Assuming you have a mass of 70 kg, what would your weight be on the equator of PSR J1748−2446ad?
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2$\begingroup$ Have you attempted to work this out for yourself? Have you done any calculations of weight on rotating bodies (eg the Earth)? $\endgroup$– James KCommented Jul 16, 2022 at 21:02
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1$\begingroup$ I have not. It is beyond my very limited knowledge. $\endgroup$– A.J PerezCommented Jul 16, 2022 at 21:19
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2$\begingroup$ One does not simply stand on the surface of a neutron star! Especially not a millisecond pulsar. $\endgroup$– PM 2RingCommented Jul 17, 2022 at 2:11
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1$\begingroup$ Using the numbers from en.wikipedia.org/wiki/PSR_J1748%E2%88%922446ad of 16 km radius, 2 solar masses, frequency 716.36 Hz, the gravitational acceleration is ~3.2 times the centrifugal, and the difference is ~ 72.7 billion g. That's using a naive Newtonian calculation, so it's probably quite wrong. $\endgroup$– PM 2RingCommented Jul 17, 2022 at 2:22
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1$\begingroup$ Pulsars are fascinating, but you really don't want to visit them. You can't even get close without getting seriously messed up, even if you somehow avoid the radiation beams. The tidal force will snap a freefalling 1 m steel bar at 100 km. The magnetic field alone will kill you long before you reach the surface. The surface is extremely hot, and of course the gravity will squash you to atomic / subatomic flatness. $\endgroup$– PM 2RingCommented Jul 17, 2022 at 2:46
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