0
$\begingroup$

In Larry Niven's Ringworld, where there is a rigid rotating ring with a radius of 1AU with the Sun at the centre, he introduces a fictional material "scrith", which apparently is strong enough to resist the mechanical stresses around the order of 10^16 pascals induced by the high rate of rotation to induce a centrifugal acceleration of 9.81 to maintain Earth-like artificial gravity. Realistically, no material currently exists that can withstand that amount of stress and the ring would break up.

After extensive research the only things I could find that was barely relevant was the calculation of the Roche Limit, however my case is breakup due to self-rotation instead of tidal forces between the primary body and itself. How would the ring break up into? What are the parameters of this new "debris" ring after each fragment of the original ring has occupied new orbits around the Sun? As in, how does one calculate the outer and inner bound of the rings, what would be its thickness and density?

$\endgroup$
1
$\begingroup$

The ring velocity needed to maintain the standard gravity is $\sqrt{g R}\approx 1200 \,\text{km}/\text{s}$ (actually slightly larger since the Sun's gravity has to be compensated). That is more than 25 times the velocity needed to leave the Solar system. So, there would be no 'debris ring', all the fragments leave the Solar system. Within 50 days they would be beyond Neptune orbit, and continue flying away indefinitely.

$\endgroup$
  • $\begingroup$ How about the fragments which were flung towards the Sun after the ring broke up? $\endgroup$ – Dt roid Mar 4 '18 at 10:48
  • $\begingroup$ @Dtroid: For a fragment to end up in any sort of bound orbit around the sun it has to receive a 'kickback' that would almost precisely (up to a several percent) negate the momentum this fragment had when the ring was rotating. I would imagine that during breakup various fragments could acquire/lose speed on the order of speed of sound in the ring material but not in excess of 1000 km/s. $\endgroup$ – A.V.S. Mar 4 '18 at 15:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.