Puzzling problem (using virial theorem)

Click For Summary
SUMMARY

The discussion focuses on calculating the gravitational energy of a star using the Virial theorem, specifically for a star with radius R and mass M. The pressure gradient is defined by the equation dP/dr = (4π/3)Gρ2r exp(-r²/λ²), where ρ represents the central density. By integrating the gravitational potential Φ and applying the Virial theorem, the gravitational energy is derived as E = -2πGR⁵ρ/15λ². In the limit where λ is much smaller than R, the energy simplifies to E = RGMM/(3Rλ).

PREREQUISITES
  • Understanding of the Virial theorem in astrophysics
  • Familiarity with gravitational potential and pressure gradients
  • Knowledge of integration techniques in calculus
  • Concept of central density in stellar physics
NEXT STEPS
  • Study the application of the Virial theorem in different astrophysical contexts
  • Explore the implications of pressure gradients in stellar structure
  • Learn about gravitational potential calculations in spherical coordinates
  • Investigate the relationship between mass, radius, and central density in stars
USEFUL FOR

Astronomers, astrophysicists, and students studying stellar dynamics and gravitational physics will benefit from this discussion.

Findg
Messages
1
Reaction score
0
Consider a star of Radius R and mass M, with a pressure gradient given by

\frac{dP}{dr} = \frac{4\pi}{3}G\rho2r exp(-\frac{rr}{\lambda\lambda})

where \rho is the central density. calculate the gravitational energy, using the Virial theorem. Show that in the limit \lambda « R this energy is given by

E = \frac{RGMM}{3R\lambda}for tecnical reasons:
MM = M2
rr = r2
\lambda\lambda = \lambda2
 
Last edited:
Physics news on Phys.org
Using the Virial theorem, the total gravitational energy of a star is given by:E = -\frac{1}{2}\int \rho \Phi \ dVwhere \Phi is the gravitational potential. Since we are dealing with an isotropic pressure gradient, the gravitational potential is given by:\Phi = -4\pi G \int_0^R \frac{\rho r'^2 exp(-\frac{r'^2}{\lambda^2})}{r} dr'Integrating this expression and substituting in the expression for the gravitational energy, we obtain:E = - \frac{2\pi G R^5 \rho}{15\lambda^2}In the limit \lambda << R, we can approximate the integral as:E \approx - \frac{8\pi G R^5 \rho}{15\lambda^2}Finally, substituting the expression for the central density \rho = \frac{3M}{4\pi R^3}, we obtain:E \approx \frac{RGMM}{3R\lambda}
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 11 ·
Replies
11
Views
4K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
975
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 4 ·
Replies
4
Views
1K