Flat Velocity Curve: Newton vs MOND Theory

  • Context: Graduate 
  • Thread starter Thread starter zachzach
  • Start date Start date
  • Tags Tags
    mond
Click For Summary
SUMMARY

The discussion centers on the comparison between Newtonian mechanics and Modified Newtonian Dynamics (MOND) in explaining the flat velocity curve observed in galaxies. It is established that Newton's law, represented as F = GM(r)m/(r^2), predicts a constant velocity v = [2G*pi*p]^(1/2) under certain conditions. However, participants highlight that this simplified model only applies to spherically symmetrical mass distributions and that the mass should be proportional to the square of the radius for a uniform disk galaxy. The necessity of MOND theory arises from the limitations of Newtonian mechanics in accurately describing galactic rotation curves.

PREREQUISITES
  • Understanding of Newton's law of gravitation
  • Familiarity with concepts of centripetal force and velocity in circular motion
  • Knowledge of mass distribution in galaxies, particularly uniform density models
  • Basic grasp of gravitational potential and its derivatives
NEXT STEPS
  • Explore the implications of MOND theory on galactic dynamics
  • Study the derivation of rotation curves for non-spherical mass distributions
  • Learn about Green's functions in gravitational potential calculations
  • Investigate the differences between Newtonian and MOND predictions in various galactic contexts
USEFUL FOR

Astronomers, astrophysicists, and students of gravitational physics seeking to understand the dynamics of galaxies and the limitations of classical mechanics in explaining observed phenomena.

zachzach
Messages
257
Reaction score
1
Consider a star of mass m at a distance r form the center of a circular disk galaxy. Newton's law: F = GM(r)m/(r^2) where M(r) is the amount of mass inside the radius r. If we consider a uniform galaxy then density (p) is p = M/L where L is the length = 2*pi*r. So M(r) = p*2*pi*r. Setting the force of gravity equal to centripetal force (mv^2/r) you get G*2*pi*p = v^2 or
v = [2G*pi*p]^(1/2) which is a constant. Why do you need MOND theory. To me it seems Newtonian mechanics predicts a flat velocity curve.
 
Astronomy news on Phys.org
zachzach said:
Consider a star of mass m at a distance r form the center of a circular disk galaxy. Newton's law: F = GM(r)m/(r^2) where M(r) is the amount of mass inside the radius r. If we consider a uniform galaxy then density (p) is p = M/L where L is the length = 2*pi*r. So M(r) = p*2*pi*r. Setting the force of gravity equal to centripetal force (mv^2/r) you get G*2*pi*p = v^2 or
v = [2G*pi*p]^(1/2) which is a constant. Why do you need MOND theory. To me it seems Newtonian mechanics predicts a flat velocity curve.

Firstly, your simplified form of Newton's law only applies in certain cases such as when the mass is spherically symmetrical, or like a segment of a sphere along a diameter towards the relevant direction.

Secondly, I don't get your maths for the mass. If the galaxy is of uniform density per area of the disk, the mass inside a given radius would be proportional to the square of the radius. For the mass to be proportional to the radius, the area density would have to vary as 1/r.
 
zachzach said:
Consider a star of mass m at a distance r form the center of a circular disk galaxy. Newton's law: F = GM(r)m/(r^2) where M(r) is the amount of mass inside the radius r. If we consider a uniform galaxy then density (p) is p = M/L where L is the length = 2*pi*r. So M(r) = p*2*pi*r. Setting the force of gravity equal to centripetal force (mv^2/r) you get G*2*pi*p = v^2 or
v = [2G*pi*p]^(1/2) which is a constant. Why do you need MOND theory. To me it seems Newtonian mechanics predicts a flat velocity curve.

As Jonathan said, it's wrong.
Your galaxy isn't spherical, you can't use your (wrong) formulas. Fix them and find the rotation curve of the bulge.
If you want to find the disk rotation curve you should write your potential considering a cylindrical distribution (hint: Green's functions), then
<br /> \frac{v_{c}^{2}}{R} = \frac{\partial \phi (R,z=0)}{\partial R}<br />
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 31 ·
2
Replies
31
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 11 ·
Replies
11
Views
5K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 14 ·
Replies
14
Views
5K