Why do we rotate along with the earth's rotation?

[asdofindia]

And I'm also asking if gravity is not centripetal force, what's keeping us glued to Earth on the equator where we ARE doing a curved motion. You saying friction?

 

[JaredJames]

And I'm also asking if gravity is not centripetal force, what's keeping us glued to Earth on the equator where we ARE doing a curved motion. You saying friction?

GRAVITY!

Gravitational force is far greater than the opposing force trying to "fling us off" at the equator.

But this does not mean gravity is centripetal force.

Gravity: http://en.wikipedia.org/wiki/Gravitation

Centripetal Force: http://en.wikipedia.org/wiki/Centripetal_force

 

[D H]

Centripetal force is a bit of a misnomer. A better term is centripetal acceleration. From the perspective of a non-rotating observer, a person standing still on the surface of the Earth undergoes uniform circular motion. From a kinematics perspective, there is a centripetal acceleration toward the center of rotation (which in general is not the center of the Earth) associated with this observed uniform circular motion.

Multiplying this observed centripetal acceleration by mass yields the net force on the person. This net force cannot be attributed to anyone force because multiple real forces act on the person: gravitation, the normal force, buoyancy, etc. Note that centrifugal force is not one of these. Centrifugal force is not a real force. It simply does not exist from the perspective of our non-rotating observer.

What about a non-inertial point of view, for example, a frame rotating with the rotating Earth? From the perspective, the person standing still on the surface of the Earth is (duh) standing still. To give the appearance that Newton's first and second laws still apply, the net apparent force acting on the person must be zero. We're looking at things from the perspective of a rotating frame, so there is a centrifugal force at play here. This in turn means there must be some other forces whose sum is exactly counter to this centrifugal force.

Not surprisingly, this net non-centrifugal force calculated by the rotating observer is exactly the same as the net force calculated by the non-rotating observer.

 

[RedX]

Not to beat a dead horse, but I read somewhere that the centrifugal force is strong enough that NASA likes to launch things near the equator. By my calculations you are lighter by 3/1000 of your weight at the equator than at the poles due to the centrifugal force.

Also, the sum of gravity and the normal force is in the radial direction. The centripetal acceleration is perpendicular to the axis. Therefore only at the equator can:

$$F_g+F_N=m\omega^2r$$

otherwise it must either be friction:

$$F_g+F_N+F_f=m\omega^2r$$

or if there is no friction then people will be slung towards the equator, like a centrifuge I guess.

 

[Drakkith]

Not to beat a dead horse, but I read somewhere that the centrifugal force is strong enough that NASA likes to launch things near the equator. By my calculations you are lighter by 3/1000 of your weight at the equator than at the poles due to the centrifugal force.

I think its more to get the extra velocity to get into orbit since the velocity of the surface of the Earth is greater the closer you are to the equator. But both are probably a reason.

 

[D H]

Not to beat a dead horse, but I read somewhere that the centrifugal force is strong enough that NASA likes to launch things near the equator. By my calculations you are lighter by 3/1000 of your weight at the equator than at the poles due to the centrifugal force.

It's ESA, not NASA, that launches from near the equator. NASA launches from the Cape (28°30′ north latitude) and from Vandenberg (34°44′ north latitude). The reason ESA does so isn't centrifugal force (which isn't a real force). It's inertia. Launch eastward from the equator and you have a velocity of about 465 m/s right off the bat.

Also, the sum of gravity and the normal force is in the radial direction. The centripetal acceleration is perpendicular to the axis.

The sum of gravity (roughly downward) and the normal force (upward) is not radial if, by radial you mean directed away from the center of the Earth. What can be said is that the sum of gravity and the normal force has no z component, and this is true for any flat spot anywhere on the surface of the Earth except at the poles. (At the poles the sum of gravity and the normal force is the zero vector.)

Therefore only at the equator can:

$$F_g+F_N=m\omega^2r$$

This equation is true everywhere, including the poles, with r being the distance between the point on the surface of the Earth and the Earth's rotation axis.

otherwise it must either be friction

Friction is needed to explain why someone can stand still on an inclined ramp. It is not needed to explain why someone can stand still on a flat surface.

 

[RedX]

The sum of gravity (roughly downward) and the normal force (upward) is not radial if, by radial you mean directed away from the center of the Earth.

