A Plan To Achieve Light Speed – Relatively Easily
Consider that the Earth is about 7,926 miles in diameter. The velocity of any given point on the equator is about 1,000 miles per hour. The circumference of 24,901 revolves once every 24 hours so points along the equator make the round trip at a clip of 1,037 miles per hour.
Now imagine if the diameter increased. At a diameter of 20,000 miles the point along the equator zips along at 2,618 mph. At a diameter of 100,000 the equator’s rate is 13,090 miles per hour or 3.6 miles per second!
You should now be able to see where all this is heading and make the calculations to arrive at a diameter of 5,120,000,000 to achieve an equatorial spin rate of 186,163 miles per second – which is pretty much the speed of light.
Of course no planet could survive the stress of spinning at such speeds. No planet could support that mass without collapsing into… well, into pieces or whatever – luckily that’s not the plan. My plan calls for 2.56 billion miles of rigid wire secured into an airless sphere that is rotating somewhere in deep space where it won’t smack into anything as it swings the tip of the wire around and around.
I’d only need 2.56 billion miles instead of the full 5.12 billion because if the structure is properly counterbalanced by a corresponding mass then I only need to extend the wire out the length of the radius which is in this case 2.56 billion miles.
I am aware of the potential for comedy this plan presents and that is definitely part of its appeal. Sometimes I substitute tinker toy parts in lieu of wire or a near infinite number of drinking straws connected with duct tape. But that’s just me. The bottom line is that it would work.
I do see some possible trouble spots. It would important to have the spherical base object already spinning before extending the wire or tinker toy construct out 2.56 billion miles.
One of the advantages of using wire or some other sturdy material is that it would allow you to launch objects at wicked velocities. Wicked may not be the correct technical term but then consider what would happen as an object was sent outward along the wire! By the time it reached the tip it too would be moving at light speed. Wicked.
No doubt there are (real) physicists among us who see the precise point where this scheme violates fundamental natural laws. I urge you to e-mail with these logical objections so I can develop a work around. I promise all civil submissions will be posted here.
Amassing the mass
It always pays to try and figure out in advance where the trouble spots lie. This is especially true in a project of this magnitude. Is there enough mass handy to create the structure as described above? Let’s say that the “wire” has a mass of 100 grams per linear foot. That’s very conservative figure if something like a carbon-fiber composite extrusion is used. How much do we really need?
Here’s how I figure it:
2,560,000,000 mile radius
1.35168 x 1013 number of feet in radius
1.35168 x 1015 Total grams required for radius
2.70336 x 1015 Total grams required to balance system
2.70336 x 1012 Total kilograms
The mass of the Moon is 7.36 × 1022 kilograms. Now I’m not saying I’d use the Moon but just for discussion’s sake it’s a useful size. While we only need 1.35168 x 1012 kilograms required for the wire we need to counterbalance that with an equal mass. Then there’s the central, spinning sphere and that’s got to hold everything together so you definitely don’t want to scrimp any there. Let’s say that’s good for 1010 kilos. There’s always waste and scrap. So the 7.36 × 1022 kilograms that the Moon represents is probably just about right.