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Cooking a turkey by dropping it

Sorry, but I just couldn’t pass this one up. Here is a great question from The Last Word.


First, some assumptions.

  • The turkey is made of water and starts off as ice at 0 degrees C.
  • The turkey is a sphere with a radius of 15 cm. (or generically, I will call it r)
  • When the turkey falls through the atmosphere, half of the energy dissipated goes into the turkey and half goes into the atmosphere.
  • I am going to ignore the increase in thermal energy when the turkey hits the ground. Just because.
  • A turkey is ready if it is at 180 F (82 C)

Maybe that is enough assumptions. How about a diagram to go with those assumptions?

So, the turkey is dropped from a height h above the surface. When someone talks about distance, you should think about the work-energy principles. As a bonus, I can use the turkey-Earth as the system and include thermal energy in there. Here is the work-energy principle for this falling turkey.

Before I go further with the work thingy, let me find the final velocity of the turkey (right before it hits the ground). Of course, if there was no air then the turkey would just keep speeding up the whole time. But this would leave no energy going into thermal energy and an uncooked bird. I will assume that the turkey eventually reaches terminal velocity. How fast is this? Time for another diagram.

At terminal velocity, the net force would be zero. Right before it hits the ground, I can use the simple form of gravity (mg) and I will use the usual model for air resistance:

Where ρ is the density of air, A is the cross sectional area, and C is some drag coefficient (for a smooth sphere this is 0.47). Using the density of water and the volume of a sphere of radius r, the following should be true at terminal velocity:

Some important notes: first the variables. I am calling ρa the density of the air and ρt the density of the turkey. I only solved for the terminal velocity squared because that is what I will use in the kinetic energy term. Also, it is interesting that the terminal velocity depends on both the density of the material and the radius. A larger turkey would have a greater terminal velocity (this is just like meteorites of different size landing in different locations).

Back to energy. I know the initial velocity is zero, I know the final velocity. Let me assume that the turkey is high enough that I should use the better gravitational potential energy:

Putting this together, I get:

I guess since I have the radius and the density of the turkey, I can get rid of the mass. Also, let me solve for the change in thermal energy of the turkey.

This increase in thermal energy has to do two things. It has to change the phase of my water-turkey to a liquid (how come no one drinks turkey anymore?) and it has to increase the temperature to 82 C. In terms of the radius of the turkey (and using the specific heat of water and the latent heat of fusion for water) I get:

Now I just have to solve for h, the initial height. I feel like a loser, but I am going to go ahead and put in some numerical values to make this a little easier (otherwise the algebra gets messy).

Skipping the boring parts (because you will see it doesn’t matter), I get an initial height of 142,000 meters. Hint – that is pretty high. It is useful to recall that the space shuttle orbits at about 300 km above the surface. Oh, but what about the changing air density? That doesn’t matter because I am only looking at the change in energies. I don’t need to know stuff about the acceleration.

Maybe the frozen turkey part as frozen water was a bad idea. Maybe it is more like liquid water (or maybe it is nothing like water). In this case, I can just take out the energy need to make that phase transition and I get a height of 72 km.

Either way, I suspect there is a serious problem with this method. My assumption about the thermal energy distributions might really be wrong. Oh well, this is the best answer I have.