>>5209652>>5209698Using these, and a terminal velocity of 125 mph,
assuming the steak is at a refrigerator temperature of 40 degrees. We would have to raise the steak 86 degrees Fahrenheit. This is 47.77 degrees Kelvin/Celsius.
Since steak is 1.59KJ/kg*C, it means we have to get 47.77*1.59KJ/kg=76KJ/kg.
Energy/kg it would get from Earth's gravity is G*M(1/R-1/x) where M is mass of earth, and R is radius of earth. x is what we are solving for.
since terminal velocity is 56 meters/sec, this leads to our final equation of 76,000+.5*56^2=G*M(1/R-1/(x+R))
Solve for everything,
http://www.wolframalpha.com/input/?i=%286*10%5E24%29*%286.67*10%5E-11%29%281%2F%286.35*10%5E6%29-1%2F%28%28x%29%2B6.35*10%5E6%29%29%3D76000%2B.5*56%5E2
And you get that you would have to drop the steak from at least 7825 meters. Or about 4.9 miles. Of course this is assumes all of the energy went into the steak, so the actual figure would be a bit higher.