Oh, I see what you're saying. If the Earth were a perfect sphere, then objects would flow to the equator, since the normal force and gravity are only in the radial direction (to the center of the earth), and can't add to give you a centripetal acceleration except at the equator where the radial direction to the center of the Earth coincides with the radial direction from the rotation axis.

In fact the Earth was a perfect sphere at one point, but due to this flow of material to the equator, it changed the shape of the Earth so that it bulges at the equator. This bulge causes the normal to the Earth to be more vertical rather than radial (i.e., tilted more in the direction of the rotation axis rather than perfectly radial from the center of the earth), and if you resolve this new normal force in the radial direction of the Earth and the tangent to the radial direction, then tangent piece will always point towards the poles, thereby stopping the flow of material towards the equator.

So only gravity and the normal force is needed, since the Earth bulges at the equator.

That's correct, and I inadvertantly contributed to the stereotype that physics students simplify things too much by assuming everything is a sphere.

As for the centrifugal force lessening gravity, I can't see why it doesn't. The idea being that some of gravity is not just used to pull you towards the earth, but to rotate you in a circle. So you don't have to counteract gravity in its entirety - you just need to be able to provide the amount that the normal force used to provide before you jumped off the earth. Of course there is the question what about energy conservation? How can it take less energy to change your distance from the center of the earth, depending on whether the Earth is spinning or not? I think the answer to that is kinetic energy. You have to provide not only thrust in the lateral direction to overcome the Coriolis force (assuming you launch in a way as to always maintain geosynchonity throughout your trajectory), but more than that to increase the speed appropriate for your orbital altitude. So there is no trade-off in energy, so if you want to escape the earth, it's not the centrifugal force that helps.

 

[cepheid]

Oh, I see what you're saying. If the Earth were a perfect sphere, then objects would flow to the equator, since the normal force and gravity are only in the radial direction (to the center of the earth), and can't add to give you a centripetal acceleration except at the equator where the radial direction to the center of the Earth coincides with the radial direction from the rotation axis.

I don't know what you mean. There would still be a component of the gravitational force that points towards the centre of the latitude circle that you're moving on. This component would just fall off as the cosine of your latitude.

 

[RedX]

I don't know what you mean. There would still be a component of the gravitational force that points towards the centre of the latitude circle that you're moving on. This component would just fall off as the cosine of your latitude.

Right, so assuming a perfect sphere, gravity+normal force is in the radial direction. There is a component of the sum of the two in the direction towards the center of the latitude circle. However, this component cannot be used to provide centripetal acceleration, because then nothing balances the vertical component (which goes as the sine of latitude).

Anyways, it turns out most planets are spherical because of tremendous gravity. But if planets are spinning, they develop a bulge due to material flowing towards the equator.

Here is a link that explains it:

http://www-istp.gsfc.nasa.gov/stargaze/Srotfram1.htm

a quarter of the way down from the top.

This flow of material stops after the bulge is formed, because then the normal and the radial directions are not the same due to the changed shape of the planet, and the normal force+gravity can now give a force in the direction towards the center of the circle of latitude.

 

[cepheid]

Right, so assuming a perfect sphere, gravity+normal force is in the radial direction. There is a component of the sum of the two in the direction towards the center of the latitude circle. However, this component cannot be used to provide centripetal acceleration, because then nothing balances the vertical component (which goes as the sine of latitude).

Anyways, it turns out most planets are spherical because of tremendous gravity. But if planets are spinning, they develop a bulge due to material flowing towards the equator.

Here is a link that explains it:

http://www-istp.gsfc.nasa.gov/stargaze/Srotfram1.htm

a quarter of the way down from the top.

This flow of material stops after the bulge is formed, because then the normal and the radial directions are not the same due to the changed shape of the planet, and the normal force+gravity can now give a force in the direction towards the center of the circle of latitude.

I think I am beginning to understand you now. Thanks for the link.

 

[RedX]

One way to look at is like this:

$$(mg) e_\rho+(N) e_\rho=-(m \omega^2 )e_r$$

where rho is the radial direction in spherical coordinates, and r is the radial direciton in cylindrical coordinates.

