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0 pointsnickjarboe9y ago0 comments

Nit: The energy of an object going 2,000 m/s is 1/16th of that going 8,000 m/s, but since this is a rocket launching from Earth the energy for the rocket to get to 2,000 m/s is much greater than the energy to go from 2,000 m/s to 8,000 m/s [1].

Just look at how much larger the first stage (mass almost all propellant) is compared to the second stage (large fraction of the mass is the payload).

[1] https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation

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vkou9y ago· 2 in thread

That has nothing to do with the rocket launching from Earth (A rocket in space follows the same rules), and everything to do with the fact that rockets expend most of their energy to push their own fuel.

That fact is incredibly relevant for getting to orbit - however, it is not at all relevant for surviving re-entry, where all that matters is your kinetic energy.

nickjarboeOP9y ago

The Earth part was just shorthand for the fact that the atmosphere and gravity of Earth add about 1.3–1.8 km/s of delta v to achieving the pure kinetic delta v of 7.8 km/s needed to achieve LEO [1]. The first stage adds almost all of this "extra" delta v.

I was mostly just correcting your "one-sixteenth of the energy required for orbit" statement. I agree that reentering at 7.8 km/s is much more difficult than 2.0 km/s, but the second stage is smaller and more spherical than the first stage. That might make reentering it a bit easier than if it had the shape of the first stage.

[1] https://en.wikipedia.org/wiki/Low_Earth_orbit

stcredzero9y ago

A better approximation to the 2nd stage's shape would be that of a cow.

(If that made no sense, it's a variation on an old freshman physics joke.)

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