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0 pointsaDyslecticCrow5mo ago0 comments

> Going fast enough sideways so you stay up there is the tricky bit.

nah, thats the simple part. getting up there efficiently is the difficulty. once we're up, its just a matter of force over time to create a nice orbit.

The faster you go, the more friction you face, and the more heat and vibration your equipment must endure.

Going slower reduce friction and stress but use more energy just negating gravity. Slow rocket is inefficient rocket.

So we wanna leave the atmosphere as soon as possible, but not so fast that the rocket melts or engines collapse. Prefferably just below the sound barrier.

once we're up, its pretty chill... until you wanna go down again. Slow rocket is alive rocket.

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advisedwang5mo ago

Energy for 1kg to reach LEO (800km * 1kg * 9.8m/s2) ~ 8MJ

Energy to reach LEO velocity ~ (1/2 * 1kg * (8km/s)^2) ~ 32MJ

nroets5mo ago

The rocket fuel needed to produce that 40 MJ weighs close to 1 kg, especially when you include the oxidiser. So the energy needed to accelerate 1kg of payload to LEO velocity is much more.

_carbyau_5mo ago

That whole "tyranny of the rocket equation" thing is why I am surprised the actual first stage for launching a rocket is NOT a ground based reusable "up-chucker".

Basically, I would have thought that any momentum that can be imparted to the rocket before it has to rely on its self propulsion would be a huge help. Not talking about eliminating self propulsion, just an assist so the rocket could carry a larger payload or be smaller or whatever.

IE like a variation on Jules Verne's big gun for throwing the payload up there but engineered to be plausible and having the rocket still be self propelled. And safe.

But we don't seem to do this. So why?

Edit: First part of video [0]. Apparently it's not completely dumb. Just stupid-hard/impossible to do practically at the size required for big rockets and payloads. But small ones might work. Maybe.

[0] https://www.youtube.com/watch?v=lWYn5hl4QWg

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aDyslecticCrowOP5mo ago

But the energy needed is not an indicator of what is difficult or dangerous. Leaving the atmosphere intact is the most difficult part of launching a rocket going by failure rate. Of those that reach space, those that still fail often took damage from the launch.

Once you're in space, force over distance until your fuel runs out.

jojobas5mo ago

Failures rate is even less a suitable indicator. Going up 100km is achievable by a simple single stage solid fuel rocket. Going to orbit requires way way more complexity, including a giant first stage that can fail in atmosphere.

hinkley5mo ago

The best way I’ve seen it described is that the first 9.8 m/s² of thrust only makes you hover in place. So the more g forces you generate the higher the efficiency of the engine - as long as thrust per kg of fuel doesn’t dip too far.

This all changes in outer space, where 0.01g is a valid propulsion mechanism for long duration missions.

LorenPechtel5mo ago

Yup. Look at the launch trajectory of the Webb telescope. The upper stage engine was too weak for the orbital insertion and the booster had to waste energy putting it higher than need be so the upper stage engine had altitude to trade for time to keep the telescope from hitting the atmosphere.

But it was the optimum solution because that engine had a long burn to take it to L2. Hauling less engine to L2 was worth more than the loss of the engine not being powerful enough to fight gravity.

hinkley5mo ago

God bless the Webb team but if a manager came to me to pitch a software project with as many sequential events that had to go perfect I would have told him to fuck right off. After I stopped laughing so hard I couldn’t breathe.

I made a pikachu face when they returned the first image.

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literalAardvark5mo ago

It's not that simple though. The rocket equation still applies so it's almost as hard to do (you just get rid of atmospheric drag), and failed launches are also extra catastrophic.

Even more, your delta v required is still huge. I can't be bothered to run the numbers right now but most of the delta v is in the orbital velocity, not in the altitude.

BobaFloutist5mo ago

Delta v feels reductive, since your fighting negative acceleration to go up and a fraction of that to go laterally, no?

JumpCrisscross5mo ago

> since your fighting negative acceleration to go up and a fraction of that to go laterally

They’re both acceleration. At high thrust, virtually equivalent.

lazide5mo ago

They’re the same thing. Either you go sideways really fast, or go straight up really far (geo-synchronous distance).

Either way, you need the same total velocity delta.

