# Sudden Earth Rotation Deceleration Theory

1. Aug 9, 2012

### Cyghost

Just throwing this out there (I'm uncertain if this has been posted before and apologise in advance if I'm just covering old topics) but I have been thinking lately about how a massive change in rotational speed caused possibly by the asteroid presumed to have wiped out the dinosaurs may help explain certain aspects regarding the Cretaceous-Tertiary mass extinction and also the rise of mammals and other species.
Firstly I want to ask of what the environment and conditions of the Earth were like prior to the impact, and the consequences that would result from such an event "If the Earth was rotating at twice it's current speed before the strike".
The Earth rotates at approximately 1070 miles/hr which means if the Earth was rotating twice as fast we would be spinning at approx 2140 miles/hr and an Earth day would be 12 hours. So from what I understand (with only a basic understanding of astrophysics) gravity holds us to the Earth but is also compensated by centrifugal force which is created from the spinning action of the planet and it is this balance of forces which creates the effects that keep us from not flying off the planet as we spin and also not having crushing gravitational forces upon our bodies. So if the mass of the Earth was basically the same (prior to the event) as present time but the rotational speed was doubled prior to the event as I have suggested, in theory would the effect of gravity be far less? So basically theorising the impact not only suddenly slowed the rotational speed of the Earth but also inflicted massive gravitational forces upon the Earth environment in a short time frame.
Now looking at the body shapes of massive dinosaurs and how seemingly massive cumbersome designed creatures were so prolific during the Cretaceous Period makes me question whether an environment with less gravitational forces (Prior to the Impact) could explain their design and when they were subjected to sudden massive gravitational forces as well as the massive ash clouds soon after the impact occurred they were unable to sustain ease of mobility to escape the disaster (Including ocean species). They would simply be to heavy to maintain their previous speed and strength and in most cases would basically fall where they stood or perish soon after (Which also may explain archaeological discoveries locations and positioning). This could then lead to the survival of less cumbersome species such as mammals and also this may explain why species prior to the impact were often led to evolve for flight as with less gravitational force would encourage animals to the sky and our modern birds are simply evolutionary adaptations from those early flighted species. Perhaps the sudden gravitational forces inflicted also is what invited our early primates ancestors from the trees to the land as it was no longer as simple to jump from tree to tree. Ouch!!! On a metaphysical aspect perhaps now having a day lasting twice as long also inspired early primates to creatively think more as they now had twice as much time during the day to utilise as they were no longer restricted to a 12 hour day night cycle and also they didn't have those pesky dinosaurs to worry about
*I am only using the 1/2 speed rotation measure purely as just a reference and I don't claim to estimate what the possible rotational speeds may have been prior to the impact according to the theory just the consequences.
Anyway it is just a theory and I am by no means stating this is fact but I am looking forward to your input and your ideas and hope at the least it is some food for thought. Again I apologise if this has been covered or it is in the wrong forum section.

2. Aug 9, 2012

### Filip Larsen

Welcome to PF!

Not really. The centrifugal acceleration [2] equals 2 where r is the radius of the circular motion (equal to 6378137 m at the equator) and ω is the rotational speed (equal to 2∏/T, with T being the sidereal rotation period of Earth at around 86164 s). If you calculate its value you can see that the magnitude of the centrifugal acceleration is around 0.3 % of the gravity at equator, so even though the centrifugal acceleration quadruples if the rotational speed doubles, it will still remain small compared to the acceleration due to gravitation [1].

[1] http://en.wikipedia.org/wiki/Gravity_of_Earth
[2] http://en.wikipedia.org/wiki/Centripetal_force

3. Aug 9, 2012

### Cyghost

Thanks for your reply Filip as I didn't begin to know the calculations required.
I was also just wondering does inertia also factor into the equations, and what rotational speed can you calculate may induce a centrifugal acceleration to essentially produce a 50% reduction in gravitational forces upon the Earth environment as I suggested?
And thanks for the welcome and sorry for my poor mathematical skills.

4. Aug 9, 2012

### Staff: Mentor

If the calculator I used is correct, then it would take a rotational period of 1.9 hours per revolution to produce approximately 0.5 g of acceleration.