This equation can never be true since the LHS and RHS are in different directions no matter the value of N, so the assumption that the acceleration is purely centripetal is incorrect. But what other acceleration can there be? Assuming that the particle stays on the sphere, the equation must be:$$(mg) e_\rho+(N) e_\rho=-(m \omega^2) e_r+(x)e_\theta+(y)e_\phi$$

but there is no way to make this equation work unless y=0, since nothing points in the azimuthal/longitudinal direction (you can dot both sides with e_\phi to get this result), so only the polar (latitude) direction is left. So the equation is:

$$(mg) e_\rho+(N) e_\rho=-(m \omega^2) e_r+(x)e_\theta$$

You can solve for x, since there is only one value of x such that the RHS adds to something in the e_rho direction. Once you solve for x, you can solve N.

Anyways, this exercise has been interesting. I originally thought that friction had to be the key, so that you could have the equation:

$$(mg) e_\rho+(N) e_\rho+(F) e_\theta=-(m \omega^2 )e_r$$

In fact, the value of F would just be -x: i.e., you can either have friction to stop the movement, or have the movement.

However, if friction is really the reason we don't slide into the equator, then it should be true that ocean and air currents point towards the equator, since fluids are flowy and aren't affected by friction as much. But looking at air/water current maps, I didn't see such a flow towards the equator. I then reasoned that the equator is hotter since the sun shines perpendicularly on the equator, and hotter objects have more pressure so this would counter the current towards the equator: hence the reason there is no general trend of currents towards the equator.

Anyways, poor reasoning, as it turns out the answer is that the vector in front of the (N) is not e_rho. In fact, D_H already mentioned this on page 3 of this thread:

From the perspective of a non-rotating observer moving alongside the Earth, a person standing still on the surface of the Earth is undergoing uniform circular motion. A net force is needed to maintain that circular motion. This net force is normal to and directed towards the Earth's rotation axis. The forces acting on this person are gravitation, directed downward, and the normal force, directed upwards. Due to the Earth's non-spherical shape the angle between these forces is not quite 180 degree. The net sum of these two forces is exactly equal to the net force needed to make the person keep following that uniform circular motion.

But it's nice to know that on a perfect sphere, everything flows to the equator, but then stops when the bulge is sufficient to unalign the normal force and gravity. This is the reason planets bulge at the equator and why there is no friction on us. So physics works.

 

[singh94]

I think I am beginning to understand you now. Thanks for the link.

the Earth is spinning so the centripetal force that should be present to stop us moving at the tangent should be m*v*v/r and it is only be due to gravity. But then if the surface of Earth on which we are standing is removed we fall under the influence of graviy. but if a body is revolving then the centripetal force only keeps it in its orbit and doesn't cause the body to come close to the center. Therefre according to me the real gravity should be mg+the centripetal force. please explain...Thanks.

 

[D H]

Anyways, it turns out most planets are spherical because of tremendous gravity. But if planets are spinning, they develop a bulge due to material flowing towards the equator.

Here is a link that explains it:

http://www-istp.gsfc.nasa.gov/stargaze/Srotfram1.htm

Yech. That link oversimplifies things to the extent of being wrong.

The Earth almost certainly never has been a perfect sphere. Shortly after the Moon formed, the Earth's rotation rate was considerably large than it is now (some estimate one rotation every four hours). The Earth's equatorial bulge was considerably larger 4.5 billion years ago than it is now.

Planets are massive enough so that they self-gravitate and achieve something very close to hydrostatic equilibrium. (This is the characteristic that distinguishes planets and dwarf planets from small solar system bodies.) A body in hydrostatic equilibrium such as the Earth will have a surface that is very close to being an equipotential surface -- but not that of gravitation alone. The potential field is instead the sum of the gravitation potential and the centrifugal potential.

When a planet forms, it does not become spherical first and then have material flow toward the equator to become an oblate spheroid. Think of Hamilton's principle. For a rotating body, it is the oblate spheroid shape is the shape that minimizes energy rather than a spherical shape.

The Earth underwent a rather cataclysmic event shortly after it formed. Per the currently favored hypothesis regarding the formation of the Moon, a Mars-sized object smacked into the Earth about 30 to 100 million years after the formation of the solar system. We don't know what about the pre-impact Earth's rotation was. For the sake of argument, assume it was much slower than four rotations per day. The post-impact Earth had a very fast rotation rate, some saying as fast as one rotation per four hours.

