A little pedagogy today, we will approach astronautics...
Calculations for a Martian launch:
A little physics, let us remember these data:
Postulate:
Let's take 2 digits after the comma
Terre = Earth = blue
Mars = red
A real case: Mars Pathfinder
Therefore, with the hypothesis of a circular orbit and in the same plane (Less precise approximation for Mars than for Earth).
Departure on December 4, 1996
Picture from Celestia
Arrival 7 months after (212 days) on July 4, 1997
Instant of launching :
But where should Mars be on launch day ?
Mars achieves a full 360° orbit in 1.88 years (or 687.67 days).
Formula: Circle × ((DS / DE) / PM) where:
Circle (in Deg) = 360
DS = Travel time in days
DE = One terrestrial year in days = 365.25 days
PM = Relative Martian period in terrestrial year = 1.88
this gives :
=> 360 x ((212 / 365.25) / 1.88) = 111°14 for Pahtfinder
=> 360 x ((180 / 365.25) / 1.88) = 94°40 for Mars II
We can calculate the relative position of the Earth and Mars.
180° - 111°14 = 68°85 for Pathfinder (validated by first picture)
180° - 94°40 = 85°60 for Mars II
Thus, launches are only possible when the Earth is located at 85°60
Behind the planet Mars (for Mars II in 2039).
For Mars II
Departure around june 10th 2039 (+/- 1 day)
The picture from Celestia gives the positions of Earth and Mars at the time of departure.
Arrival around Dec 07th 2039 (+/- 1 day) near 2040...
The picture gives the positions of the Earth and Mars at the moment of arrival.
The calculations will be the same for Mars III.
When do you think that Mars III will arrive on Mars ?
I wait your answers (with average of + or - 2 days)...
The winner will become one of passengers of the next mission.
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Evil progresses when good people do nothing!
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SH3D 7.1 and nothing else - W11 64b in 4K
