# Calculate the Kinetic Energy of an Impactor (Fun in Physics)

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So, just for fun.

I play Eve Online - earlier this year, they had a live event that resulted in a Leviathan-class Titan plowing into a planet from orbit. According to the game lore, very little damage was sustained on the planet - I'm just trying to find out how much impact energy such a collision would actually have.

For this purpose, I am assuming the planet itself is roughly equivalent to the Earth in terms of size, density, and chemical composition.

Ship Stats:

$Mass= 2,430,000,000 kg$

$Volume= 132,500,000 g/cm^3$

Given that the ship was in orbit around the planet at the time, and was subsequently forced to crash, I'll assume the initial velocity was the orbital velocity.

In order to calculate the orbital velocity we can set the centripetal force and the gravitational force equations equal to each other, and solve for the velocity in m/s

$\frac{mv_o^2}{r} = G\frac{Mm}{r^2}$

Where:

$m$ = Mass of the ship

$v_o$ = orbital velocity

$r$ = radius of the orbit

$G$ = Gravitational constant

$M$ = Mass of the planet

Solving for the velocity gives us the equation:

$v_o = \sqrt{G\frac{M}{r}}$

I had to make some assumptions about the orbital radius, but since the lore tells us that these ships cannot orbit too close to a planet's surface, I chose to use a high earth orbit for the orbital radius.

Setting $r = 101,000km$ gives us a final equation of

$v_o^2 = \frac{6.67384\times10^{-11}m^3}{kgs^2}\frac{5.97219 \times 10^{24} kg}{101\times10^6 m}$

$v_o^2 = \frac{(5.97219 \times 10^{24}) (6.67384\times10^{-11})}{101\times10^6 }\frac{m^2}{s^2}$

$v_o^2 = 3,946,281.2386 \frac{m^2}{s^2}$

$v_o = 1,986.5 m/s$ which seems fair, given the orbital velocity of satellites at that altititude.

(continued...)

So my first question is: is it fair to use the orbital velocity or do we need to calculate an actual impact velocity somehow, and if so, how would we go about doing that?

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How can it crash if it has orbital velocity - or is it deliberately driven into the planet? I think you are talking over 8 km/s for actually hitting the earth otherwise your orbit will just get a little smaller and in reality about minumum 11km/s. If you are in orbit you need to do something pretty energetic to get lower - much as it is tempting to think you don't just fall.

the online estimators of this sort of thing tend to break when you try something that is so light (as in un-dense) as your ship. It has a density of about 18 kg/m^3 - that's only ten times more that CO2 and much much lower than anything we expect might hit the earth (ie ice is 1000kg/m^3 and iron 8000 kg/m^3) . I think it would probably break up in the atmosphere - wouldn't be pleasant - it would be a huge bang in the order of megatons of tnt

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Yeah, this is why I posted the thread. Orbital mechanics and I are not good friends. From what I understand of the lore from the event, the ship was crashed into the planet semi deliberately (through a combination of battle damage and bad luck). It left a city sized crater in the surface and a good majority of the ship can be seen sticking up over the city.

My guess is they simply didn't let physics get in the way of a good story - they wanted the end result, so they "magicked" it in.

Let's remove the atmosphere for a minute. Assuming you pointed this thing at the planet with 0 orbital velocity and just let it free fall in under gravity alone, is there a way to calculate the final impact velocity?

Actually, it looks like there is - from our good friend conservation of energy.

$gmh = \frac{1}{2}mv^2$

where:

$g = acceleration\; due\; to\; gravity\; (-9.8 m/s^2)$

$h = height$

Solving for v we get

$v = \sqrt{2gh}$

So the velocity just prior to impact would be

$v^2 = 2\times 9.8 m/s^2 \times 101,000 m$

$v^2 = 989,798 m^2/s^2$

$v = 994.89 m/s$ or just under Mach 3.

Edited by Greg H.
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If your planet resembles Earth, the speed on low orbit is near 8000m/s. Crashing from a low orbit take a minimal brake, like 50m/s only. The ship gains some speed by losing altitude, it's like 200m/s. Put together, you can take the orbital speed as the impact speed, which is then nearly horizontal - unless the ship made a hard manoeuvre that takes very much propellant with present human technology.

1000m/s downwards is reasonable for 100km height, but this would demand to stop the 8km/s orbital speed first.

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If your planet resembles Earth, the speed on low orbit is near 8000m/s. Crashing from a low orbit take a minimal brake, like 50m/s only. The ship gains some speed by losing altitude, it's like 200m/s. Put together, you can take the orbital speed as the impact speed, which is then nearly horizontal - unless the ship made a hard manoeuvre that takes very much propellant with present human technology.

1000m/s downwards is reasonable for 100km height, but this would demand to stop the 8km/s orbital speed first.

Thanks for the input. I think I am going to stick with the "straight-down" approach for now, as that may be the simplest for me. I'll probably come back and revisit the horizontal impact later.

So kinetic energy before impact can be found by

$KE=\frac{1}{2}mv^2$

Substituting our known values we get:

$KE = \frac{1}{2} \times 2,430,000,000 kg \times (994.89 m/s)^2$

$KE = 1,202,614,426,201,500\;joules$

If my math is right, this is roughly equivalent to about 280 kilotons of TNT, according to Wikipedia. In comparison, the bombing of Hiroshima was done with a 16 kiloton weapon.

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Yeah, this is why I posted the thread. Orbital mechanics and I are not good friends. From what I understand of the lore from the event, the ship was crashed into the planet semi deliberately (through a combination of battle damage and bad luck). It left a city sized crater in the surface and a good majority of the ship can be seen sticking up over the city.

My guess is they simply didn't let physics get in the way of a good story - they wanted the end result, so they "magicked" it in.

Let's remove the atmosphere for a minute. Assuming you pointed this thing at the planet with 0 orbital velocity and just let it free fall in under gravity alone, is there a way to calculate the final impact velocity?

Actually, it looks like there is - from our good friend conservation of energy.

$gmh = \frac{1}{2}mv^2$

where:

$g = acceleration\; due\; to\; gravity\; (-9.8 m/s^2)$

$h = height$

Solving for v we get

$v = \sqrt{2gh}$

So the velocity just prior to impact would be

$v^2 = 2\times 9.8 m/s^2 \times 101,000 m$

$v^2 = 989,798 m^2/s^2$

$v = 994.89 m/s$ or just under Mach 3.

Gravity varies with distance. 200000 km away your gravity has dropped by about a factor of a thousand. You need to work in "gravitational potential" with decent distances -GM/x which is in joules per kilo.

It if is in orbit at 100,000 km it will not fall unless you add a vast amount of energy to speed it up to 8km/s.

In my opinion - you either have a suicidal kamikaze run with the ship impacting at close to 90 degrees and constantly correcting to stay on course and not go into orbit, scoot past or impact very horizontally. Or you have a decaying orbit - ie vessel was meant to be put into a low earth orbit and the but just cannot wipe off enough speed.

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Gravity varies with distance. 200000 km away your gravity has dropped by about a factor of a thousand. You need to work in "gravitational potential" with decent distances -GM/x which is in joules per kilo.

It if is in orbit at 100,000 km it will not fall unless you add a vast amount of energy to speed it up to 8km/s.

In my opinion - you either have a suicidal kamikaze run with the ship impacting at close to 90 degrees and constantly correcting to stay on course and not go into orbit, scoot past or impact very horizontally. Or you have a decaying orbit - ie vessel was meant to be put into a low earth orbit and the but just cannot wipe off enough speed.

That makes sense. I'll have to give this some more thought.

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