This would have meant that the Earth just after impact would not have been in hydrostatic equilibrium. Would this have meant that material flowed toward the equator? Not really. For a better model, imagine squashing a sphere of clay or spinning a ball of pizza dough. There is little if any flow along the surface toward the equator. Instead the shape changes as a whole. Material at the poles move inward while at the equator material bulges outward.

 

[ashishsinghal]

The answer is friction. Neither the Earth nor the Moon is coated with teflon.

That seems a bit odd to me! Imagine a vacuum space where you have levitated a charge by balancing electric and gravitational fields. Remember oil drop experiment. In that case there is no contact whatsoever and hence no friction at all. In that case according to your argument the charge will remain at rest wrt space and hence will collide with the container at great speeds. I really don't think that happened during the experiment!

It is the gravitational force of the earth.

 

[Lsos]

That seems a bit odd to me! Imagine a vacuum space where you have levitated a charge by balancing electric and gravitational fields. Remember oil drop experiment. In that case there is no contact whatsoever and hence no friction at all. In that case according to your argument the charge will remain at rest wrt space and hence will collide with the container at great speeds. I really don't think that happened during the experiment!

It is the gravitational force of the earth.

If it was JUST the gravitational force, but 0 friction, then an outside body such as a meteor would NOT rotate with the Earth after hitting it. It would just sit there, with the ground whizzing beneath it at thousands of mph. Friction definitely plays its part in bringing the meteor to the same speed as the spin of the earth.

 

[Mute]

See, if we draw a free body diagram. We'd draw an arrow from the man to the centre, calling it centripetal force. And another opposite to it calling it centrifugal force, right?
I thought they'd cancel, but I don't think I've clearly finished that thought process, I made a quick reply...

And of course they wouldn't have any component tangential to the surface.

But a body already moving with a velocity tangential do not need a force to keep it moving along the tangent.
But that's along the tangent...

Oh... I think I'm confused. Let me think for a while...

To elaborate a bit on DH's answer to your confusions: Your free body diagram would have neither centripetal nor centrifugal forces drawn on it.

Centripetal force is a net force pointing in the radial direction towards the centre of the rotation, and net forces never appear on a free body diagram! In the intro physics example of a satellite traveling around the Earth in a circular orbit, the only force acting on the satellite is gravity, so gravity does provide the net centripetal force in this case. (but that's a simplified example).

Now, the above all takes place from the point of view of someone in an inertia reference frame watching the rotating object. In an inertial frame there is no such thing as a centrifugal force. If you go to the point of view of the rotating object, however, then you "feel" a centrifugal force on you, but you are now in a non-inertial reference frame, and so you have to introduce "fictitious forces", of which the centrifugal force is one, in order for Newton's second law to continue to work in the non-inertial reference frame. It turns out that in the non-inertial reference frame the centrifugal force is the opposite of the centripetal force, but neither exists in the same frame at the same time, so they don't "cancel out".

 

[Stcloud]

'QUOTE=RedX;3359983]You have it the opposite. If the moon does not spin on its axis, after half an orbit around earth, you would see the opposite side of the moon. '

Suppose someone is walking in a circle around you, always facing you. For example if they are walking clockwise around you, each footstep they take they need to rotate a little bit in order to remain facing you, using their right foot as their pivot foot (hence they are rotating clockwise about their axis, while walking clockwise about you). Suppose the person begins to the north of you. If they don't use their right foot as a pivot to turn while they walk, then when they get south of you, their back will be to you.[/QUOTE]''>>>>

Suppose someone IS walking in a CIRCULAR DIRECTION around you, that person's FACE will ALWAYS be facing the direction in which he/she is travelling. Just like the face of a greyhound racing around a circular track is ALWAYS facing the direction it is travelling. If that greyhound started to spin around on its OWN axis then at some time during its spin it would be moving in the same direction around the track, tail first. ''

 

[Doc Al]

Suppose someone IS walking in a CIRCULAR DIRECTION around you, that person's FACE will ALWAYS be facing the direction in which he/she is travelling. Just like the face of a greyhound racing around a circular track is ALWAYS facing the direction it is travelling. If that greyhound started to spin around on its OWN axis then at some time during its spin it would be moving in the same direction around the track, tail first. ''

Yes, both you and the greyhound always face the direction that you travel. But that direction changes. At first you face north, then east, then south... you're rotating!

