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# 5.26 Calculating Tidal Forces

If the Sun keeps the Earth in its orbit why is it the Moon that causes tides? To understand this, we need to compare the strength of the gravitational force of the Sun and the Moon on the Earth. The force of gravity is proportional to the mass of two bodies and inversely proportional to the square of the distance between them. In this equation, there is also a numerical constant G. The subscripts S, E, and M represent the Sun, Earth, and Moon, respectively. The force of gravity caused by the Sun on the Earth is:

FSE = (G MS ME)/RSE2

The force of gravity caused by the Moon on the Earth is:

FME = (G MM ME)/RME2

Tidal force as apparance of Gravity force. Click here for original source URL.

Some quantities will cancel out if we take the ratio of the Sun’s force on the Earth to the Moon’s force, or, FSE/FME. In general, when you are doing algebra problems, you should wait until you have simplified the relations as much as you can before plugging in numbers and solving the equation.) The ratio of these forces is:

FSE/FME = MS/MM (RME/RSE)2

Now we can insert the values to get the answer. The masses of the Sun and Moon are MS = 2.0 × 1030 kilograms and MM = 7.4 × 1022kilograms. The distances from the Earth are RSE = 1.5 × 108 kilometers (1 Astronomical Unit or A.U., by definition) and RME = 3.8 × 105kilometers. We get the result FSE/ FME = 173. So the Sun’s attractive force on the Earth is over a hundred times the size of the Moon’s attractive force. There is no question that the Sun controls the orbit of the Earth.

So how can the Moon cause the tides on the Earth? The gravitational force depends on the inverse square of distance. So the gravitational force on the near side of a large object is larger than the gravitational force on the far side – the result is a stretching force called a tidal force. Tides are caused by the difference between the gravitational force on one side of an object and the other side.

We can approximate the strength of the tidal force by taking the gravitational force we just calculated and multiplying it by the ratio of the front-to-back distance of the Earth divided by its distance from the Sun or Moon. (You’d need calculus to derive a precise result.) Let’s call the Earth’s diameter DE. For the stretching of the Sun on the Earth, we get:

DE/RSE = 12,700 / 1.5 × 108 = 8.5 × 10-5

For the stretching of the Moon on the Earth, we get:

DE/RME = 12,700 / 384,000 = 0.033

The ratio of these two numbers is 390. The size of the Earth is a much larger fraction of the Earth-Moon distance than it is of the Earth-Sun distance.

In other words, the difference between the Moon’s gravitational force at the near side of the Earth and the far side is a much larger fraction of the total force exerted on the Earth. This difference causes tides.