How does centrifugal force causes tidal bulge on the other side of the moon?

Question

I didn't understand why the opposite end gets tidal bulge as well, all reading points to centrifugal force, but HOW does centrifugal force cause that? Can anyone explain in simpler terms please?

Answer 1

^ What Stardust said is absolutely correct.

Imagine you've got one of those little bungee cords.  If you pull on one end, that end gets a LOT closer to you.  The middle of the cord only gets a LITTLE closer to you.  The opposite end of the cord stays where it is.  But if you were an ant standing in the middle of the cord, you would think that BOTH ends of the cord had got further away from you.  You THINK that you, in the middle, are sitting still, so you THINK both ends are moving away from the middle.

Tides are the same way.  The moon pulls the ocean on the close side towards it a LOT.  And it pulls the rocky middle of the earth towards it a LITTLE.  Then the ocean on the far side stays where it was, and what you get is two tidal bulges.

That's greatly oversimplified, but the point is to visualize that the moon isn't PUSHING the ocean away on the far side to make a bulge.  It's actually pulling the rock right out from under the water to make a bulge.

Answer 2

the moon doesn't rotate.it shows the same face to the earth all the time.the earth however rotates on it's axis ever 24 hours ,and it bulges sightly at the equator as a result..the tidal forces exerted on the moon serve to cause some minor (moon quakes) and some shifting and settling and probably bulges toward the earth .i doubt centrifical forces factor in much ,it takes the moon 28 days to circle the earth.........tom

Answer 3

The centrifugal force DOES NOT cause tides.

Tides have nothing to do with the centrifugal bulge, it is a different bulge!!

The tidal bulge occurs because the force of gravity is distance dependent. The pull from the moon acting on earth is strongest on the nearside (moonside) of earth and weakest on the far side.

So the ocean on the side of earth poiting towards the moon is pulled more than the solid earth, and the solid earth is pulled more than the ocean on the far side. So we get two bulges!

Look up tides on BadAstronomy.com for more...

Answer 4

The centrifugal force is a pseudoforce (or fictitious force) that can be useful in rotating reference frames, but it gets us into trouble when we try to conceptualize something as complex as the tides using a force that doesn't really exist.  Let's try to re-imagine the scenario using only real forces and acceleraion.

The first thing we have to consider is that the Moon is NOT revolving around a stationary Earth.  Instead, the Moon and Earth BOTH swing around an imaginary point called the barycenter.  Here's an analogy: lock arms with a small child and swing them around (they seem to love that!).  You'll notice that you have to move in circles as well, although your circles are much smaller than the child's circles.

It's no different between the Earth and the Moon.  The Earth and Moon BOTH revolve around this imaginary barycenter, which lies about 1700 km below the center of the Earth.  Why is it so much closer to the Earth's center than it is to the Moon's center?  Because the Earth is about 80 times as massive as the Moon.

Now, what does this have to do with the tides?  Everything!  Let's divide the tidal bulge into two parts: the near bulge (meaning the end of the bulge nearest the Moon) and the far bulge (which is what you want to have explained).  The near bulge is easy: the Moon's gravity is stronger on the near bulge than it is on the far bulge or the Earth itself, so it rises more toward the Moon.  

The far bulge is attracted LESS toward the Moon.  When the Earth is swinging around the barycenter, the far bulge wants to keep going off in a straight line.  However, it can't do that because (1) the Earth's gravity is holding it "down", and (2) the Moon's gravity is pulling it in the same direction as the other bulge.  It can't fly off in a straight line, but it can lag behind a bit and take a bit longer to accelerate toward the Moon.  So there you have it.  The "far bulge" is more appropriately called the "lagging bulge" because the force pulling it into this spinning mess is smaller than the force acting on the Earth or the "near bulge".

I know that's not terribly simple, but it works, and it does so without having to invoke any fictitious forces.  Keep in mind that fictitious forces can seem perfectly real when you're in a rotating reference frame.  Also keep in mind that fictitious forces are sometimes more useful for making certain calculations than real forces are, but every fictitious force (such as centrifugal force) is just a rotating analog of very real forces.

I hope that helps.  Good luck!