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Gravitational And Tidal Forces

Created: Mon Jan 10 2000
Last updated: Sat Jun 23 2001
 

Orbital Forces

The dominant effect of gravity is first and foremost the orbital forces that keep planets and their moons in their relative orbits around each other and around the Sun. The forces experienced by planet 1 and planet 2 are the same but their accelerations are in inverse proportion to their masses.
    FOrbital = m1a1 = m2a2 = GNm1m2/(d1-2)2
GN is Newton's gravitational constant (big G), and d1-2 is the distance between the centers of planet 1 and 2.

Weight

Another common effect of gravity is the phenomenon of weight. Weight is a force arising from the acceleration towards the center experienced by any mass (m) at the surface of the Earth. It is commonly denoted g (small g).
    FWeight = mg = mGNmEarth/rEarth2
Here rEarth is the radius of the Earth.

Tidal Forces

Tidal forces are a third and more subtle class of gravitational effects resulting from the actual finite size of gravitating objects (planets, not points!). In the case of the Moon's attraction on the Earth, The acceleration of the Earth'center towards the Moon is different than that of a point on the Earth's surface. As the accelerations are different, the surface points are going to feel a relative acceleration with respect to the center of the Earth. For an object of mass m on the surface, this is the tidal force.
Earth Mass 5.95×1024 kg
Earth Radius 6370 km
Moon Mass 7.35×1022 kg
Moon Dist 384400 km
Sun Mass 1.99×1030 kg
Sun Dist 150×106 km
    FTide = matide = mGNmMoon((1/dEarthCenter-Moon)2 - (1/dEarthSurface-Moon)2)
To first approximation, this is at most
    FTide = matide = 2mGNmMoonrEarth/(dEarth-Moon)3

Taking into account the finite radius of the Earth, rEarth, the tidal force is effectively an inverse cube law. The overall factor of 2 is another measurable reminder of the original inverse square law. We see that in magnitude,

     FTide/FWeight = 2mMoon/mEarth(rEarth/dEarth-Moon)3 = 1.12×10-7
Similarly, for the tidal force due to the Sun
     FSunTide/FMoonTide = mSun/mMoon(rEarth-Moon/dEarth-Sun)3 = 0.455

On those rare occasions where the Moon and Sun tidal accelerations are aligned

     FTide/FWeight = 1.12×10-7(1 + 0.455) = 1.63×10-7

Tidal forces on Earth are very, very small. Moreover they vary over time depending on the position of the Moon and the Sun. Today we can calculate the position of the Moon and the Sun with great accuracy, but historically, calculating the Moon's position from first principles was a difficult challenge. More than 60 years elapsed between Newton's Principia (1687) and Clairaut's Theorie de la Lune (1752).

It is important to realize that the Earth does not take tidal forces passively. In addition to ocean tides, atmospheric and earth tides are observed. The Earth strains in response to tidal forces and the actual effects measured at any given point are a combination of these forces and the Earth's response. Even though the driving tidal forces can be calculated accurately, the combined result is hard to predict accurately, to this day.