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1

To answer your question, both conditions are satisfied. Yes it is best (most efficient delta-v wise) to perform a maneuver at the line of nodes (at the equator). It is also a characteristic of GEO orbits that the solar gravity interacts with a equatorial (GEO) orbit to drive the line of nodes to the 90/270 line in the inertial frame. (the tilt of the earth ...


0

Your error is in the calculation of the alteration of the semi-major axis. If you are going from 250 km circular altitude, and perform a maneuver to achieve an apogee of 35,786 km altitude, your new post-burn semi-major axis will be (250 + 35,786 )/2 + 6,371 km = 24,389 km. Your change in semi-major axis is then 18,018 km.


0

I think one can combine the various answers and comments thusly: Moon It is clearly within our general technological ability to reach the Moon. It's not that we have technologically regressed since the successful missions in the 1960s; but like with automobiles and bicycles we haven't made much progress either in the core technology in the past 80 years or ...


8

Return trips are harder The main "problem" with crewed trips is that we generally want those people to return back. This means that we don't just need to accelerate the manned part to the required velocity, but we also need to accelerate a sufficient amount of fuel and engines for the return trip, which is a significant increase - as the other ...


33

It's not hard, it's just expensive. We know exactly how to do it. Compare this to building computer processors with 1nm transistors, or making reliable self-driving cars. Those are both things that we currently don't know how to do, and we don't even know exactly how to get better at doing them. Even going past low Earth orbit to another planet, like Mars, ...


45

Delta-V to LEO is about 10 km/s. From there to C3 (Earth escape) is another 3.2 km/s. It's just another 30% delta-V. The problem is the Tyranny of the Rocket Equation. More delta-V means more fuel. More fuel means more mass. More mass means more fuel. How much more? Fuel costs scale according to $e^{\frac{\Delta V}{v_e}}$, that is e to the power of the ratio ...


5

I have also been unable to implement PEG and struggled to find an adequate approach in my orbital launch sim. I have tried simply steering proportional to fraction-of-orbital-altitude-reached in the orbital insertion phase. I have tried various forms of quadratic guidance (inspired by the the Apollo LM's landing guidance). None have gotten good results. At ...


6

The equation you're looking for is: $$T = 2 * \sqrt {D/A} $$ Where T is time in seconds, A is acceleration in m/s^2 (~9.81 for 1 g), D is distance in meters. Note that this is dead-stop to dead-stop, whereas real interplanetary travel involves initial and final velocities which are usually very different from one another, and it also doesn't account for the ...


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