# Is it possible to apply Newton's equation of gravitation to attractive force between the moon and the ocean? [closed]

Someone was asking me how much force is acting on the waves when the moon is closest to the earth. Of course the first thing I tried to do is apply Newton's equation for universal gravitation

$$F = G \frac{m_1 m_2}{r^2}$$ but the formula does not seem to take into consideration the geometry of either the ocean or the moon. Further more, it seems intuitive that most of the force is experienced on a patch of the ocean and negligible elsewhere.

I am confused whether we should take into consideration of the geometry of these objects. If so, should the equation be modified to calculate the attractive force for this scenario? If not, how can we calculate this attractive force?

• If you make the approximation that the oceans are uniform and therefore spherically symmetric, you can just treat the oceans as a single point mass at the centre of the earth (proof here). – lemon Jun 6 '15 at 1:03
• Waves are mainly generated by the wind, so force of gravity from the Moon will have negligible effect on them. Or are you referring to the tides, which are mainly caused by the Moon, and are a more gradual change oscillation of sea level. – fibonatic Jun 6 '15 at 2:51
• "How much force is acting" is a very fuzzy concept. The earth is always pulling on the water... that is a force acting too. There are several questions / answers about tides etc on this site - see for example physics.stackexchange.com/q/118460/26969 or the links given in physics.stackexchange.com/q/161454 – Floris Jun 6 '15 at 4:42

The mass of all the oceans on the surface of the Earth is estimated to be 1.35 * 10^18 metric tons, or 1.35 * 10^21 kilograms.

The mass of the Moon is estimated to be 7.35 * 10^22 kilograms.

The distance from the Moon to the Earth at perigee (the closest distance) is 363,104 kilometers, or 363,104,000 meters.

If you assume the oceans cover the entire earth (rather than 71%) in a uniform layer, you may consider their gravitation to originate from a point in the center of the Earth (see the link in lemon's comment for a proof). Likewise, the gravitation of the Moon can be taken to originate at a point in its center.

The gravitational field of the Moon applies to the entirety of Earth's oceans, not just to a patch. Tidal forces are caused by the difference in gravitational force distance between oceans facing the Moon and oceans on the far side of the Earth.

Adding the interior radii of the Moon (1,737 kilometers) and of the Earth (6,371 kilometers) to the distance at perigee, changing kilometers to meters, and plugging the other amounts into the formula for universal gravitation, with Newton's gravitational constant expressed as force in newtons:

F = (6.67 * 10^-11)*(1.35 * 10^21)*(7.35 * 10^22) / (363,104,000 + 1,737,000 + 6,371,000)^2 = 4.80 * 10^16 newtons

But 4.80 * 10^16 newtons gravitational attraction between the Moon and the oceans means little by itself, because the Earth's gravitational field also acts on the oceans. Obviously, the Earth's attraction for the oceans is much greater than the Moon's, since the Earth is larger and the oceans are closer to the Earth than to the Moon. Therefore, the Moon's pull on the oceans could be represented as a fraction of the Earth's pull on the oceans, and you'd get an idea of the magnitude of the power that causes tides.

The Earth's mass is estimated to be 5.97 * 10^24 kilograms. I'm going to use the acceleration of gravity at the Earth's surface to compute the Earth's force on the oceans: F = (5.97 *10^24) * 9.81 = 5.86 * 10^25 newtons.

So the Moon's pull on the oceans is (4.80 * 10^16) / (5.86 * 10^25) = 8.19 * 10^-10 of the gravitational force holding the oceans on the Earth's surface. A very tiny percentage of the total combined gravitational force on the oceans causes the tides.

The statement of Newton's law you have given is specifically about the gravitational attraction between point masses. It also comes to mean, through the Shell theorem the force between a spherically symmetric body and another outside the first body: we replace the spherical body with a point at its center.

For other geometries you calculate the force through linear superposition. That is, you consider them to be made up of points, and then sum up all the point mass contributions through a superposition sum or integral.