Re: Would an object burn up if dropped at the top of the atmosphere with zero velocity?

It depends on what you mean by "the top of the atmosphere". Let's discuss your question in terms of different heights. But first, a few assumptions.
1. For the first part of this discussion we will neglect air resistance. It is air resistance (and/ or the compression of the air) that causes the heating of objects at speed, but for calculations of the falling speed "above the atmosphere" we will neglect it. The atmosphere's density at a height of 50 kilometers is only about 1/1000 of what it is at sea level. At 100 kilometers the density is about 1/330000 of what it is at sea level.
2. The speed needed to burn up in the atmosphere is in the ballpark of 1000 meters per second. This varies quite a bit depending on the shape of the object, its composition, etc., but for discussion purposes consider that the SR71 Blackbird aircraft was capable of flying at about Mach 3, which is roughly 915 meters per second, and some of its surfaces glowed red with heat.
3. For this first part we will let our objects fall from R_height down to R_final, and we will assign to R_final the value 50 kilometers (50000 meters) above the Earth's surface. That is about five times as high as commercial aircraft ever fly, and the air is pretty thin up there, so we will assume that the falling object doesn't meet much atmosphere until it gets to that height. The radius of the Earth will be taken to be 6.371E6 meters.

Okay, to calculate the fallling speed. You can derive, using energy considerations, the equation giving the falling speed of an object released at height R_height and falling to a height R_final. It is
v = sqrt(2GM_E*[1/R_final - 1/R_height])
and remember to keep all the units consistent. We will use SI units (meters, kilograms, seconds). The values we need are
G = 6.6726E-11 (the gravitational constant in SI units)
M_E = 5.975E24 (the mass of the Earth in kilograms)
R_final = 6.421E6 meters (50000 meters above the surface)

I put together an Excel spreadsheet using all this, varying the height of the object above 50 km above the Earth's surface. (If you would like a copy of the spreadsheet, email me.) That is, if I drop the object at a height of 50 km above the surface I calculate zero velocity, because I dropped it from zero height above the "top of the atmosphere". Some of the heights and speeds are:

The value 1E20 meters is very very large, and we can take the resulting speed (11140 m/s) to be the escape velocity from the Earth, or the speed an object falling to the Earth from "infinity", or very far away, obtains. Notice that our object must be between 50000 and 60000 meters above our 50000 meter "top of atmosphere" to obtain a speed over 1000 m/s, so, if that is indeed the speed at which an object will burn up in the atmosphere then we must drop the object from roughly a height of 50000 to 60000 meters (above the "top"). Please keep in mind that this is based on some rough assumptions. You can play with the numbers and assign the assumptions you want.

Now for the second part of this rather long answer! If we call the "top of the atmosphere" 50000 meters above the surface, what would the object's speed be if we drop it at that height? That takes us into much more complicated territory because we need to know the shape of the object and it's density, and the orientation in which it falls. We must compute the "terminal velocity" at each stage of its fall.

To start, though, let's simplify things a bit (A favorite trick of professors everywhere!) and ask what the object's velocity would be at the surface after falling from a height of 50000 meters if there were no atmosphere. The answer: 987 meters per second. So, if the object could fall in vacuum to the surface of the Earth from a height of 50000 meters it would be going almost fast enough to heat up substantially if it were in an atmosphere. The terminal velocity in the atmosphere, though, would keep the object way below this speed, so an object dropped "at the top of the atmosphere" (50 km up) would not burn up.

We can actually go into a bit more detail and find the terminal velocity at various altitudes. At this excellent site is a page that allows calculation of terminal velocity of objects of various types and masses, and using different densities for the air. There's also a good page about terminal velocity here at NASA.

Using the first site, and the second column in the following table, which I found in the book Smithsonian Physical Tables(Smithsonian Institution Press), the final column gives the terminal velocity at various heights in the atmosphere. I am assuming a 1-cm radius plastic ball (density about 1.5 grams per centimeter cubed).

So, this plastic ball dropped from 50000 meters (50 km) would quickly reach its terminal velocity of roughly 877 meters per second, which is less than our assumed "burnup" speed. However, if one were to drop this ball from 110000 meters (110 km), which is 60 km above our "50-km-top of atmosphere" it will have obtained a velocity of 1070 m/s by the time it reaches our "top of atmosphere". It would slow down quickly to terminal velocity, and would probably become quite warm. Warm enough to burn up? I don't have a really definitive answer, but it would certainly be at a speed fast enough to consider it in danger of burning up. But it's easy to see why things burn up if they come from higher altitudes: they can obtain a velocity much faster than both the terminal velocity and our assumed burnup speed.

Objects of various other materials and shapes behave differently. You can find "C" values of various shapes, and play with the calculations at the site above.

John Link, MadSci Physicist