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Physics LibreTexts

5.25 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 gravity 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. We will use the subscripts S, E, and M to represent the Sun, Earth, and Moon. The force of gravity caused by the Sun on the Earth is:

FSE = G MS ME / (RSE)2

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

FME = G MM ME / (RME)2

Some quantities will cancel out when we take the ratio of the Sun's force on the Earth to the Moon's force, 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 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? Gravity depends on the inverse square of distance. So the gravity on the near side of a large object is larger than the gravity on the far side. The tidal force is a stretching force. Tides are caused by the difference between the gravity force on one side of an object and the other side.

We can make a good approximation for the strength of the tidal force by taking the gravity force we have 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. (Calculus is needed to derive the result precisely.) 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.

While the Sun has a larger force on the Earth than the Moon, the Moon has a larger stretching force. Larger by what factor? It is larger by the ratio of 390 to 173, or roughly a factor of two. Even though the Moon controls the Earth’s tides, the Sun is a significant contributor. This is the reason that tides are more extreme near a full Moon or a new Moon, when the stretching forces due to the Moon and Sun line up in the same direction.

The tidal force is a universal consequence of gravity. The force that causes our oceans to move operates elsewhere in the Solar System, and beyond. Even when there is no water to respond to the force, the solid mass of a planet feels the stress caused by this force. Large objects in close proximity exert the strongest tidal forces. Since gravity is a long range force, tides exist on larger scales in the universe. Galaxies experience tidal forces and their shapes can changes as a result. On all scales, gravity shapes the universe.



Schematic of the tidal forces of the Moon on the Earth's oceans. Click here for original source URL.