precisely to scale, of course…
I may have written about this before, and I know I’ve talked to people about it, but the topic is fascinating and counter-intuitive to most people, including myself, so I’ll try to ‘rise above myself and grasp the world’ yet again, and hopefully I won’t mess it up as I did with LaGrange points recently. Must have another go at that, by the way.
A recent video by SpaceMog (Dr Maggie Lieu), linked below, has inspired me to make this effort, and I’ll just be repeating closely her account, at least for the first part of this post. She’s an astrophysicist after all, and I’m just as autodidactic dilettante. Ma vie est un désert, malheureusement.
It takes between 6 and 9 months to get to Mars, and we have to time it carefully, for there are certain ‘windows of opportunity’ – best, or perhaps only, times for setting out. Every two years, apparently.
We can’t just go directly to Mars, the fourth planet from the Sun (at their closest, Earth and Mars are some 56 million kilometres away). More on that later perhaps. Instead we go by a precisely calculated fuel-efficient route for transferring between two orbits, known as the Hohmann transfer orbit. Fuel efficiency is absolutely key, as fuel for space travel is astronomically expensive, some thousands of dollars per kilo according to Dr Maggie. She worked on the Exomars 2016 mission, particularly the Trace Gas Orbiter, which had a total launch mass of 4300 kilos, of which only 700 kgs were science instruments.
So, some calculations. Picture our solar orbit as circular. The radius to the Sun is 1 AU (astronomical unit), while that of Mars is 1.5 AU. The elliptical Hohmann transfer orbit connects Earth’s orbit to that of Mars. So if we picture Earth at 0° on the circle, the trick is to get our spacecraft to Mars at 180° on its orbit – that’s to say, on the other side of the Sun. This kind of orbit, between any one planet and another, has been calculated as the most fuel-efficient route. So the spacecraft is first launched into a low-Earth orbit, building up enough speed to put it into this transfer orbit, at a precisely calculated trajectory, which enables it to get to Mars, where it will slow down, via its engines, for the Mars orbit.
So far so good, but we’ve only just begun. Dr Maggie takes us through the calculations for the journey. First we calculate the semi-major axis (a) for the transfer orbit. To be clear (for me) the semi-major axis for a precisely circular orbit would be the radius. In a non-circular case it is the average distance from a point on the ellipse to the centre. In this context it’s half the average distance between Earth and Mars (on the other side of the Sun), that’s
a = 1/2(REarth + RMars), which is 1/2(1AU + 1.5AU), or 1/2(2.5AU), which is 1.25AU.
Now to Kepler’s Third Law, which describes the relationship between the orbital periods of planets and their average distance from the Sun. That is, that the square of a planet’s orbital period is directly proportional to the cube of its semi-major axis. The formula is T² = ka³, T being the orbital period of the planet, in years, a being the semi-major axis of the orbit, and k is a constant of proportionality, which is the same for all the planets of the solar system.
So, calculating k from Earth’s orbit, the period of which is of course one year, and its distance from the Sun is 1AU. To isolate k, in T² = ka³, we divide both sides by a³, giving
k = T²/a³
So T, the orbital period, is 1, and a, the semi-major axis, is 1. So k is equal to 1. Pretty simple, no worries about the squaring and cubing. Now, the semi-major axis for the transfer orbit has already been calculated as 1.25AU. So we get T (being the transfer orbit) is equal to
√1 x 1.25³
which comes out as approximately 1.4 years or around 511 days. That’s the whole orbit, but getting to Mars requires only half an orbit. That’s around 8.5 months (256 days).
So I think I get this, thanks entirely to Dr Maggie. The next issue she takes us through is the launch timing. The orbital period of Mars (TMars) is 687 days. Dividing 360° by 687 days gives us a movement of 0.52° per day. After half the orbit, 256 days, Mars will have moved
256 x .52° = 133°
So, this key. In order for Mars to be 180° from Earth when the spacecraft enters the Mars orbit, the craft must be launched at 180°minus the 133° of Mars’ orbital movement. That leaves 47°. In other words, Mars must be 47° ahead of Earth’s orbit at launch-time to ensure that it will be in the ‘right place’ when the craft arrives.
Now, apparently, that’s the mathematically easy part of this exercise (and to be honest it’s been pretty straightforward so far). So what happens if we miss the ‘launch window’ as described above? Earth is moving, in terms of its orbit, at 360° every 365.2 days or so, which makes about .99° per day. The location of Earth at any given time follows this equation
E = t.vEarth – (360°.z)
where E is the position of Earth (in degrees), t is the time (in days), vEarth is the Earth’s orbital speed, measured in degrees per day, and z is an ‘arbitrary integer’, ‘because we know that, at 360°, you’re already going back to the start, so it’s the same as zero degrees, so by subtracting 360 you can get numbers only between 360 and zero and nothing above that…’ I quote directly from SpaceMog because I’m not sure I understand it, though I know it all goes round in ellipses. Anyway, it’s all about distance being velocity multiplied by time.
So we do the same calculation for Mars
M = t.vMars – (360°.y) + M0
where things are as above for Mars and time, with y as the arbitrary integer, and M0 as the initial position of Mars, which isn’t zero, but 47° ahead of Earth. So subtracting the two equations from each other, we get this:
(M – E) = t(vMars – vEarth) + 360°(z – y) + M0
which is plugged in as
(47) = t(0.52 – 0.99) + 360° (x) + 47
and rearranging the time factor
t = -360°x/(0.52 – 0.99)
t = 766x …… t being thus about every two years. Orbits, as mentioned, are in fact slightly elliptical, so the calculations would need to be adjusted for that.
So I understand some of this, but I still don’t understand some of the most basic stuff, e.g. about gravity. Like, how does the Sun’s gravity keep the planets in orbit (if in fact it does)? Why doesn’t the sun, with its hefty gravitational force (yes, I know it’s not a force but a curvature of space-time – which is easy enough to write…), draw the planets closer and closer in, so that they burn up? Same with black holes and galactic stars. Which leads to the question, how did the planets get to form in the first place? Something explosive happened, that defied gravity?? And then the planets formed by a kind of gravitational accretion? I know that the Moon is spiralling away from Earth, very very slowly. It’s being measured precisely, due to a kind of reflective mirror planted on the Moon during the Apollo days. But wouldn’t it make more sense if it was spiralling towards Earth, due to gravity? Learning equations is one thing – here’s one:
F = G.m1m2/r²
– not clearly put, but I’m told that G here is the gravitational constant, which is a constant of proportionality, and which is directly proportional to the product of mass1 and mass2 and inversely proportional to distance between their centres of mass. But I could be getting it wrong. The point is, I might be able to read equations, eventually, but will I ever get to really understand? It’s not the equations that are the problem, it’s the why of it all…
I’ll keep trying though, so more on gravity in the future.
References