Nonspherical earth (split from centrifugal forces)

[MigL]

Holy crap !

Do you guys have those books memorised or what ?

( bunch of geeks and nerds )

[Janus]

Janus, doesn't all this assume a homogenous density function?

 

As we know the actual density/depth function is concave down starting from the surface. It rises quite sharply just down from the surface, with the rate of increase steadily decreasing with depth, finally becoming asymptotic to some max value near the middle.

 

Have you any working to show the effect of this?

To work out the exact effect would take a bit of calculus, but we can get a general idea. If the density of the spheroid increased with depth, then I think it is pretty safe to assume that the density of the sphere in the solid black would be, on average higher than that of the bulges or the spheroid as a whole. This would have the effect increasing the mass of this sphere to greater than 1/4 of the mass of the spheroid and make the polar gravity even higher when compared to the equatorial gravity.

[swansont]

Sorry Strange, I thought I made myself clearer, but I guess not.

Things farther away actually weigh less because weight is a force due to gravity and changes with the inverse of the radius squared. In effect, the equatorial surface is feeling less gravitational force

Mass does not change, and the two are not the same.

 

That's not the whole story.

 

If "weight" is "what a scale reads", then it's the normal force rather than mg that is being measured. Then rotation plays a role, too.

 

Consider a rigid sphere — no deformation. If you spun it such that v^2 = GM/r, and were on the equator, you would be in orbit – in freefall, at the surface. You would not register on a scale. All of the gravitational force is being "used" as the centripetal force. There's nothing "left over" so there is no normal force needed.

 

At a smaller rotational speed, the centripetal force is the difference between the gravitational force and the normal force. Because the centripetal force is always the net force on something moving in a circle. So it's true that a lower value of g from a greater r will reduce your weight. But rotation lowers the scale reading, too.

[MigL]

Sure, I never claimed otherwise.

'Weight' is the net force an object feels.

[Robittybob1]

 

That's not the whole story.

 

If "weight" is "what a scale reads", then it's the normal force rather than mg that is being measured. Then rotation plays a role, too.

 

Consider a rigid sphere — no deformation. If you spun it such that v^2 = GM/r, and were on the equator, you would be in orbit – in freefall, at the surface. You would not register on a scale. All of the gravitational force is being "used" as the centripetal force. There's nothing "left over" so there is no normal force needed.

 

At a smaller rotational speed, the centripetal force is the difference between the gravitational force and the normal force. Because the centripetal force is always the net force on something moving in a circle. So it's true that a lower value of g from a greater r will reduce your weight. But rotation lowers the scale reading, too.

You would have to shorten the day length down to around 1.5 hours to get to that sort of velocity, but at that sort of rotation rate the Earth would be even more oblate in shape so the radius increases so the period can come down as a result.

I wonder how the shape would change with increased rotation rate?

Edited April 17, 2015 by Robittybob1

[Strange]

Holy crap !

Do you guys have those books memorised or what ?

 

Radio programme, actually.

 

 

( bunch of geeks and nerds )

 

Thank you.

[Robittybob1]

You would have to shorten the day length down to around 1.5 hours to get to that sort of velocity, but at that sort of rotation rate the Earth would be even more oblate in shape so the radius increases so the period can come down as a result.

I wonder how the shape would change with increased rotation rate?

This from Wiki http://en.wikipedia.org/wiki/Equatorial_bulge#The_equilibrium_as_a_balance_of_energies

Something analogous to this occurs in planet formation. Matter first coalesces into a slowly rotating disk-shaped distribution, and collisions and friction convert kinetic energy to heat, which allows the disk to self-gravitate into a very oblate spheroid.

As long as the proto-planet is still too oblate to be in equilibrium, the release of gravitational potential energy on contraction keeps driving the increase in rotational kinetic energy. As the contraction proceeds the rotation rate keeps going up, hence the required force for further contraction keeps going up. There is a point where the increase of rotational kinetic energy on further contraction would be larger than the release of gravitational potential energy. The contraction process can only proceed up to that point, so it halts there.

As long as there is no equilibrium there can be violent convection, and as long as there is violent convection friction can convert kinetic energy to heat, draining rotational kinetic energy from the system. When the equilibrium state has been reached then large scale conversion of kinetic energy to heat ceases. In that sense the equilibrium state is the lowest state of energy that can be reached.

The Earth's rotation rate is still slowing down, but gradually, about two thousandth of a second per rotation every 100 years.[1] Estimates of how fast the Earth was rotating in the past vary, because it is not known exactly how the moon was formed. Estimates of the Earth's rotation 500 million years ago are around 20 modern hours per "day".

The Earth's rate of rotation is slowing down mainly because of tidal interactions with the Moon and the Sun. Since the solid parts of the Earth are ductile, the Earth's equatorial bulge has been decreasing in step with the decrease in the rate of rotation.

 

 

I wonder if there is something more scientific?

[swansont]

You would have to shorten the day length down to around 1.5 hours to get to that sort of velocity, but at that sort of rotation rate the Earth would be even more oblate in shape so the radius increases so the period can come down as a result.

I wonder how the shape would change with increased rotation rate?

 

The first doesn't matter, and the second is moot, since I specified a rigid body. The point was to show that rotation matters.

 

As it stands, it affects weight by 0.3% at the equator.

[Robittybob1]

 

The first doesn't matter, and the second is moot, since I specified a rigid body. The point was to show that rotation matters.

 

As it stands, it affects weight by 0.3% at the equator.

I did calculate its rate of rotation for a rigid body with Earth's radius and it would need to spin once per 1.5 hours to have orbital speed at its surface.

I then considered what would happen if the Earth wasn't so rigid, for the rigidity seemed to be unnatural.

 

I might have done the calculations wrong for I had a higher figure for the ratio of velocity : orbital velocity.

According to Wikipedia a person on the equator is doing 465.1 m/s (1,040 mph) and needs 7.9 km/s (17,672 mph) to get to orbital speed. Does this mean he has already 1/17 th of the required orbital speed just from the rotation of the Earth?

http://en.wikipedia.org/wiki/Orbital_speed#Tangential_velocities_at_altitude

 

Edit: I get it now because the angular acceleration is calculated by v^2/r even though it is 1/17 of the velocity compared to orbital velocity, it is only 0.3% of the acceleration compared to gravity of 9.8 m/sec^2

Edited April 17, 2015 by Robittybob1

[imatfaal]

!

Moderator Note

 

Hijack including "a novel idea" regarding earth's formation has been split off to the trash. Novel ideas belong in speculations - not in a mainstream thread.

 

[Robittybob1]

....

You cannot just eliminate Earth's angular momentum by snapping your fingers. If you could dissipate it and 'brake' the Earth, it would take time, and in the end, you'd find that the Earth would be close, if not spherical.

Just like, as it was coming together 4.5 Bi yrs ago, it was probably extremely flattened ( accretion disc ), and slowly rotating. It then speeded up ( conservation of ang. momentum ) and reached an equilibrium of forces in the somewhat flattened spherical shape it has today. The very fact that the surface of the Earth sits appreciably higher in the gravity well, at the equator, is because it feels less net force. In effect, it weighs less. Even rock feels the effects of buoyancy, given several billion years to react, and so the surface has 'floated' up higher at the equator than the poles.

 

.....

That was an interesting idea.

Edited April 19, 2015 by Robittybob1