By Joshua Winn, Princeton University
Tidal forces can act on entire galaxies. There’s one called the Porpoise Galaxy. It used to be a normal spiral galaxy, but it got twisted and stretched out into an arc during a close encounter with the more compact, white galaxy to the right. Over the next billion years, the Porpoise will come apart completely, perhaps making a ring around the other galaxy, before they eventually merge together.
Ocean Tides
A more down-to-earth example of tidal forces is ocean tides. The Earth’s gravity pulls on the Moon, and the Moon’s gravity pulls back on the Earth. That means the Moon exerts tidal forces that, left unopposed, would tear the Earth apart by squeezing it along the direction to the Moon. The Earth’s own gravity prevents that from happening. But there’s more to it than that.
Let’s draw little vectors representing the gravitational force of the Moon at different points in the space surrounding the Earth. Close to the Moon, the vectors are longer because the force is stronger. There’s also some variation in direction, as all the vectors point straight at the center of the Moon. But importantly, the Moon is orbiting the Earth. To see the Earth and the Moon sitting still, we must be in a frame of reference that’s rotating along with the orbit, once a month.
That’s okay, we are allowed to do physics in a rotating frame of reference. But the price we pay is that we must insert a fictitious force—the centrifugal force. In this case, the centrifugal force on the Earth points away from the Moon, with a strength such that at the center of the Earth, the centrifugal force cancels the gravitational force exactly. That’s why the Earth is sitting still, in this frame of reference.
This article comes directly from content in the video series Introduction to Astrophysics. Watch it now, on Wondrium.
What If There Were No Ocean Tides?
Let’s add the centrifugal and gravitational forces, and then replot the net force vectors. There’s no net force at the center of the Earth. The net force points toward the Moon, on the near side, and away from the Moon, on the far side, where the centrifugal force is larger than the gravitational attraction. And in between there are sideways forces.
Now, imagine the Earth is a frictionless sphere surrounded by a thin layer of water. What would happen to that water? It would feel these net forces and flow around the surface, to form two bulges, one on the near side, and one on the far side. And then, if that frictionless Earth were to rotate, sliding underneath that layer of water, an observer on the surface would see the ocean rise in height, then fall, rise, and fall again, over the course of a full day. That’s why we observe two high tides, and two low tides, in one day.
Now, the Earth is not a frictionless sphere. There’s lots of friction, and there are continents, underwater mountains, and all kinds of things that make the picture more complicated. That’s why we need tide tables.
Does the Sun Cause Tidal Forces?
One interesting complication is that the tidal forces from the Sun are also significant. That’s why the maximum height of the tide varies with the phase of the Moon. When the Sun, Moon, and Earth are along a line, the Sun and Moon work together and produce unusually high tides; these are called spring tides. When the Sun and the Moon are at right angles, the contrast between high and low tides is reduced, and we call those neap tides.
From the relative heights of the spring and the neap tides you can tell the Sun’s tidal forces are not quite as strong as the Moon’s; they’re only about half as strong. Tidal forces are proportional to the mass of the attracting body, over the cube of the distance. So, the ratio of solar to lunar tidal forces is equal to the Sun’s mass divided by the Moon’s mass, times the cube of the distance to the Moon over the distance to the Sun.
Tidal Force: The Sun vs. the Moon
Let’s pretend we don’t already know the masses. Nor we do know the orbital distances. It’s 1660 again, and we’re in Isaac Newton’s shoes. Can we learn something interesting about the Sun, or Moon, by observing the spring and neap tides?
Yes, we can, if we also know about total solar eclipses. The stunning thing about total eclipses is that the Moon blots out the Sun almost exactly, rim to rim, so they have the same angular radius in the sky. Since the angular radius, Delta-theta, is equal to the true radius divided by the distance, the observation of total eclipses tells us that R_sun over d_sun equals R_moon over d_moon. Or, equivalently, the ratio of distances is equal to the ratio of radii. So, in our tidal force equation, we can replace the cube of the distance ratio by the cube of the radius ratio.
And now look, we have M over R-cubed for the Sun divided by M over R-cubed for the Moon. That’s the ratio of the average densities of the Sun and the Moon!
So, the fact that the Sun’s tidal force is about half of the Moon’s tells us the Sun’s average density is half that of the Moon. The Moon looks like a rock, so its density is about 3 grams per cubic centimeter, from which we can deduce the Sun’s average density is around 1.5 grams per cubic centimeter.
Common Questions about Tidal Forces and How They Act
Ocean tides are produced by combining the gravitational pull of the sun and the moon, while the spring tides are when the Sun, Moon, and Earth are aligned. Neap tides, on the other hand, occur when the Sun and the Moon are at right angles.
Yes, ocean tides are created by the natural rise and fall of tides caused by the gravitational force exerted between Earth, the Sun, and the Moon.
The Moon’s density is about 3 grams per cubic centimeter, while the Sun’s average density is around 1.5 grams per cubic centimeter, so the Sun’s average density is half that of the Moon. This can also be concluded from the fact that the Sun’s tidal force is about half of the Moon.
