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

[asdofindia]

You do NOT need to invoke friction at any point in history (not even going back to Earth's formation) to understand why everything that makes up part of the Earth is rotating. Only conservation of angular momentum need be applied.

I thought conservation of angular momentum was applicable only to a rigid body rotating on its own axis. In that case we'd have to be glued to Earth's surface from time immemorial.

(But that's what the gravitational force and friction do. Gluing us to the surface)

So, the answer is a complex interplay of conservative and non-conservative forces, and conservation of motion.

 

[Drakkith]

I agree with asdofindia. It is a result of several forces over a long period of time. But in the short term, we don't need friction to keep us rotating with the earth.

 

[asdofindia]

Imagine a magnetic sphere. Rotating.
Imagine a sticky man on it with iron legs. Let's say he already has acquired the velocity of the point right under his legs.
At every point, the man's velocity wants him to go tangential to the surface, to be thrown away from the sphere.
But the magnetic force pulls him onto the sphere.
(Centrifugal and centripetal forces)
Agreed till now?
The centrifugal force and centripetal forces are exactly opposite and cancel each other. But they've got no component along the surface of the sphere!
(So there's got to be friction?!?!)
I don't think I've understood my answer.

 

[Drakkith]

The centrifugal force and centripetal forces are exactly opposite and cancel each other. But they've got no component along the surface of the sphere!

What do you mean by this?

 

[asdofindia]

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...

 

[Drakkith]

The force holding the man to your sphere is the magnetic force between his legs and the sphere. This force is greater than the upward force trying to push him away. If it wasn't any small simply movement by the man would send him moving upwards and away.

Because of this, the man is constantly being pulled down towards the sphere while also moving...tangently??...through space around the sphere. The magnetic force on him is similar to the gravitational force on a satellite in orbit. The satellite is constantly falling towards the earth, but also moving, resulting in an orbit.

 

[asdofindia]

That makes me wonder whether or not to subtract mv²/r from GMm/r² when calculating the weight of a body

Edit:http://en.wikipedia.org/wiki/Centrifugal_force_(fictitious)#Earth"
Appears like there are effects due to centrifugal force

 

[Drakkith]

There are most definitely effects due to the centrifigal force. They are usually so small as to be irrevelent in day to day stuff. However, for space launches we do calculate that effect as well as the velocity of the Earth's rotation from the launch site in order to put something into orbit correctly.

 

[asdofindia]

Centripetal force, F_g=GMm/r²
Centrifugal force, F_r=mv²/r
So, when v is > v_max where v_max> \sqrt{GM/r} we do fly away! If it's \sqrt{2} times the v_max in the above equation, it'd escape Earth's gravity too (because that'd be 11.2 km/s)
If v/v_max is between 1 and \sqrt{2}, it'd fly away and fall down back.
(But luckily on earth, v is approximately 1/17 times v_max)
That's why we're not flying away from earth.

Now, let me do something I've never done before.

Since the centrifugal force on Earth is canceled out by the gravity (289 times stronger), we can safely assume the body to be at rest with respect to the reference frame of earth. (I don't know if that's right, because as I said, I've never done this before) (There'd be no longer any effects that'd be observed due to rotational motion of Earth with respect to the space around it)
So, since Earth and the body are both at rest (in that inertial frame) there wouldn't be the need of any frictional force.

We have to view this in the inertial frame of earth.

(I have understood my point)

 

[Lsos]

i mean a person who is in space. if everything was not moving(not at respect to each other but in actual) then he would be at rest with respect to them. now do u get it

I'm not sure what you are asking, but when you are in orbit in space then we can confidently say that you are NOT rotating with the earth. You are moving much, much faster than the Earth is spinning.

Now what happens when you re-enter the atmosphere? The atmosphere WILL slow you the fuk back down to it's own speed, and it will use extreme force to do so. The force is so great in fact, that whatever is re-entering must use some kind of heat shield, otherwise it will burn up with a brightness of the sun.

So you see what happens whenever something is not rotating with the earth. The Earth WILL slow or speed it up to match its own rotation, and violently so if need be. If you don't believe me go jump out of a moving car, and you'll see first hand how it feels when something is NOT rotating with the earth.

 

[JaredJames]

asdofindia centripetal force is not gravity. Your equation above defines gravity not centripetal force.

Note, your equation is virtually constant wherever you are on the planet, which is gravity, but the centripetal force is almost zero at the poles and maximum at the equator.

 

[asdofindia]

asdofindia centripetal force is not gravity. Your equation above defines gravity not centripetal force.

At the poles there is virtually no centrifugal forces compared to at the equator.

So the centripetal force is...?

 

[JaredJames]

So the centripetal force is...?

Centripetal force occurs due to a body moving on a curved path.

Gravity occurs simply because a body has mass.

We have both on earth, one is down to the mass, the other the rotation.

Gravity, aside from the ellipsoid shape of the Earth is relatively constant all over the planet.

Centripetal force is at its maximum at the equator and tends to zero towards the poles.

 

[asdofindia]

Centripetal force is at its maximum at the equator and tends to zero towards the poles.

yeah, i was talking at the equator.

 

[JaredJames]

yeah, i was talking at the equator.

Doesn't matter.

You are describing gravity as centripetal force which is incorrect.

 

[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.