# Would a moon shatter if it collided with an unmovable barrier?

For story purposes, I'm trying to determine what would happen if a small moon-sized object crashed into an unmovable barrier surrounding a planet. My goal here is to eventually get a ring surrounding the planet, but I don't know if that would happen, or if fragments of the moon would achieve escape velocity, or the moon wouldn't shatter at all. So that's my question: Would a moon shatter if it collided with an unmovable barrier surrounding a planet? Details are below.

The moon isn't technically a moon, as it is not orbiting a planet. I'm not even sure if it's orbiting a sun, or is a 'rogue dwarf planet'. Disregard its origins for now. I will call it a 'moon' for simplicity's sake.

Consider the moon to be Mimas, divided by 95. So it has a mass of approximately $3.95 \cdot 10^{17}$ kg (Assuming I did the math right). The moon is primarily icy, just like Mimas. I don't have the faintest idea how fast the moon needs to be moving to shatter when it hits the barrier, so I will leave that up to you. I want it to shatter, not just bounce off. Remember I'm trying to get a ring. For all purposes, assume the moon is the exact composition of Mimas, just 95 times smaller.

The barrier would be some sort of energy barrier, though it's exact composition is unknown at this time. It is surrounding the entire planet and is unmovable. For all purposes, just assume that the moon has run into a material barrier which is unmovable, and indestructible. The moon hits this barrier head-on, at a perfect ninety-degree angle. The barrier will not give at all, so all force is transferred back to the moon.

Given these details, and assuming the speed of the moon is sufficient to shatter it, as opposed to just ricocheting back into space, what would happen to the moon? Please let me know if you need additional details.

Disregard for now the chances of such an event actually occurring.

• How did the moon come to collide with the planetary barrier? If it is from without the solar system, it will impact at solar-system-escape velocity (at minimum). That might be 40km/s. If it's a real moon, it will impact as somewhere between 8 and 11 km/s (all values for earth) - independent of the weight of the moon. Because energy is ~ velocity squared, that's a 16x energy-difference right there. Is the barrier translucent for radiation? If it is, the side of the planet that watches the impact most likely gets grilled. At any appreciable velocity the moon will behave like a water droplet. – bukwyrm Jul 11 '18 at 5:55
• The physics from this question are so far away from reality that every answer is possible, so is, they are all meaningless. – Rekesoft Jul 11 '18 at 10:15
• If you want a ring around your planet, why not ask how to get a ring around a planet? This smells badly of a XY question to me. – a CVn Jul 11 '18 at 12:51
• @MichaelKjörling Asking how to get a ring around my planet would be far too opinion based, as there are several potential ways to do it. – Thomas Reinstate Monica Myron Jul 11 '18 at 14:41
• @Rekesoft Could you enlighten me on what I'm missing so that I can add it in? Preferably in layman's terms... – Thomas Reinstate Monica Myron Jul 11 '18 at 14:42

# Yes it will break, spectacularly

For this demonstration I will use the Earth and the Moon to have some solid numbers to use.

The Moon's orbital velocity is 1 022 m/s. Let us first assume a dead stop. The Moon just crumbles in place... no splashing, and all the energy is converted to heat.

The kinetic energy of the moon is... $E = \frac{mv^2}{2} = \frac{7.342×10^{22} kg * (1.022×10^3 m/s)^2}2$

This in turn translates to approximately 500 000 J per kg, or 500 kJ / kg.

Most rocks and stone have a fairly poor specific heat capacity, usually below 1 kJ per kg and K.

So this collision heats up the material of the Moon by 500 degrees Kelvin.

The material closest to the impact point will take the brunt of it and be heated the most. We can expect that to become several thousands of degrees. That will be vapourised. The outer layers of the Moon will then be blasted away from the hot core like a water melon with a fire-cracker in it.

The side of the Earth that is facing the Moon will then be subject to several Extinction Level Impact Events.

Mimas is a lot worse, because it has an orbital velocity of 14 620 m/s. And while Mimas is assumed to be mostly water, which has a higher specific heat capacity of 4.1 kJ/kg, that still means a dead stop will translate to a heating of about 26 000 Kelvin per kg.

So will it break? Oh yeah... it will vapourize into something that is even hotter than vapour. Mimas will be converted to plasma... not even gas, but it will be a rapidly expanding cloud of plasma.

And if I am not completely guessing wrong, on Earth we will see a bright new star flare up for a few seconds, probably visible even during daylight.

• What happens if Mimas is moving much slower? Say around 1000 m/s? Can I avoid the plasma? – Thomas Reinstate Monica Myron Jul 13 '18 at 14:35

The moon would certainly shatter, not unlike a dirt clod against a brick wall. if the barrier were perfect (inflexible, immovable, etc.), much (if not most) of the moon's mass would bounce away, most in a circular pattern within 45° of the plane of the barrier. Some would bounce back. The rest would sit there, stopped.

• "The rest would sit there, stopped." In which reference frame? I'm having a hard time imagining anything "sitting, stopped" in space. – a CVn Dec 26 '18 at 12:28
• @aCVn, I thought that was obvious: the unmovable barrier's reference frame. – JBH Dec 26 '18 at 15:13

I was trying to find an equivalent model of an icy moon hitting an unmovable object. The closest thing that came to mind was a snow ball launched on a brick wall.

If you live in those parts of the world having snow and brick walls, you know that when you launch a snow ball on a wall you have a part of the balls that shatters, spreading snow around, and the remaining part that holds onto the wall.

The collision between the wall and the ball is basically inelastic, and I guess that the amount of snow remaining on the wall depends, among other things, on how effectively one compacts the snow before throwing it (someone should fund a study on this). It should be therefore possible to make a snow ball which accurately models the density of the moon in the question for a more accurate experiment.

• I suspect the amount of snow that sticks to the wall is also, possibly primarily, a function of how wet or humid the snow is. My experience is that wet/humid "heavy" snow tend to stick better than the dry stuff to most surfaces, including clothes. – a CVn Jul 11 '18 at 21:12
• @MichaelKjörling, that can make sense: the more the humidity/wetnesse, the easier to make ice when compacting it – L.Dutch - Reinstate Monica Jul 12 '18 at 3:33

The moon would of course be reduced to rubble, a lot of hot gasses, and everything in between.

The real issue is that your barrier robbs the mass of all of its orbital velocity, so you get an apocalyptic rain of a lot of nasty stuff on the planet below (or somewhere else, if the planet is protected by the "unmovable barrier surrounding a planet"; i'm not sure of the layout here since the moon is expected to crash into it head-on) instead of a ring.