# 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.

• I think it would be good to rewrite it to focus on the "ice cube from orbit" issue, that's somewhat interesting. – Volker Siegel Jul 23 '14 at 15:05
• Can you as a physicist share your thoughts on problem? We are non experts and have a hard time grasping the scope of the issue. Thanks! – parceval Jul 23 '14 at 15:46
• I was thinking it will get less down votes and better answers if you remove the sentence with the climate conditions (a completely different question) and maybe other parts of the "back story" - it should not cover the physical part too much. But don't worry, as it got upvoted and has a good answer already. – Volker Siegel Jul 23 '14 at 16:14

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.

• @user2012797 no, he's saying for $N$ grams of ice in your cup, you need to dissipate $10*N$ grams' worth of ice-energy, so you can never catch up. – Carl Witthoft Jul 23 '14 at 16:57
• @user2012797: no, because the 10kg of ice will produce as much energy as would melt 100kg of ice. But it you add an extra 100kg then that ice would produce as much energy as needed to melt 1000kg of ice, and so on. It can't be done by just melting the ice. You need to find some other way of disposing of 90% of the energy. – John Rennie Jul 23 '14 at 16:57
• So you are basically saying if a big ice meteor is heading earth, we could just sit back and watch the stars? – parceval Jul 23 '14 at 17:44
• @parceval John's calculations assume an equilibrium of sorts, and deals only with thermodynamics. A large enough meteor will hit the ground long before thermally equilibrating, and (for km-sized objects and above) even before a sound wave can cross from the front of the object to the back, and so kinetics becomes more important. – user10851 Jul 23 '14 at 18:50
• @parceval: my calculation is based on the fact you need to bring the ice to a stop (unless you want it smash your glass). All the energy needs to be dissipated to slow the ice from its 20,000 or so mph orbital velocity to zero. However the ice meteor won't be coming to a stop - it'll be hitting us at something like 20,000 mph. This wouldn't be pleasant. – John Rennie Jul 23 '14 at 19:19