Question

The Sun revolves around the galaxy with speed of $250\dfrac{{km}}{s}$ around the center of the milky way and its radius is $3 \times {10^4}$ light year. The mass of milky way in kg is
(A) $6 \times {10^{41}}$
(B) $5 \times {10^{41}}$
(C) $4 \times {10^{41}}$
(D) $3 \times {10^{41}}$

Answer 1

Before solving this problem we need to know Newton's Law of gravitation. Also we should have a clear concept of rotational motion because here the idea of centripetal force has to incorporate to solve this problem. The gravitational force between two bodies with masses ${m_1}$ and ${m_2}$ separated by distance $r$ is given by, ${F_{gra}} = \dfrac{{G{m_1}{m_2}}}{{{r^2}}}$, where G is the universal constant.
If a body of mass $m$ rotating in circular path of radius $r$ with a velocity $v$, then the centripetal force acting on the body is given by, ${F_{cen}} = \dfrac{{m{v^2}}}{r}$

Complete step by step answer:
The attractive force between the sun and the milky way can be calculated by considering the sun as a point mass and also the total mass of the milky way is concentrated at the center which is taken to a center of the sun’s orbit.
The radius of the orbit can be taken as the distance between them. And as the sun revolves around the galaxy there will also create an extra force which is called centripetal force due the orbital motion. The sun will stay in the same orbit if these two forces become equal and opposite in direction. Because gravitational force is attractive in nature and centripetal force is repulsive in nature.
Now let m be the mass of the sun and M is the mass of the milky way and radius of the orbit is r .

$${F_{gra}} = {F_{cen}} \\\ \Rightarrow \dfrac{{GMm}}{{{r^2}}} = \dfrac{{m{v^2}}}{r} \\\ \Rightarrow M = \dfrac{{{v^2}r}}{G} \\\$$

Now $v = 250km/s = 250 \times {10^3}m/s$
$r = 3 \times {10^4}light{\text{ }}year = 3 \times {10^4} \times (24 \times 60 \times 60 \times 365) \times 3 \times {10^8}m \\\ \\\$
$G = 6.67 \times {10^{ - 11}}(SI)$
Now,

$$M = \dfrac{{{v^2}r}}{G} = \dfrac{{{{(250 \times {{10}^3})}^2} \times 3 \times {{10}^4} \times (24 \times 60 \times 60 \times 365) \times 3 \times {{10}^8}}}{{6.67 \times {{10}^{ - 11}}}} \\\ \Rightarrow M = 2.66 \times {10^{41}} \approx 3 \times {10^{41}} \\\$$

So the mass of the milky way is $3 \times {10^{41}}kg$
So option D is correct.

Note: The sun will stay in the same orbit if these two forces become equal and opposite in direction. Because gravitational force is attractive in nature and centripetal force is repulsive in nature.