Ok. Now this would be quite light. It's not pop corn which is green. Or even if it was green and wet, the cob would dry out [freeze dry] in space.So, it's low density.I would say impact on surface Earth would be somewhere around baseball pitcher's fast ball [100 mph].But Mars would much faster.So can I assume from your comments that the pop corn on cob, survives re-entry? Or are saying *even if it somehow were to survived re-entry* it would hit the ground at very high velocity [both Earth and Mars]?

(but buy the books )

No, read carefully! Take the square root of the C3. sqrt(60km^2/s^2)=7.75km/s

What books?

So if the object is escaping Earth at 7.75 km/s, the object would have a relative velocity to the Sun of 26.75 km/s?

What Robotbeat said. But make sure the value you got is C3, the same unit is also used for specific orbital energy (of which C3 is a special case) in which case you first need to multiply the value by two before taking the square root. Wiki explains these quite well;http://en.wikipedia.org/wiki/Specific_orbital_energyhttp://en.wikipedia.org/wiki/Characteristic_energy(but buy the books )

Quote from: Robotbeat on 07/30/2013 03:41 PMNo, read carefully! Take the square root of the C3. sqrt(60km^2/s^2)=7.75km/sAh, I multiplied by 2 thinking it was specific orbital energy, but it wasn't.

I understand He, but why is Xe used?

The selected configuration for the Project Prometheus Space Nuclear Power Plant was a direct coupling of Brayton energy conversion loop(s) to a single reactor heat source through the gas coolant/working fluid. A mixture of helium (He) and xenon (Xe) gas was assumed as the coolant/working fluid. Helium has superior thermal conductivity while xenon is added to increase the gas atomic weight to benefit turbomachinery design.

I've looked for this question here and couldn't find it, thought I might have skimmed over it because 56 pages is a bit much to read very well.How can you estimate the gravitational drag of a rocket? The only thing I could find was "Divide orbital dV by acceleration in planetary G's" but doing that for a few real life rockets has given me ridiculous amounts of gravitational drag.

Quote from: M129K on 09/29/2013 01:45 PMI've looked for this question here and couldn't find it, thought I might have skimmed over it because 56 pages is a bit much to read very well.How can you estimate the gravitational drag of a rocket? The only thing I could find was "Divide orbital dV by acceleration in planetary G's" but doing that for a few real life rockets has given me ridiculous amounts of gravitational drag.You mean gravity losses? That's the integral of g(h(t)).sin(rho(t)), where rho is the angle to the normal of the gravity source (or cos() if you take to the tangent to the Earth's surface) and h it the distance to the gravity source. You can assume that in Earth's case, at 6,371km distance the gravity is 9.8m/s.

Quote from: baldusi on 09/29/2013 06:06 PMQuote from: M129K on 09/29/2013 01:45 PMI've looked for this question here and couldn't find it, thought I might have skimmed over it because 56 pages is a bit much to read very well.How can you estimate the gravitational drag of a rocket? The only thing I could find was "Divide orbital dV by acceleration in planetary G's" but doing that for a few real life rockets has given me ridiculous amounts of gravitational drag.You mean gravity losses? That's the integral of g(h(t)).sin(rho(t)), where rho is the angle to the normal of the gravity source (or cos() if you take to the tangent to the Earth's surface) and h it the distance to the gravity source. You can assume that in Earth's case, at 6,371km distance the gravity is 9.8m/s.If I understand it correctly, t is total time? If so, thank you! Seems very accurate.