Universal Gravitation and neutron stars

Click For Summary
SUMMARY

The discussion centers on calculating the minimum mass of a neutron star with a radius of 15 km and a rotation rate of 6 revolutions per second, ensuring that material on its surface remains in place. The gravitational constant used is G = 6.67 x 10-11 m3 kg-1 s-2. The user derived the centripetal acceleration formula and equated it to gravitational acceleration, leading to the mass calculation of approximately 7.29 x 1024 kg. A correction was noted regarding a missing "pi" in the calculations, which is critical for accuracy.

PREREQUISITES
  • Understanding of centripetal force and gravitational force equations
  • Familiarity with the gravitational constant (G)
  • Basic knowledge of neutron star properties
  • Ability to perform algebraic manipulations and unit conversions
NEXT STEPS
  • Study the relationship between centripetal acceleration and gravitational acceleration in astrophysical contexts
  • Explore the properties and behavior of neutron stars in more detail
  • Learn about the implications of mass and rotation on stellar stability
  • Investigate advanced gravitational physics concepts, such as general relativity
USEFUL FOR

Astronomy students, astrophysicists, and anyone interested in the dynamics of neutron stars and gravitational physics will benefit from this discussion.

1st2fall
Messages
21
Reaction score
0

Homework Statement


Certain neutron stars (extremely dense stars) are believed to be rotating at about 6 rev/s. If such a star has a radius of 15 km, what must be its minimum mass so that material on its surface remains in place during the rapid rotation?


G=6.67*10-11m3 kg-1 s-2

Homework Equations


F_c{}=\frac{mv^{2}}{r}
F_g{}=\frac{GM_1{M_2{}}}{r^{2}}

The Attempt at a Solution


30\pikm*6rev/s=180\pikm/s
which gives the linear velocity of something on the surface of the neutron star... but I'm clueless as to how to arrive at a mass of the star from it. I could the Centripetal acceleration but I'm not sure how that's related here. the only thing I can think of there is setting the centripetal acceleration equal to the gravitation acceleration which gives

Gm/r2=v2/r
m=v2r/G
which yields something like 7.28635682 × 10^24(kg?)

I'm just looking for advice on what I'm actually looking to do. I don't know what I should be looking for...
 
Physics news on Phys.org
Your formula is correct, but check the calculation.

ehild
 
Thank you very much! I realized a dropped a "pi" at some point when I went back over it.
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 0 ·
Replies
0
Views
3K
  • · Replies 23 ·
Replies
23
Views
5K
Replies
1
Views
3K
Replies
2
Views
3K