Question

How will you ‘weigh the sun’, that is, estimate its mass? The mean orbital radius of the earth around the sun is \[1.5 \times {10^8}\,km\].

Answer 1

Hint: To solve this question we have to know about the orbit of the Earth. We realize that, the bent way of a divine article or shuttle cycle a star, planet, or moon, particularly an occasional circular upset. We additionally realize that, the gravitational fascination of our Sun for the Earth is the centripetal power causing the Earth's roundabout movement around the Sun, we can utilize Newton's law of all-inclusive inclination toward tracking down the mass of the Sun without visiting the Sun. This is a similar strategy you would use to decide the mass of Cygnus \[x - 1\] with a test. The Earth, circling the Sun, assumes a similar part as the test shipped off circle Cygnus \[x - 1\].

Complete step by step answer:
Orbital radius of the Earth around the sun, \[r = 1.5 \times {10^{11}}\,m\].
Time taken by the Earth to complete one revolution around the Sun, \[T = 1\,year = 365.25\] days which is equal to \[365.25 \times 24 \times 60 \times 60\,s\].
Universal gravitational constant, $G$ is equal to \[6.67 \times {10^{ - 11}}N{m^2}k{g^{ - 2}}\].Thus, mass of the Sun :
$M = \dfrac{{4{\pi ^2}{r^3}}}{{G{T^2}}} \\
\Rightarrow M = \dfrac{{4 \times {{3.14}^2} \times {{(1.5 \times {{10}^{11}})}^3}}}{{6.67 \times {{10}^{ - 11}} \times {{(365.25 \times 24 \times 60 \times 60)}^2}}} \\
\therefore M\approx 2 \times {10^{30}}\,kg$

Hence, the mass of the sun is \[2 \times {10^{30}}\,kg\].

Note:We can say, the sun powered mass (M☉) is a standard unit of mass in stargazing, equivalent to roughly It is frequently used to show the majority of different stars, just as heavenly groups, nebulae, worlds and dark openings. It is around equivalent to \[2 \times {10^{30}}\,kg\] the mass of the Sun. we need to keep this in our psyche.