The Cosmos

How Long Would It Take Sir Bradley Wiggins To Cycle To The Planets Orbiting Trappist-1?

If Wiggo chooses to accept this epic space mission, he could be gone for a very long time indeed.

An artist impression of Sir Bradley Wiggins cycling in space (Image by Jack Clayton).

So, you might be aware that NASA have just announced the very exciting discovery of seven Earth-like planets orbiting nearby star Trappist-1. The potentially game-changing news has massively increased the chances of us discovering alien life in the next decade, and has put the human race one step closer to answering that age old question: are we alone in the universe?

The new planets are roughly 40 light years from Earth (meaning it would take someone travelling the speed of light 40 Earth years to reach them). Now, here on Earth, we don’t have anyone or anything capable of moving at that speed. We do, however, have cycling legend and sideburn champion Sir Bradley Wiggins who, while not capable of light-speed, can move pretty quickly when he wants to.

This morning we got our thinking hats on and decided to work out exactly how long it would take Wiggo, travelling at an average speed of 61km/h to reach Trappist-1 and the planets that surround it. This maths equation was calculated on the basis that a specially-built cycle path would go in a straight line from Earth to the Trappist-1 solar system, and that the path could be engineered in such a way that it would have the same gravity you’d find here on our own planet. Here’s how our maths mission played out.

Firstly, and we feel it’s important to tell you this incase you get the wrong idea, nobody at the Mpora office would label themselves a mathematician. We can do our basic times-tables, and can count up to 100 without hesitation, but yes we’re not really familiar with maths at the type of level where billions and trillions are the main ingredients.

After discovering that one light year is equivalent to nine trillion kilometres, it was relatively simple for us to conclude that 40 light years equalled 360 trillion kilometres (9 x 40 – easy). Then things got super complicated. We had to look at how many hours are in a year, and how far Wiggo could cycle in that time at an average speed of 61km/h. Then we did some stuff, and some things, that all seem like a confused blur now and ended up with this answer.

673,703,121.491 Years = Time Needed For Wiggo To Reach Trappist-1

Rounding to the nearest year then, the time we calculated it would take for Wiggins to cycle from Earth to Trappist-1 is 673,703,121 years (well over 673 million years, but less than 674 million years). We looked at this number with our heads vertical, we looked at this number with our heads horizontal, and we looked at the number with our heads at jaunty angles and…yes…it really did look like a plausible answer to the initial question.

Screenshot of the Mpora Skype chat (provided by Lou Boyd), taken just after we’d come up with an answer.

To confirm that we were definitely in the right ballpark with our calculation, we decided to approach a proper mathematician. I dropped my girlfriend’s brother, who studied maths at a university-level, a message on WhatsApp and he agreed to look into it. He got back to us with the following figure.

708,136,837 Years = Time Needed For Wiggo To Reach Trappist-1

Although there was a 35 million year difference between our answer and his answer, we were generally quite happy with how close our answer had been (relatively speaking, when you consider the absolutely enormous numbers involve) to someone who actually knows about maths. Of course, neither of the calculations take into account things like Wiggo’s toilet breaks and need to sleep, eat, drink etc.

How long would it take for Sir Bradley Wiggins to cycle 360 trillion kilometres to the Trappist-1 solar system then? A really, really, really, really, really, really, really, really, really, really long time. Better get pedalling now, Mr Wiggo. You’ve got an extremely long journey ahead of you.

Screenshot of my WhatsApp conversation with an actual mathematician.

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