Getting to Earth Orbit: vertical dV?

[daniu]

So in this what-if http://what-if.xkcd.com/58/, I stumbled upon the statement "Getting to space is easy [...] just by going fast and then steering up [...] The problem is staying there".

Well yeah, I'm sure reaching the orbital velocity of ~7km/s at 200km is the larger part of getting into earth orbit; but how much dV is needed for the vertical part exactly?

I can't find data on it, and my math fails me...

In other words: if I were to use a non-gravity turn lift to go up 200km before doing the 7km/s orbital velocity burn, how much energy/dV would that take? I can't imagine it being that little.

And how large a part do gravity and atmospheric drag play in it, respectively?

[Deathsoul097]

Gravity is the real killer, and it eats most of your DV. As for drag, it's bad but not so bad as you would expect. Considering you need ~9 - 10 Km/s of DV to get to earth orbit, I would say that gravity losses and drag losses account for ~2-3Km/s of DV usage when getting to orbit. (It's been a long time since RSS)

So really, yeah, getting to space is easy. Staying there, not so much. (Trust XKCD on this one, he worked at NASA and he plays KSP...)

[Pursuedtank]

You'll not want to go streight up and then burn horizontally, any benefit you might gain from this would be tiny compared to the extra delta v you would need.

The best method is a smooth turn as you go up, if you are using FAR.

[cantab]

He mentioned it in another what-if, http://what-if.xkcd.com/24/

About 2 km/s, and that factors in drag.

You can determine the figure neglecting drag by the SUVAT equations. 100 km is small enough compared to the Earth's radius that treating gravity as constant should be fine.

[Z-Man]

Neglecting air resistance, the minimal dv requirement can be calculated in reverse: start at 200 km, fall down, calculate the speed as you hit the ground. The formula for that is v = sqrt(2gh) and gives 2 km/s.

If you don't want to get shot out of a cannon and prefer to be gently accelerated, say feeling 2g, for the first 100 km and then free fall for the second 100 km, that number needs to be multiplied with sqrt(2). Air resistance can be neglected for these kinds of napkin calculations, but if you must, make that a factor of 1.5, which yields 3 km/s.

Sooo... Deathsoul097's intuition is 100% correct.

[xcorps]

He mentioned it in another what-if, http://what-if.xkcd.com/24/

If you try to produce an orbital rocket using the same design math we used for the suborbital rocket, it spits out a description of a pancake-shaped mountain of model rocket engines over a mile wide. It would taper to a 10-meter-high spire in the center and would weigh about as much as the Great Pyramid.

I think Whackjob needs a go at this?

[cantab]

When I briefly mucked around building stuff powered just by Sepratrons, I thought of that xkcd.

But a model rockets mod, with tiny radial sizes and engines that match the specs of the Estes ones and so on, would be fun. Wonder how many model rocket engines you'd need for a Gilly lander?

[Deathsoul097]

Neglecting air resistance, the minimal dv requirement can be calculated in reverse: start at 200 km, fall down, calculate the speed as you hit the ground. The formula for that is v = sqrt(2gh) and gives 2 km/s.

If you don't want to get shot out of a cannon and prefer to be gently accelerated, say feeling 2g, for the first 100 km and then free fall for the second 100 km, that number needs to be multiplied with sqrt(2). Air resistance can be neglected for these kinds of napkin calculations, but if you must, make that a factor of 1.5, which yields 3 km/s.

Sooo... Deathsoul097's intuition is 100% correct.

Woo! Estimations and RSS, FTW! (I guess RSS is pretty accurate after all. Nice work, NK.)

[maccollo]

The Saturn V had a fairly low lift off TWR, only 1.17, and the gravity losses only amounted to 1740 m/s, and a mere 50 m/s of drag losses.

http://www.braeunig.us/apollo/saturnV.htm

Edited May 22, 2014 by maccollo