# Ok, so we still need more boosters.

Begin LogEntry_0005 StarDate 197709.03

Ok, so this hasn't gone too well. Remember that "dust particle into low orbit" thing we wrote about two logs ago? Yeah, that turned out to be closer than we thought. The calculations we fed to our robots gave us results that looked promising, but when we told them to validate it, they just shrugged and showed hand sings we think mean "rocket crashing and everyone dies". We're not sure, but what else could a hand rising, then falling followed by explosion noises mean. When we asked for further explanation, they simply wrote "I'm sorry Dave, I'm afraid nature can't do that", we don't know who this Dave fellow is, but we're fairly certain he won't be pleased with the results. They're not built to communicate well, just calculate. It's not their fault.

This is to say, that from our calculations we reached escape velocity, and managed to enter interplanetary space, however the robots seem to disagree.

We feel it relevant to describe our method for these simulations, first of all we set up a coordinate system with or planets center in the origin, and then we placed our rocket at the coordinates \((r_0,0)\), where \(r_0\) is the planets radius.

From there we had to find our acceleration \(\vec{a}\) which is a combination of our thrust \(F\) and the gravitational acceleration of the planet \(g\), combining all this our acceleration is: \(\vec{a} = \Big( \frac{F}{M_{rocket}+M_{fuel}} - \frac{GM_{planet}}{r^{2}} \Big) \hat{u}_{r}\), where \(r\) is the distance to the planets center and \(\hat{u}_r\) is the unit-vector pointing in the direction of our rocket.

This has to be recalculated for each timestep in our simulation, because the distance to the planet \(r\) increases(hopefully), and the fuel mass \(M_{fuel}\) also decreases(probably).

Our planet is (luckily) not tidally locked with the star, and because of this we also have to account for our planets rotation that we bring with us as we fly off.

How this adds to our total velocity is not that hard to find out, from the illustration on the left you can easily see that \(\vec{v} = \vec{\omega} \times \vec{r}\) and solving this with \(\vec{omega} = (0,0,\omega_0)\) and \(\vec{r} = (x, y, 0)\) gives us \(\vec{v} = (-\omega_0y, \omega_0 x,0)\) .

Now we simply add this to our velocity at each time step(multiplied by the time step), as well as the rest of the normal calculations.

Oh and as all "good" physicists do, we're ignoring drag from air resistance(because it's hard).

Anyway, our new approach will be trying to understand how they work on this, and maybe just extract their methods and use them. It won't teach us how it's done, but we can at least move along quickly enough.

For now, our failed experiments are appended here.

End LogEntry_0005

Fig 2 is sourced from: https://www.omnicalculator.com/physics/angular-velocity