Question

Consider a spherical galaxy of total mass \( M \) and radius \( R \), having a uniform matter distribution. In this idealized situation, the orbital speed \( v \) of a star of mass \( m \ll M \) as a function of the distance \( r \) from the galactic centre is best described by \[ (G \text{ is the universal gravitational constant}) \] 

• A. A
• B. B
• C. C
• D. D

Hint:

The orbital velocity \( v \) of a star in a uniform mass distribution follows \( v \propto \sqrt{\frac{GM}{r}} \). The further the star is from the center, the slower its velocity.

Answer 1

The Correct Option is A

In the case of a spherical galaxy with a uniform matter distribution, we can apply the concepts of gravitational force and orbital mechanics. The force on a star of mass \( m \) at a distance \( r \) from the galactic center is due to the gravitational attraction from the mass inside that radius. 1. Gravitational Force: According to Newton's law of gravitation, the gravitational force acting on the star is: \[ F = \frac{G M_{\text{enclosed}} m}{r^2} \] where \( M_{\text{enclosed}} \) is the mass inside the radius \( r \), which for a uniform distribution is proportional to \( r^3 \). Since \( m \ll M \), the effective mass inside radius \( r \) is given by: \[ M_{\text{enclosed}} = M \cdot \left( \frac{r^3}{R^3} \right) \] where \( M \) is the total mass of the galaxy. 2. Centripetal Force: The centripetal force required for circular motion is: \[ F_{\text{centripetal}} = \frac{m v^2}{r} \] 3. Equating the Forces: For stable orbital motion, these two forces must be equal: \[ \frac{m v^2}{r} = \frac{G M_{\text{enclosed}} m}{r^2} \] Simplifying the equation, we get: \[ v^2 = \frac{GM}{r} \] Thus, the orbital speed \( v \) is proportional to \( \sqrt{\frac{GM}{r}} \). Since the question asks for the relationship at a specific radius \( R \), we substitute \( r = R \) to obtain: \[ v \propto \sqrt{\frac{GM}{R}} \] This matches the expression in option (A). Therefore, the correct answer is option (A).