Thoughts on the ice cube from orbit problem Let's say we have a really exquisite cocktail party somewhere in New Mexico, and we just ran out of ice cubes. To the rescue comes this new service provided by Orbital Glacier Inc.
They provide ice cubes around the world within only 5 minutes (!). How do they do it?
How big must an ice cube be when you drop it from the orbit, so that it will have the size of a typical ice cube when it hits your glass of scotch that you're holding in your hand.
Assuming the ice cube in the orbit needs to be of an enourmous size, would Orbit Glacier Inc. inevitably impact the climate conditions on planet earth when celebrating several such cocktail parties?
Disclaimer: We are in no way related to our competitor Moon Ice Now, Inc.
 A: I don't think the question can be answered because you don't say how the orbital energy is to be dissipated. However it's quite interesting to compare the orbital energy with the energy required to boil the ice.
Let's suppose our ice supplied is aboard the International Space Station, so they are at an altitude of $h = 300\ \mathrm{km}$ and moving at an orbital velocity of about $v_\mathrm o = 7.7\ \mathrm{km/s}$. At the latitude of New Mexico (34° N) the Earth's surface is moving at about $v_\mathrm e = 370\ \mathrm{m/s}$. So the change in kinetic energy is:
$$\begin{align}
 \Delta T &= \tfrac{1}{2}m v_\mathrm o^2 - \tfrac{1}{2}m v_\mathrm e^2 \\
          &= 29.6\ \mathrm{MJ/kg}
\end{align}$$
The change in potential energy is:
$$\begin{align}
 \Delta U &= \frac{GM}{r_\mathrm e} - \frac{GM}{r_\mathrm e + h} \\
          &= 3.1\ \mathrm{MJ/kg}
\end{align}$$
So the total energy change in bringing $1\ \mathrm{kg}$ of ice from the ISS to New Mexico is:
$$ \Delta E = \Delta T + \Delta U = 32.7\ \mathrm{MJ/kg} $$
Could we use this energy to boil off some of $1\ \mathrm{kg}$ of ice and leave the rest available for cooling drinks? Well suppose we start with the ice at absolute zero (it's cold in space) and see how much energy it takes to boil it. The constants we need are:
$$\begin{align}
\text{Specific heat of ice}\ (-10\ \mathrm{^\circ C}) &= 2\,000\ \mathrm{J/(kg\ K)} \\
\text{Latent heat of fusion} &= 334\,000\ \mathrm{J/kg} \\
\text{Specific heat of water} &= 4\,200\ \mathrm{J/(kg\ K)} \\
\text{Latent heat of vap.} &= 2\,257\,000\ \mathrm{J/kg}
\end{align}$$
Assuming these constants don't change with temperature$^1$ the energy required to turn $1\ \mathrm{kg}$ of ice at absolute zero to a $\mathrm{kg}$ of steam at $100\ \mathrm{^\circ C}$ is:
$$\begin{align}
\Delta E &= 2\,000\ \mathrm{J/(kg\ K)}\times273\ \mathrm K + 334\,000\ \mathrm{J/kg} + 4\,200\ \mathrm{J/(kg\ K)}\times100\ \mathrm K + 2\,257\,000\ \mathrm{J/kg} \\
         &= 3.56\ \mathrm{MJ/kg}
\end{align}$$
So the energy required to bring $1\ \mathrm{kg}$ of ice to rest in New Mexico is about ten times the amount of energy needed to boil away the ice even starting from absolute zero. You're going to have to find some other way of dissipating the energy.

$^1$ the specific heat of ice decreases with falling temperature so the energy calculated to boil the ice is a slight overestimate.
