Delta V and orbital bombardment

[paltrysum]

WARNING: I'm about to expose myself as the math-challenged individual that I am. Not that anyone needs encouragement around here, but please feel free to disavow me of my illusions.

In the Core Rulebook (p. 153) we are given transit times that include acceleration and deceleration for several arbitrary distances. I'm trying to figure out how much delta V one could accumulate given various amounts of time and acceleration. Interpolating from the table data:

• At 1G acceleration the game supposes that you can accelerate and decelerate to a distance of 1,000 km in 633 seconds.
• So at 500 km, the ship turns around and slows down. So let’s cut the time in half, too: 316.5 seconds to that halfway point. (1.58 km/sec.)
• To get the delta V, can we just accumulate the 1.58 km/sec figure? For example: https://docs.google.com/spreadsheets/d/ ... sp=sharing

I suppose we could make it more complex and accurate using a proper delta-V calculation: https://en.wikipedia.org/wiki/Delta-v

But I'd be content with a hand wavey Traveller rules-based solution that achieves some modicum of realism.

This all comes of my players' possible use of a planetoid or starship as a missile launched at a terrestrial target on a world with minimal orbital defenses. I'm trying to achieve some level of verisimilitude and step one is figuring out how fast they could accelerate said missile.

[AnotherDilbert]

Forget ∆v, that is defined in the context of Tsiolkovsky's rocket equation that we don't have to deal with since we have reaction-less drives.

We just have constant acceleration, regardless of petty concerns like Conservation of Energy.

Basic formula:
So with a few integrations (assuming we start at position 0 with speed 0),
Acceleration [m/s²] = A __________ (constant, say 1 g ≈ 10 m/s²)
Velocity [m/s] ____ = A × t _______ (constant times time in seconds)
Position [m] ______ = A × t² / 2 ___ (constant times time squared)


Example:
So if we accelerate at A = 1 g ≈ 10 m/s² our velocity will increase by 10 m/s every second.
After 1 hour = 3600 s we have attained a velocity of 10 × 3600 = 36 000 m/s.
After 1 day = 24 hours we have velocity 36000 × 24 = 864 000 m/s.
After 1 month = 30 days we have velocity 864000 × 2 4 = 25 920 000 m/s.

Around that velocity relativity will start to rear its ugly head, so we will not go higher for now...

During that month we have moved A × t² / 2 = 10 m/s² × (30×24×3600 s)² / 2 ≈ 33 600 000 000 000 m ≈ 224 AU (the distance between the Earth and the Sun).


Meaning:
This represents, say, a micro-jump away from the planet, and accelerating back.

We arrive with a velocity of 25 920 000 m/s (from above). Let's say we do this in a small starship with a mass of, say, 1000 tonnes = 1 000 000 kg. We then have a kinetic energy of mv²/2 = 1000000 × 25920000² / 2 ≈ 338 000 000 000 000 000 000 J ≈ 80 000 megaton.

This is 1000 times a large nuke, but 1000 times smaller than an extinction level asteroid impact like Chicxulub.

I'm guessing here but it might make a continent uninhabitable, but not the entire world?


End Note:
Sorry for the rambling, but if we know the formulas in bold above it's relatively simple to find the velocity and position after accelerating for a given time. Run the formula in reverse to find the time it took to accelerate to a given velocity or position.

It's not quite this simple in the real world, but it's a reasonable estimate.

I hope this made at least a sliver of sense to you?

[paltrysum]

Nicely stated, AnotherDilbert. Sliver of sense has been achieved. You have given me enough to create a nice, reasonably realistic event for the adventure. Thanks for taking the time!

[steve98052]

The ability to wipe out life on entire continents with kinetic bombardment is one of those Traveller ideas that are often discussed, often with results that violate laws of social science, as well as conservation of momentum.

Take the example of the starship that does a microjump, then a month long full power acceleration toward a planet. If it's done as an act of war, it likely leads to a total war retaliation, along the lines of the "family atomics" agreement in Dune. Besides the deterrent effect of retaliation, empires don't use such attacks because they want to conquer worlds, not destroy them.

But a murderer intent on going out in a blast of massive destruction is not deterred by the possibility of massive retaliation. He kills a continent full of people, and because he's killed himself, there's no one to retaliate against unless it's proven that he was working for some enemy power, rather than solely nihilistic destruction.

[phavoc]

NASA worked out how much damage their probe orbiting Mercury will do when they crash it into the surface. Basically something weighing the same as a Harley impacting from orbit will make a crater about 50m wide (didn't provide depth), or the equivalent of about a ton of TNT. All the formula's are there if you were curious and wanted to try your hand at the math.

Mass would be very important to determine the destructive energy. Since the destructive force is essentially mass x energy (e.g. speed), the more mass the impacting vessel has the bigger the boom will be.

[paltrysum]

I guess I'm kind of alarmed that we're littering on the surface of Mercury, but hey, science must march on! Interesting stuff, phavoc. Thanks for the share.

[AndrewW]

I guess I'm kind of alarmed that we're littering on the surface of Mercury, but hey, science must march on! Interesting stuff, phavoc. Thanks for the share.

Wait till the Martians send us the bill.

[Moppy]

If the planet has an atmosphere, you do not to hit it particularly hard with an asteroid - the atmosphere will make any impact much worse. Explosions in vacuum don't actually do anything unless you get hit by shrapnel. Explosions in atmosphere cause a pressure wave that will kill people, knock over buildings, start tidal waves etc . All you need is a rock big enough to not burn up in the atmosphere.

A high tech planet can probably deal with the effect of tidal waves and dust blocking out the sun.

While you're on the subject, there's a fusion reactor in your starship you can blow up whenever you want.

edit: Although if it wasn't moving that fast, the coastguard will probably see it from a long way off. Assuming they had one. There's a lot of under-developed worlds in Traveller.