Of course, you're not just rotating. You're also moving in a circle.

 

[Drakkith]

'QUOTE=RedX;3359983]
Suppose someone IS walking in a CIRCULAR DIRECTION around you, that person's FACE will ALWAYS be facing the direction in which he/she is travelling. Just like the face of a greyhound racing around a circular track is ALWAYS facing the direction it is travelling. If that greyhound started to spin around on its OWN axis then at some time during its spin it would be moving in the same direction around the track, tail first. ''

This is the last time I will explain this, as this is extremely fundamental.
The moon completes one revolution (orbit) around the Earth in about 27.3 days.
If it was NOT rotating then the same face of the moon would always face the same direction. But, it does not. Because it is revolving around the Earth and the same face of the moon always shows to earth, that requires that it rotate around its axis. If I were looking down upon the Earth and the Moon from up above the solar system somewhere, I could easily see this. From that frame of reference it is obvious that the moon rotates, otherwise how could its direction of facing change? However, because we are here on Earth the situation can be slightly confusing.

The reason that someones face is always facing the way they are walking is because it is natural to face they way you are walking. However objects in space aren't walking, they simply coast through space. The greyhound in your example IS rotating on it's axis as it moves around the track. Just like I am every time I run a lap around the track at the gym. I rotate once per lap. If I did not, I would end up running sideways after the first turn, backwards after the 2nd, sideways again on the 3rd, and back to forwards after the 4th turn.

Again, I refer you to this article on tidal locking. I suggest you read it through thoroughly. Specifically look at this sentence: A tidally locked body takes just as long to rotate around its own axis as it does to revolve around its partner.http://en.wikipedia.org/wiki/Tidal_locking

 

[cepheid]

Supose someone IS walking in a CIRCULAR DIRECTION around you, that person's FACE will ALWAYS be facing the direction in which he/she is travelling. Just like the face of a greyhound racing around a circular track is ALWAYS facing the direction it is travelling. If that greyhound started to spin around on its OWN axis then at some time during its spin it would be moving in the same direction around the track, tail first. ''

Perhaps the attached diagrams will help visualize the situation. In the first one, entitled, "What would happen if the moon didn't spin", I've depicted the moon at various points in its orbit around Earth. On each moon, I've drawn an arrow that is fixed to a specific point on the moon's surface, and always points radially outward from that point. We know that the moon is not spinning in this diagram, because the arrow never changes direction. As a result, as the moon moves around Earth in a circle, the portion of its surface that it presents towards Earth changes continuously. To aid in seeing that, I've drawn lines through the centre of each moon, dividing it into two hemispheres, the one that is visible to Earth, and the one that is invisible to Earth. If we start with the moon at the top of the diagram, it's clear that half an orbit later, when the moon is at the bottom of the diagram, people on Earth are looking at exactly the opposite side of the moon as they were at the start. If the moon didn't spin, then over the course of one orbit, we'd see its entire surface.

In the second diagram, entitled, "What actually happens", we can see that the moon is spinning. We know it's spinning, because the arrow continously changes direction. It does so in such a way that the same hemisphere of the moon is always presented towards Earth. In order for this to occur, the arrow has to do a full 360 over the course of one orbit. In other words, the time required for the moon to spin once on its axis is equal to the time it takes to complete one orbit around Earth. The moon's rotation period is the same as its orbital period. As has been mentioned, this is not a coincidence, but is due to something called tidal locking.

 

[Stcloud]

You didn't look at the article called Tidal Locking that I linked did you? Here:

Not if it wasn't rotating around its own axis. If the Earth stopped rotating, the stars would never change. We would see the same ones all the time. (Other than the sun, which would appear to slowly move through the sky to make one day last a full year.)

Huh? If the body (Earth in this case) stopped spinning around as it orbited the sun the people on Earth would see stars that 'never changed'? You mean always see the same stars? All the time? Of course they would see all the stars surrounding them, as the moved along their circular path around the sun!... And if people lived on the far side of the moon from the sun, they wouldn't ever see the sun, if the Earth stopped spinning... Just as the case with the moon.. if the moon stopped spinning on its own axis, any people on the far side of the moon from earth, would never see Earth... their 'other side of the moon' would always face earth.

