Gravitation

Question : 5 of 25

Marks: +1, -0

Let us assume that our galaxy consists of

2.5 \times 10^{11}

stars each of one solar mass. How long will a star at a distance of 50,000 ly from the galactic centre take to complete one revolution? Take the diameter of the Milky Way to be

10^{5}

ly.

Solution:

Using,

M = \frac{4 \pi^{2} R^{3}}{G T^{2}}

T = \left( \frac{4 \pi^{2} R^{3}}{G M} \right)^{1/2}

Given,

M = 2.5 \times 10^{11}

Solar mass

= 2.5 \times 10^{11} \times (2 \times 10^{30}) \text{kg}

= 5.0 \times 10^{41} \text{kg}

and

R = 50,0001 y

= 50,000 \times 9.4 \times 10^{15} \text{m}

= 4.73 \times 10^{20} \text{m}

\therefore T = \left[ \frac{4 \times (3.14)^{2} \times (4.73 \times 10^{20})^{3}}{(6.67 \times 10^{-11}) \times (5.0 \times 10^{41})} \right]^{1/2}

T = 1.12 \times 10^{16} \text{s}

= 3.54 \times 10^{8} \text{years}

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