Question:

A very long (length L) cylindrical galaxy is made of uniformly distributed mass and has radius $R (R << L)$. A star outside the galaxy is orbiting the galaxy in a plane perpendicular to the galaxy and passing through its centre. If the time period of star is $T$ and its distance from the galaxy's axis is $r$, then :

Updated On: Oct 10, 2024
  • $T^{2} \,\propto\,r^{3}$
  • $T \,\propto\,r^{2}$
  • $T \,\propto\,r$
  • $T \,\propto\,\sqrt{r}$
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The Correct Option is C

Solution and Explanation



Let the linear mass density of the cylindrical galaxy be $\lambda\, kg / m$.
Gravitational field $=\frac{2 G \lambda}{r}=E_{r}$
Therefore, gravitational force $F=m E_{g}=m \omega^{2} r$
Hence, $m\left(\frac{2 G \lambda}{r}\right)=m \cdot\left(\frac{2 \pi}{T}\right)^{2} r$
$\Rightarrow T^{2} \propto r^{2} \Rightarrow T \propto r$
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Concepts Used:

Gravitation

In mechanics, the universal force of attraction acting between all matter is known as Gravity, also called gravitation, . It is the weakest known force in nature.

Newton’s Law of Gravitation

According to Newton’s law of gravitation, “Every particle in the universe attracts every other particle with a force whose magnitude is,

  • F ∝ (M1M2) . . . . (1)
  • (F ∝ 1/r2) . . . . (2)

On combining equations (1) and (2) we get,

F ∝ M1M2/r2

F = G × [M1M2]/r2 . . . . (7)

Or, f(r) = GM1M2/r2

The dimension formula of G is [M-1L3T-2].