5. Aug 9, 2012

### Cyghost

Ok I have done a few quick calculations and I now understand that it is a minimal increase of centrifugal acceleration, but I must admit the mathematics still eludes me as I was thinking if a person was standing at true north in the Arctic they are theoretically not subjected to the same centrifugal force as a person at the equator. So does that mean a person at the north pole would feel more of a gravitational force than at the equator if the planet rotated enough to include the result of stronger centrifugal acceleration?
Sorry if I'm hijacking my own thread :)

6. Aug 9, 2012

### Cyghost

Oh I now realise inertia is already a fundamental factor in the formula..
Again I must apologise for my limited knowledge.
In fact the last time I saw a numerator it was standing over a denominator.. Ok that was poor humour :)

7. Aug 9, 2012

### Staff: Mentor

Yes, a person standing at the poles feels a greater net force due to there being not centrifugal force and because the poles are slightly closer to the center of mass of the Earth than the equator is.

8. Aug 9, 2012

### Cyghost

Thanks Drakkith for your input and I appreciate the help.

9. Aug 10, 2012

### Filip Larsen

Your theory also has an issue with the size of impactor needed to create a significant change in rotational speed. A simple impact model could be to have a spherical non-rotating impactor of radius r, density $\delta_i$ and mass m impacting and "sticking" (all ejecta eventually settles on Earth) tangential to the equator of a spherical, solid Earth of radius R, density $\delta_E$ and mass M. If we introduce the density-normalized relative radius $\rho$, such that

$$\rho^3 = \frac{\delta_i}{\delta_E}\frac{r^3}{R^3} = \frac{m}{M}$$

then I get that the change in rotational speed of Earth after impact is

$$\omega_1 = \frac{\frac{2}{5}+\nu\rho^3}{\frac{2}{5}+\rho^3+ \frac{2}{5}\delta\rho^5}\omega_0$$

which results in a change in rotational period as

$$\Delta T = T \rho^3 \frac{\nu - 1 - \frac{2}{5}\delta\rho^2}{\frac{2}{5}+\nu \rho^3}$$

where T is the rotational period, $\delta = \delta_E/\delta_i$ is the relative density and $\nu = v/v_e$ is the speed of the impactor relative to the surface speed at equator, with $\nu$ being negative when the impactor comes in from east. Depending on trajectory of the impactor, $\nu$ can be expected to be in the range 20 to 150 (for example, the Chicxulub impactor seems to have had a relative speed of around 36). For very small impactors the change can be approximated as

$$\Delta T \approx \frac{5}{2} T (\nu-1)\rho^3$$

Inserting numbers, it seems that you need a rather large impactor to get a significant change in rotational period, far larger than any impactor that have hit Earth in (geological) recent times. For an impactor of such size I think it will be safe to say that the changes to the Earth rotational period will have no consequence on life on Earth compared to the effect from the release of impact energy and disruption of lithospheric and atmospheric structures.

Last edited: Aug 10, 2012
10. Aug 10, 2012

### Filip Larsen

I can see that I have started out with a poor model in my previous post.

If we assume the impactor mass is distributed over the surface of the Earth after impact (instead of the also unrealistic assumption that it simply sticks to Earth as an unbroken sphere), I get the much simpler relationships

$$\omega_1 = \frac{\frac{2}{5}+\mu\nu}{\frac{2}{5}+\mu}\omega_0$$

and

$$\Delta T = \frac{\mu(\nu-1)}{\frac{2}{5}+\mu\nu} T_0$$

where $\mu = m/M$ is the relative mass. No need for relative radius and density with this model. This equations are simple enough that we can solve for $\mu$ to get

$$\mu = \frac{2}{5}\frac{\omega-1}{\nu-\omega}$$

where $\omega = \omega_1/\omega_0$. This means, for instance, that a reduction to half angular speed, $\omega = 0.5$, for a very fast retrograde impactor, $\nu=-150$, gives $\mu \approx 1.3\cdot 10^{-3}$, which should give an impactor diameter of around 1400 km.