the Earth's tides. I am thinking that the difference in force from the front side to the back side of the Earth is greater for the Moon's pull but I have not done this calculation. Basically if you think of an inverse square graph and go out further into the field than the difference in force is less for same distance (Earth's diameter).[/quote]
The Sun's pull on the Earth is a lot more than ~18 times that as the Moon's pull on the Earth. (Maybe more like 180 times.)
But the difference in force from the front side to the back side of the Earth is indeed greater for the Moon's pull as you state.
Tidal forces drop off with the cube of the distance not the square. Since volume also increases with the cube of the distance and the moon just covers the sun during a solar eclipse it turns out that the difference between lunar tides & solar tides is just due to their different densities.
The moon is 2.4 times denser than the sun on average; hence, it's tidal effect is 2.4 times larger.
[quote="hstarbuck"]If the moon and Earth were removed from the Solar system and placed in empty space with the same velocities and distance a month would still be a month right? I calculate 27.4 some odd days with formula from my physics book. So what does any of this mean? What effect does it have?[/quote]
Under that situation the moon does take a few minutes shorter time to orbit the earth than it currently does (due to the orbiting around the sun).
If such a 'free range' moon were out at a distance of 1.5 million km then it would take about 211 days to orbit the earth.
However, artificial satellites out at 1.5 million km in the [b]L1 & L2 Lagrange points[/b] actually take 365 days to orbit the earth due to the influence of the sun.