LorenPechtel5mo ago

Actually, not quite. You tolerate a bit of extra fire in exchange for reduced gravity loss. Going straight up until you clear atmosphere is not the minimum energy trajectory.

raducu5mo ago

> once we're up, its just a matter of force over time to create a nice orbit.

It depends what you mean by "up there". ChatGpt tells me you'd free fall from 1000 km to 100km in about 8 minutes. It also did the math that you'd need 1.65G of sideways thrust to reach orbital speed. That's quite a bit of force for spacecraft sized objects.

If you have an actual space elevator, sure, you can go to close to geosynchronous altitude and by that time you'd have enormous sideways velocity just by being dragged sideways by space elevator and indeed it would be easy to propel yourself to orbit (above a certain altitude my intuition tells me you could let go of the rope and while you'd end up on an eliptise you'd still be in orbit)

hermitcrab5mo ago

>ChatGpt tells me you'd free fall from 1000 km to 100km in about 8 minutes

You trusted an LLM to do the maths when it is just s = 5t^2?

kaonwarb5mo ago

Not quite; g drops materially between 100 km and 1000 km. 8 minutes I believe is quite close, whereas 5t^2 (or even 4.9t^2) will lead to underestimating time.

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advisedwang5mo ago

Energy for 1kg to reach LEO (800km * 1kg * 9.8m/s2) ~ 8MJ

Energy to reach LEO velocity ~ (1/2 * 1kg * (8km/s)^2) ~ 32MJ

nroets5mo ago

The rocket fuel needed to produce that 40 MJ weighs close to 1 kg, especially when you include the oxidiser. So the energy needed to accelerate 1kg of payload to LEO velocity is much more.

_carbyau_5mo ago

That whole "tyranny of the rocket equation" thing is why I am surprised the actual first stage for launching a rocket is NOT a ground based reusable "up-chucker".

IE like a variation on Jules Verne's big gun for throwing the payload up there but engineered to be plausible and having the rocket still be self propelled. And safe.

But we don't seem to do this. So why?

Edit: First part of video [0]. Apparently it's not completely dumb. Just stupid-hard/impossible to do practically at the size required for big rockets and payloads. But small ones might work. Maybe.

[0] https://www.youtube.com/watch?v=lWYn5hl4QWg

3 more replies

aDyslecticCrowOP5mo ago

Once you're in space, force over distance until your fuel runs out.

jojobas5mo ago

hinkley5mo ago

This all changes in outer space, where 0.01g is a valid propulsion mechanism for long duration missions.

LorenPechtel5mo ago

But it was the optimum solution because that engine had a long burn to take it to L2. Hauling less engine to L2 was worth more than the loss of the engine not being powerful enough to fight gravity.

hinkley5mo ago

I made a pikachu face when they returned the first image.

1 more reply

literalAardvark5mo ago

It's not that simple though. The rocket equation still applies so it's almost as hard to do (you just get rid of atmospheric drag), and failed launches are also extra catastrophic.

Even more, your delta v required is still huge. I can't be bothered to run the numbers right now but most of the delta v is in the orbital velocity, not in the altitude.

BobaFloutist5mo ago

Delta v feels reductive, since your fighting negative acceleration to go up and a fraction of that to go laterally, no?

JumpCrisscross5mo ago

> since your fighting negative acceleration to go up and a fraction of that to go laterally

They’re both acceleration. At high thrust, virtually equivalent.

lazide5mo ago

They’re the same thing. Either you go sideways really fast, or go straight up really far (geo-synchronous distance).

Either way, you need the same total velocity delta.

LorenPechtel5mo ago

Actually, not quite. You tolerate a bit of extra fire in exchange for reduced gravity loss. Going straight up until you clear atmosphere is not the minimum energy trajectory.

raducu5mo ago

> once we're up, its just a matter of force over time to create a nice orbit.

hermitcrab5mo ago

>ChatGpt tells me you'd free fall from 1000 km to 100km in about 8 minutes

You trusted an LLM to do the maths when it is just s = 5t^2?

kaonwarb5mo ago

Not quite; g drops materially between 100 km and 1000 km. 8 minutes I believe is quite close, whereas 5t^2 (or even 4.9t^2) will lead to underestimating time.

1 more reply

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