plz tell the formula to calculate the force with which moon attract very kilogram of water in sea or riven
Answers
Answer:
Explanation:
The moon (or any gravitational body, for that matter) exerts agravitational pull that is stronger the closer you are to it. Since the Earth has a physical size, the gravitational pull on the Earth is stronger on the side facing the moon than it is on the side facing away from the moon.
Because the Earth is being pulled on more strongly on one side than from the other, it is effectively being stretched by the difference in the magnitude of gravitational force from the moon. Most of the Earth, of course, is solid (it is made of rock), but the oceans are not, and, as a result, they are much easier to stretch than the solid Earth. Therefore, the liquid is pulled toward the moon more strongly on the side facing the moon, causing it to accumulate there, while the side opposite the moon, not feeling the moon's pull as strongly, stays on the side opposite the moon. Because the ocean is still held by the Earth's gravity as well, it does not come off the surface, and instead sea level on the side of the Earth facing the moon and on the side opposite the moon rises. At the same time, sea level falls due to a relative lack of water halfway around the Earth from the moon. This change in sea level is what we call tides.
Also of note: the Sun has about the same tidal influence on the Earth as the moon does.
If you are good at math, here is the equation that determines gravity:
F = G*m*M/(r2),
F = Force due to gravity,
G = a constant which has to be looked up
m = the mass of the moon
M = the mass of the Earth
r = the distance between the Earth and Moon.
G:The gravitational constant, also known as the universal gravitational constant, or as Newton's constant, denoted by the letter G, Its measured value is
6.674 08 x 10-11 m3 kg-1 s-2.
The Earth has a diameter of 12,800 kilometers.
Hope it helps you...
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Here is the mathematical explanation of the tides. Call the distance from the Earth's center to the Moon's center, D. Call the radius of the Earth R. Anything on the Earth's surface where the Moon is directly overhead is then at (D - R) distance from the Moon's center. Using Newton's gravitational equation F = G * M1 * M2 / R2 we know what the FORCE on it is. But Newton also told us that F = M * a. So the Moon attracts it (upward, there) with an ACCELERATION of F / M1 = atide = G * Mmoon / (D - R)2. THAT describes the acceleration that is caused in the water (or you!) when the Moon is exactly overhead. However, the entire Earth itself is ALSO being attracted, accelerated by the Moon at F / Mearth = aearth = G * Mmoon / (D)2. The effective tidal acceleration on our object is then just the DIFFERENCE of these accelerations, or (G * Mmoon) * ((1 / (D - R)2) - (1 / D2)). Simplifying, this becomes G * Mmoon * (2 * D * R - R2) / (D2 (D - R)2).
Since R is small compared to D (about 1/60 of it) this is nearly equal to G * Mmoon * (2 * D * R) / D4, or (a constant) / D3. Tidal acceleration is therefore approximately proportional to the CUBE of the distance of the attracting body. If the Moon were only half the distance as high above the Earth, the tidal accelerations would be EIGHT TIMES as great...