A star moves in an orbit under the influence of massive but invisible ...
The Two Possible Circular Orbits in the Given System
To understand the two possible circular orbits of the star in the given system, let's analyze the effective one-dimensional potential, V(r), and its implications.
1. Effective One-Dimensional Potential
The effective one-dimensional potential, V(r), is given by the equation:
V(r) = - 1/r - (L^2)/(2r^2) - (L^2)/(r^3)
where r is the distance of the star from the center of the massive but invisible object, and L is the angular momentum of the star.
2. Circular Orbit Conditions
In a circular orbit, the net force acting on the star is directed towards the center of the orbit, providing the necessary centripetal force for circular motion. Mathematically, this condition can be expressed as:
F_net = - dV/dr = F_centripetal
Therefore, to find the circular orbits, we need to solve the equation - dV/dr = F_centripetal, where F_centripetal is given by mω^2r, with ω being the angular velocity and m the mass of the star.
3. Solution for Circular Orbit Radii
Solving the equation - dV/dr = F_centripetal, we obtain:
dV/dr = 1/r^2 - (L^2)/r^3 - (3L^2)/(2r^4) = mω^2r
Multiplying through by r^4, we get:
r^2 - L^2/r - (3L^2)/2 = mω^2r^5
This equation can be rearranged as:
r^5 - (mω^2 - 1)r^3 + (3L^2)/2r - L^2 = 0
4. Two Possible Circular Orbit Solutions
To find the two possible circular orbit radii, we solve the equation r^5 - (mω^2 - 1)r^3 + (3L^2)/2r - L^2 = 0.
The equation is a quintic equation in terms of r, which is generally difficult to solve analytically. However, we can analyze the equation qualitatively to understand the behavior of the system.
Case 1: Stable Circular Orbit
- In this case, the star is in a stable circular orbit.
- The radius of the stable circular orbit, r_1, corresponds to the minimum of the effective potential, V(r).
- To find r_1, we can set dV/dr = 0 and solve for r.
- By differentiating V(r) with respect to r and setting it equal to zero, we obtain:
2/r^3 - (L^2)/r^4 + (3L^2)/(2r^5) = 0
Multiplying through by r^5 and rearranging, we get:
2r^2 - (L^2)/r + (3L^2)/2 = 0
This is a quadratic equation in r, and solving it gives us the radius of the stable circular orbit, r_1.
Case 2: Unstable Circular Orbit
- In this case, the star is