And a related question:Let's say that you need to launch a spacecraft to Mars on a given day, and you have no tolerance for a delay (for whatever reason). So, you cannot afford to hope that it doesn't rain at the launch site.Is it possible to launch, say, 100 days before the actual planned launch date, fly away from the Earth for 50 days, return, and then use a swingby to go to Mars on the desired date? Yes, I know this means exposing the spacecraft to space for an extra 100 days, but assuming that the 50 day out and 50 days back trajectory is feasible, what is the downside to this approach? Why hasn't it been tried as a way to ensure that a payload makes it to Mars during any particular launch window?Oh, in the event that the launch at T - 100 days is itself delayed, I assume that we have the computers today that can quickly generate a 98 day trajectory that accomplishes the same purpose and would allow a launch at T - 98 days.Also, any chances of obtaining a velocity boost during the Earth swingby?

Apologies if this isn't in the correct thread, but it seemed the best fit and I didn't want to start a new one for what I'm sure will end up being a stupid question.When a vehicle (or for that matter anything) enters the earths atmoshere and there's "loads" of heat energy generated by friction, does the Earth's rotation slow by a very, very small amount or just the rotation of the atmoshere (or are they the same thing or linked)?If so, over a very, very long time (if we continue to re-enter bigger and more numerous things (e.g. ISS)) might we eventually cause something "bad" to happen?It should be clear to anyone reading this that I know naught about physics, so I'd be grateful if anyone who takes the time to answer would be kind...Thanks.

Don't worry about it. Angular momentum is conserved, so the change in rotation is basically the same (but of opposite sign) as the amount the rotation was changed when it was launched.

Besides, the Moon changes the Earth's rotation by A LOT more than some puny station....

Quote from: Robotbeat on 06/06/2011 11:57 pmDon't worry about it. Angular momentum is conserved, so the change in rotation is basically the same (but of opposite sign) as the amount the rotation was changed when it was launched.There is a very slight amount of mass that reaches Earth escape velocity, though. Spacecraft which de-orbit fire thrusters in the direction of their motion briefly, which can give the thruster exhaust Earth escape velocity. Since all of these spacecraft have been in prograde orbits, the overall effect is to extremely slightly reduce Earth's rotation...QuoteBesides, the Moon changes the Earth's rotation by A LOT more than some puny station.......but of course, this effect is utterly insignificant compared to the Moon and other factors.

When a vehicle (or for that matter anything) enters the earths atmoshere and there's "loads" of heat energy generated by friction, does the Earth's rotation slow by a very, very small amount or just the rotation of the atmoshere (or are they the same thing or linked)?

(you edited the quotes wrong... I didn't say that, I was quoting someone else. )

Using L1/L2 as a staging point would be the natural solution. That's a good idea for a long list of other reasons too. So much so that not using L1/L2 would be a bad idea.

Recently, I contemplated a new option--staging a tanker at in a solar co-orbit perhaps 2+ million kilometers from Earth. This co-orbit is a quasi-orbit around a point roughly 2 million kilometers ahead of or behind Earth, such that it's at an angle around 30 degrees off. (This is roughly equal to the hyperbolic escape half angle.)

Quote from: IsaacKuo on 07/31/2011 09:52 pmRecently, I contemplated a new option--staging a tanker at in a solar co-orbit perhaps 2+ million kilometers from Earth. This co-orbit is a quasi-orbit around a point roughly 2 million kilometers ahead of or behind Earth, such that it's at an angle around 30 degrees off. (This is roughly equal to the hyperbolic escape half angle.)Do I understand correctly these co-orbits are roughly half way to the SEL4 and SEL5 points (which are roughly 60 degrees off)?

Let's say that time is of not great importance. How much delta-v would be required to got to a different place withing Earth's orbit? For example a troyan? Significantly more than C3=0? Because, intuitively, once escaping along the correct vector, you might be pretty much along the orbit, right?

When one describes the characteristics of a sun-synchronous orbit (SSO), it is convenient to describe the mean local time at the ascending/descending node of the satellite. What is the time system used for the MLT?