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How much fuel would be required to send a 300g satellite to space using a rockoon that can travel to a height of 32 km, and then send the satellite to a inclination of 28, Apogee: 350 km, Perigee: 280 km, using an APCP propellant?

Please provide the exact calculation if possible so I can follow.

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    $\begingroup$ I don't think that the 32 km "rockoon" altitude matters at all compared to launching from the ground. The most elaborate concept with airship to space I know of is that of JPaerospace jpaerospace.com/atohandout.pdf $\endgroup$
    – LocalFluff
    Sep 30, 2014 at 7:19
  • $\begingroup$ I guess the air density, thus drag would be drastically reduced hence the fuel requirement, if am correct ? $\endgroup$
    – Sujay sreedhar
    Sep 30, 2014 at 7:40
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    $\begingroup$ Air density is unimportant, conventional rockets move slowly through the lower atmosphere anyway. Launching is about speed, not height. You need to move at about 7,700 meters per second, or 28,000 kilometers per hour (100 times the speed of an F1 racing car), in order not to fall back to Earth. A balloon does very little to better that, it just creates a problematic environment for rocket launching. But then again, that JPaerospace airship to space concept seems to make a point. $\endgroup$
    – LocalFluff
    Sep 30, 2014 at 7:44
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    $\begingroup$ Possible duplicate of Effect of atmospheric drag on rocket launches and benefits of high altitude launch sites $\endgroup$
    – TildalWave
    Sep 30, 2014 at 9:28
  • $\begingroup$ Its more like Jpaerospace's plan, We plan to use Rockoon , what it would do is take the rocket to 32Km height and launch a small rocket to space. well that's the plan. $\endgroup$
    – Sujay sreedhar
    Sep 30, 2014 at 10:08

1 Answer 1

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Randall Munroe in his What If book says it well:

The reason it's hard to get to orbit isn't that space is high up.
It is hard to get to orbit because you have to go fast

Being $32$ km higher doesn't save much at all. Lifting a mass from $32$ km to $302$ km uses $270,000 \cdot 9.8 \approx 2650$ kjoules/kg.
Lifting a mass from sea level to $302$ km uses about $3,000$ kjoules/kg.
Accelerating to $7.73$ km/sec orbital velocity given here uses $\frac 12 (7730000)^2 \approx 3\cdot 10^{10} $ kjoules/kg, over $10,000$ times more.

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    $\begingroup$ Comparing potential and kinetic energy is an incomplete model. There is an additional expense of gravity loss during vertical ascent. The ascent takes more energy than the potential energy difference between altitudes. $\endgroup$
    – HopDavid
    Dec 30, 2014 at 21:28
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    $\begingroup$ Gravity drag and atmospheric drag should be considered. $\endgroup$
    – Erik
    Jan 29, 2015 at 22:14
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    $\begingroup$ Atmospheric density at 32km is basically nothing (en.wikipedia.org/wiki/U.S._Standard_Atmosphere#mediaviewer/…) -- so it would make a huge difference on the design of a vehicle. For one thing, you would probably not have to worry about designing a thrust bucket to reduce max Q. The amount of time you would spend fighting gravity drag would be much less too, so you could probably fit a lower thrust/weight ratio propulsion system. These are not small advantages. Of course, equipping a balloon with any reasonable payload size is problematic $\endgroup$
    – Erik
    Jan 30, 2015 at 3:31
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    $\begingroup$ The STS thrust bucket was absolutely not to protect the launch pad. "So, shortly after T zero, shortly after lift off, we throttle the main engines back down to around 64% rated power to keep that dynamic pressure on the vehicle to a minimum. If we didn't throttle down, the loads on the external tank and the solid rocket boosters and the orbiter would be too high because we'd be flying faster through this regime in the atmosphere called the maximum dynamic pressure. "tinyurl.com/asz5avd $\endgroup$
    – Organic Marble
    Mar 1, 2015 at 3:15
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    $\begingroup$ @RossMillikan The spacecraft needs to ascend to about 100 km before it does the major horizontal burn. Gravity is 9.8 meters/sec^2. Each 102 seconds of vertical ascent costs 102 seconds * 9.8 m/s^2 gravity loss. That's 1 km/s. Time for a 68 km vertical ascent would be substantially less than the time spent for a 100 km vertical ascent. Especially if you didn't have to throttle back to avoid excessive max Q. $\endgroup$
    – HopDavid
    Mar 31, 2015 at 3:23

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