I think it is time to use logic, pure logic.. The moon IF it spun around on its own axis MUST at some time during its spin face earth... Which is to say.. if the moon faced all of itself toward a central object or point.. it is spinning. Regardless as to what is or may at that central point.

THIS is unequivocally understood : The far side of the moon is aways the same side. It never faces earth. It therefore can not be spinning around on its own axis.. because it does not do other than 'keep facing the same side of itself toward Earth'.. with which claim you all who want to say the moon IS spinning on its own axis agree!

 

[WannabeNewton]

THIS is unequivocally understood : The far side of the moon is aways the same side. It never faces earth. It therefore can not be spinning around on its own axis.. because it does not do other than 'keep facing the same side of itself toward Earth'.. with which claim you all who want to say the moon IS spinning on its own axis agree!

I can't quite understand your wording but it seems to me you are saying that it is unanimously accepted that the moon does not spin on its axis because it always faces the same side in relation to us. On the contrary, it is a well known fact that the moon's rate of rotation and its orbital period are equal so that it always faces the same side.

 

[Drakkith]

Huh? If the body (Earth in this case) stopped spinning around as it orbited the sun the people on Earth would see stars that 'never changed'? You mean always see the same stars? All the time? Of course they would see all the stars surrounding them, as the moved along their circular path around the sun!... And if people lived on the far side of the moon from the sun, they wouldn't ever see the sun, if the Earth stopped spinning... Just as the case with the moon.. if the moon stopped spinning on its own axis, any people on the far side of the moon from earth, would never see Earth... their 'other side of the moon' would always face earth.

They would see the same sky all year long. Throughout the year the sky in the night changes and we can see the full celestial sphere over the course of a year. If the Earth stopped spinning this would not be the case. We would only see about 50% of the sky. The same constellations and stars would never change their positions in the sky relative to the time. If Arcturus were directly above me when the Earth stopped spinning, it would stay there forever. (Arcturus is a star that is currently overhead at night. In a few months it will not be.)

I think it is time to use logic, pure logic.. The moon IF it spun around on its own axis MUST at some time during its spin face earth... Which is to say.. if the moon faced all of itself toward a central object or point.. it is spinning. Regardless as to what is or may at that central point.

No. This is 100% wrong for the reasons already explained to you. I'm sorry you cannot understand or accept why.

THIS is unequivocally understood : The far side of the moon is aways the same side. It never faces earth. It therefore can not be spinning around on its own axis.. because it does not do other than 'keep facing the same side of itself toward Earth'.. with which claim you all who want to say the moon IS spinning on its own axis agree!

Correction: The far side of the moon is always the same side FROM THE POINT OF VIEW OF THE EARTH. From the suns point of view it sees every side of the moon every 27 days. (Also from the POV of distant stars, galaxies, ETC.)

 

[cepheid]

I think it is time to use logic, pure logic.. The moon IF it spun around on its own axis MUST at some time during its spin face earth...

Your wording is not entirely clear here, but it seems like you're saying that if the moon spins on its axis, then all parts of it must eventually turn to face towards Earth. This is simply not the case. It is possible for the moon to spin and yet for there to be parts of it that never see Earth. Did you take a look at the diagrams that I posted in my previous post? The one on the right, entitled, "What actually happens", explains how this can be so. So as to attract more attention to it, I will embed it directly into this post below, rather than as an attachment:

http://img811.imageshack.us/img811/3373/moon2l.png

Which is to say.. if the moon faced all of itself toward a central object or point.. it is spinning. Regardless as to what is or may at that central point.

THIS is unequivocally understood : The far side of the moon is aways the same side. It never faces earth. It therefore can not be spinning around on its own axis.. because it does not do other than 'keep facing the same side of itself toward Earth'.. with which claim you all who want to say the moon IS spinning on its own axis agree!

Again, it is possible for the moon to spin and yet for there to be one side of it that always faces toward Earth, and one that never does. The diagram above shows this. If you do not follow it, I posted an explanation of it in my previous post.

 

[DaveC426913]

The issue here is about choosing a frame of reference.

