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. 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.
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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.
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. Wt = Wd
Md/Mt = (9.8.034)/(9.8  .034*k*k)
____________________________________________ 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?
