<p>Assuming you have a mass of 70 kg, what would your weight be on the equator of PSR J1748−2446ad?</p>

CC BY-SA 4.0
2022-07-16T21:02:35.153Z
Have you attempted to work this out for yourself? Have you done any calculations of weight on rotating bodies (eg the Earth)?
James K
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2022-07-16T21:19:16.350Z
I have not. It is beyond my very limited knowledge.
A.J Perez
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2022-07-17T02:11:28.917Z
One does not simply *stand* on the surface of a neutron star! Especially not a millisecond pulsar.
PM 2Ring
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2022-07-17T02:22:14.650Z
Using the numbers from https://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.
PM 2Ring
CC BY-SA 4.0
2022-07-17T02:46:41.960Z
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](https://physics.stackexchange.com/a/631427/123208) 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.
PM 2Ring
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2022-07-17T06:45:07.970Z
@PM2Ring perhaps the OP is [asking for a friend](https://aliens.fandom.com/wiki/Cheela) :-)
uhoh
CC BY-SA 4.0
2022-07-19T02:25:39.150Z
My first guess would have been zero or negative, since I would have assumed that the fastest millisecond pulsar would be close to break up speed.  But it isn't.  So I searched for why not and found that the rapid rotation creates gravitational waves and that acts to limit the rotation speed to well below break up speed.
eshaya
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2023-04-12T14:48:38.953Z
If it spun too fast, it would be a disk, not a neutron star and effectively come apart.   To be compact enough to be a neutron star, which is one of the criteria of a pulsar, the gravity would have to be enormous.  Far and above white dwarf gravity.   a 70kg person would weigh as much (on Earth) as. . . the Chrystler building?   A mountain?    It'd be something ridiculously high, but a precise answer would require a good deal of math and calculation of ultra-high gravity going against the repulsive pressure of that much compacting and outward centrifugal force and a dash of relativity.
userLTK
<gravity><astrophysics><rotation><neutron-star><pulsar>
How much would you weigh at the equator of the fastest spinning pulsar?
A.J Perez
# How much would you weigh at the equator of the fastest spinning pulsar?

(Asked by: username_0 on 2022-07-16)

Assuming you have a mass of 70 kg, what would your weight be on the equator of PSR J1748−2446ad?

**Tags:** gravity, astrophysics, rotation, neutron-star, pulsar

> **Comments:**
> **username_1**: Have you attempted to work this out for yourself? Have you done any calculations of weight on rotating bodies (eg the Earth)?  
> **username_0**: I have not. It is beyond my very limited knowledge.  
> **username_2**: One does not simply *stand* on the surface of a neutron star! Especially not a millisecond pulsar.  
> **username_2**: Using the numbers from https://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.  
> **username_2**: 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](https://physics.stackexchange.com/a/631427/123208) 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.  
> **username_3**: @PM2Ring perhaps the OP is [asking for a friend](https://aliens.fandom.com/wiki/Cheela) :-)  
> **username_4**: My first guess would have been zero or negative, since I would have assumed that the fastest millisecond pulsar would be close to break up speed.  But it isn't.  So I searched for why not and found that the rapid rotation creates gravitational waves and that acts to limit the rotation speed to well below break up speed.  
> **username_5**: If it spun too fast, it would be a disk, not a neutron star and effectively come apart.   To be compact enough to be a neutron star, which is one of the criteria of a pulsar, the gravity would have to be enormous.  Far and above white dwarf gravity.   a 70kg person would weigh as much (on Earth) as. . . the Chrystler building?   A mountain?    It'd be something ridiculously high, but a precise answer would require a good deal of math and calculation of ultra-high gravity going against the repulsive pressure of that much compacting and outward centrifugal force and a dash of relativity.  

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