If I hold a tennis ball at arm's length, and turn on the balls of my feet, I will see the same side of the tennis ball at all times. From my reference point, the tennis ball is not rotating. From the reference point of my brother, on the porch, the tennis ball is rotating at the same rate it is revolving.

As with the tennis ball, so it is with the Moon. In the Earth's frame of reference, the Moon is not rotating. From an external frame of reference it is.

 

[cjl]

The issue here is about choosing a frame of reference.

If I hold a tennis ball at arm's length, and turn on the balls of my feet, I will see the same side of the tennis ball at all times. From my reference point, the tennis ball is not rotating. From the reference point of my brother, on the porch, the tennis ball is rotating at the same rate it is revolving.

As with the tennis ball, so it is with the Moon. In the Earth's frame of reference, the Moon is not rotating. From an external frame of reference it is.

However, if an object appears not to be rotating when viewed from a rotating frame of reference, the object must in fact be rotating.

 

[Stcloud]

Your wording is not entirely clear here, but it seems like you're saying that if the moon spins on its axis, then all parts of it must eventually turn to face towards Earth. This is simply not the case. It is possible for the moon to spin and yet for there to be parts of it that never see Earth. Did you take a look at the diagrams that I posted in my previous post? The one on the right, entitled, "What actually happens", explains how this can be so. So as to attract more attention to it, I will embed it directly into this post below, rather than as an attachment:

http://img811.imageshack.us/img811/3373/moon2l.png



Again, it is possible for the moon to spin and yet for there to be one side of it that always faces toward Earth, and one that never does. The diagram above shows this. If you do not follow it, I posted an explanation of it in my previous post.

What your diagram shows is that, the moon is orbiting a central point. And that is ALL it shows. Whether the moon spins on its own axis has nothing to do with any other object!.. The moon either spins or it does not. The diagram by the way needs more arrows that those that show the moon is traveling in a circular direction - if the moon is spinning it needs arrow around the moon showing the moon is spinning around its own axis.

By the way re : 'Your wording is not entirely clear here, but it seems like you're saying that if the moon spins on its axis, then all parts of it must eventually turn to face towards Earth.'. I said that the far side of the moon must at some time during its spin, face Earth.. I was specific, by the way, to ensure it was understood that the astronaut on the far side of the moon be landed on the surface of the CENTRE of the far side from Earth. Thus he WOULD see 'all of Earth' if the moon spun itself around, and he with it.

 

[Stcloud]

The issue here is about choosing a frame of reference.

If I hold a tennis ball at arm's length, and turn on the balls of my feet, I will see the same side of the tennis ball at all times. From my reference point, the tennis ball is not rotating. From the reference point of my brother, on the porch, the tennis ball is rotating at the same rate it is revolving.

As with the tennis ball, so it is with the Moon. In the Earth's frame of reference, the Moon is not rotating. From an external frame of reference it is.

As I said, it is not a question of 'frame of reference'.. from far out in space the moon always faces the same side of itself toward a central point, around which it is circling. From Earth (that 'central point) the moon always faces the same side of itself toward a central point..

Now, that ball you are holding at arm's length while YOU spin around on your own axis - Does that ball miraculously start to spin itself around its own axis in your clasping hand? uh huh?

 

[Drakkith]

What your diagram shows is that, the moon is orbiting a central point. And that is ALL it shows. Whether the moon spins on its own axis has nothing to do with any other object!.. The moon either spins or it does not. The diagram by the way needs more arrows that those that show the moon is traveling in a circular direction - if the moon is spinning it needs arrow around the moon showing the moon is spinning around its own axis.

Not only are you wrong, you are also argumentative. The issue has been explained multiple times to you. We aren't making this up, this is actually how it happens.

As I said, it is not a question of 'frame of reference'.. from far out in space the moon always faces the same side of itself toward a central point, around which it is circling. From Earth (that 'central point) the moon always faces the same side of itself toward a central point..

Now, that ball you are holding at arm's length while YOU spin around on your own axis - Does that ball miraculously start to spin itself around its own axis in your clasping hand? uh huh?

See above. Again you are wrong, unwilling to listen, and just plain rude.

 

[Doc Al]

This thread is going in circles. (But is it rotating as well? )

Closed.