Gravitation and circular motion/ disintegration of earth

[Kushal]

my physics teacher told me that: 'for an object placed at the surface of the earth, the gravtitional force must be greater than the normal contact force due to the required centripetal force'

it's alil bit difficult for me to understand this... and if smeone could explain to me what is the normal contact force. I guess what it means, but i want a more concrete description.

he also said that: 'If periodic time decreases, then the required centripetal force will be bigger an dthe reaction will decrease. For one particular angular speed, the reaction will become zero and Earth will disintegrate , objects can no longer remain in contact'

i can't grasp what he exactly means and wot is the contact force..

 

[phyind]

Normal force is the type of force exists due to two objects in contact. A type of contact force and it is always normal (perpendicular) to the surface in contact. thus two forces ie normal is having dirn radially outwards and gravitation has dirn radially inwards. thus gravitation shld be stronger so that object remains on the surface of Earth in contact.

 

[LHarriger]

When you rotate on a merry go round a "force" tries to throw you off right? Why don't you get thrown off? Because you are holding on to something or the friction between you and the merry go round keep you there, ie: some counter force is applied to keep you in place.
Now recast this:
When you rotate on the Earth the same kind of "force" is trying to throw you off. But now the counter force that keeps you firmly planted to the ground is gravity. However, if you made the Earth spin really, really fast (which would make its period, the length of a day, shorter) then gravity would become insufficient to hold you in place. Just like if someone spun you fast enough on a merry go round you would not have the strength to hold on and would fly off.

 

[LHarriger]

i can't grasp what he exactly means and wot is the contact force..

In this context, the contact force is F=mg where F is the force that some object of mass m feels due to gravity g.
g is an acceleration and equals 9.8m/s^2 at sea level.

An important point to keep in mind though is that g=9.8 is actually the effective gravity. Meaning it is the acceleration due to a=Gm/r^2 minus the merry go round acceleration.

 

[disregardthat]

But at the moment you "leave" Earth due to the spinning, won't you return at some point because of the gravity pulling you back in (assuming your velocity doesn't exceed the Earth's escape velocity.

 

[LHarriger]

But at the moment you "leave" Earth due to the spinning, won't you return at some point because of the gravity pulling you back in (assuming your velocity doesn't exceed the Earth's escape velocity.

That's a good question. In the merry go round analog, as soon as you loose your grip then no force is left to fight the centrifugal force and you will definitely fly off. However, in the Earth example once you loose your "grip" gravity is still fighting to pull you back down to the earth. So what happens.
One way to see the answer is to look at the effective force for this situation:

$F(r)=-\frac{mMG}{r^2}+\frac{L^2}{mr^3}$

The first term is the force of gravity, the second term is the centrifugal force, r is the distance from the center of the earth, and L is you angular momentum. Suppose that the Earth is spinning fast enough that these two terms result in a only a small risidual gravitaional force. Now you spin the Earth even faster which increases L and results in a net risidual centrifugal force allowing you to loose contact with the earth. Once contact is lost L becomes conserved (no torque.)

Notice however that the centrifugal force goes as $\frac{1}{r^3}$ while the gravitational force goes as $\frac{1}{r^2}$. The means that as soon as you loose contact with the ground (increasing r)your centrifugal force term will decrease quicker than your gravitational force. So that the net force quickly returns to zero.

Final result: you hover a fraction of an inch off the ground or if you were able to suddenly spike your anglular momentum then you quickly reach a new equilibrium point that you oscillate about.

Disclaimer: I assumed the Earth itself was a solid sphere with detachable objects on its surface. In reality, the liquid interior would feel an even greater centrifugal force and smaller gravitational force than objects on the surface, so in all likelyhood the Earth would bulge, resulting in earthquates, spewing magma, mass extinction, terror in the streets, etc. before you got to experience zero g

 

[Kushal]

for the merry go roung thing, what actually is this force which throws you off?? is it the centrifugal force??

Thanks, except for the above thing, i understood what you meant...

 

[Janus]

But at the moment you "leave" Earth due to the spinning, won't you return at some point because of the gravity pulling you back in (assuming your velocity doesn't exceed the Earth's escape velocity.

Another way of looking at this is that you go into orbit around the Earth. At the instant the contact force goes to zero, you enter a circular orbit. If you are able to hold on to the Earth while it spins up a little faster, and then let go, you will go into an elliptical orbit that always returns to near the surface of the Earth. If you can hold on long enough to reach escape velocityt and then let go, you will leave the Earth on a parabolic path.

If, in the first case (contact force just goes to zero), you reach out and push yourself straight away from the Earth (like you were crouching at the moment and then pushed off with your feet.) you will rise up away from the Earth, come back down and will hit the Earth, As your new orbit will intersect the Earth's body.

 

[D H]

for the merry go roung thing, what actually is this force which throws you off?? is it the centrifugal force??

There is no force that throws you off. Instead, there is a force that holds you on the merry go round. Should this force suddenly stop, you will travel in a straight line with the velocity that you had at the time the force stopped (Newton's First Law of Motion). This straight-line path is tangential to the merry go round.

 

[country boy]

An important point to keep in mind though is that g=9.8 is actually the effective gravity. Meaning it is the acceleration due to a=Gm/r^2 minus the merry go round acceleration.

Does this mean that g is latitude dependent? What happens at the North Pole?

 

[D H]

g depends on latitude because it is defined in a rotating reference frame (the spinning Earth) and because the Earth is not spherical (and this happens because the Earth is spinning). A person at the North Pole is a bit closer to the center of the Earth than someone standing on the equator, making the person at the pole experience a stronger gravitational acceleration than does the person at the equator.

A person at the North Pole is rotating about his own axis and thus doesn't undergo any centrifugal acceleration. On the other hand, the person at the equator does undergo centrifugal acceleration. Because gravity ('g') is defined in a rotating frame, the value for g at the equator is once again smaller than it is at the Pole.

BTW, the rotational ($\omega \times r$) effect is about twice that of the equatorial bulge.

 

[DaveC426913]

for the merry go roung thing, what actually is this force which throws you off?? is it the centrifugal force??

D_H's detailed explanation can be summed up as 'It is inertia'.

 

[country boy]

A person at the North Pole is rotating about his own axis and thus doesn't undergo any centrifugal acceleration. On the other hand, the person at the equator does undergo centrifugal acceleration. Because gravity ('g') is defined in a rotating frame, the value for g at the equator is once again smaller than it is at the Pole.

And don't forget the angular change with latitude. The apparent gravity vector (with gravity + merry-go-round + bulge) does not, in general point at the center of the earth.