Physics behind Dinosaur Weight Argument

 

The equations for our effective weight on a spinning globe   

W = GMm/(R*R) - v*vm/R  

the first term is force due to gravity, the second term is centripetal force. W is the effective weight (downward force felt by the object), G is the gravitational constant, M is the mass of the Earth, R is the radius of the earth, m is the mass of the object on earth, v is the velocity that the earth is spinning which is greatest for an object on the equator and zero for an object at the rotational poles.

For the earth, an object at the equator will weigh

W = 9.8*m - .034*m   

The force due to gravity is the familiar 9.8 meters per seconds squared. The second term uses the velocity v = 465 meters/sec and a radiusof the earth as R= 6,356,000 meters. The second term is very small, less that 1% of our effective weight.

But notice, if we increase the speed that the earth is rotating, the second term increases by the velocity squared.

 

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Size Limit

The largest land animal today positioned at the equator will feel an effective weight of

Wt = (9.8 - .034)Mt

where Wt is max weight today and Mt is mass today, of the largest land animal.

 


If we assume a faster spinning earth of the dinosaurs, the largest land dinosaur will will feel an effective weight of


Wd = (9.8 - .034*k*k)Md

where Wd is max weight of the dinosaur, Md is mass of the dinosaur, k is a factor increase in velocity. Think of k as an increase in the rotation speed of the earth in terms of today's rotation. If k=2 the earth would rotate twice as fast.

If we assume that the largest land animal of a given time is limited by the weight that they feel and that maximum weight is a constant we get

Wt = Wd


This equation follows immediately from the above assumption

Md/Mt = (9.8-.034)/(9.8 - .034*k*k)


The equation is a way of stating the following:

If the earth is spinning 15 times as fast (ie. k=15), we would expect the largest dinosaurs to be about 5 times the size of the largest land animals of today.



What we see today is evidence of land animals that were huge compared to the animals of today. The largest Dinosaur is estimated to be 5 or 6 times the size of the largest elephant. There is evidence of flying dinosaurs living at the same time, with wingspans 5 times the largest birds that are able to fly today. These equations show a scenario using only simple physics where this situation could arise. The only real assumption made is that the weight an animal feels puts the limit on their maximum size.

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Shape Distortion

 

When Spinning faster the globe may change shape - becoming more of an oblate spheroid. If we include this into the equations we get

W = GMm/(zR*zR) - kv*kvm/(z*R)

 Where z indicates the increase in the Radius of the earth at the equator. Rearranging terms and assuming the same mass of the earth and gravitational constants, the largest dinosaurs would feel an effective weight of

Wd = (9.8/(z*z) - .034k*k/z)Md

 

If we assume k = 10 and z = 1.5 (bulge of 50%) we get

Md/Mt = 4.667

Another way to write the equation is

Wd = (1/(z*z))*(9.8 - .034*k*k*z)

 

 

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Include Increase in Mass

 

Assuming the earth stays roughly the same density D(ie D = M/(R*R*R)) then our original equation for effective weight becomes

W = GxRDm - v*vm/xR

Where xR is the radius of the earth of the dinosaurs, x would be the fraction of today's earth and R is the Radius of today's earth.

 

So if we assume again that the max size land animal's can reach is limited by their effective weight - Wt = Wd then

Md/Mt = (9.8-.034)/(9.8x - .034*k*k/x)

Let x = .6 which is the approximate size radius at which the continents fit together in a continuous shell, then our equation becomes

Md/Mt = (9.766)/(5.88 - .057*k*k)

To get the same 5 times impact the earth of the dinosaurs would only need to spin at about 8 times what it is spinning today.

 

 

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putting it altogether we get

 

Md/Mt = (9.8-.034)/(9.8x/(z*z) - .034*k*k/(x*z))

 

Let k = 10, x = .75, z = 1.25

we get

Md/Mt = 5

This would be adding the additional mass of 20 moons. We can get a similar result if we assume k=12, x=.9, and z=1.25 which would be adding the mass of about 8 moons.

 

How many moons could the pacific ocean hold?