Question

How to calculate the mass of the Sun?

Answer 1

Hint:The gravitational attraction between the Sun and the Earth is due to the centripetal force which is causing the circular motion of the Earth around the Sun. We can use Newton’s law of universal gravitation to find the mass of the Sun.

Complete step-by-step solution:
From Newton’s Law of Gravitation, we can find the mass of the Sun without even visiting the Sun. This same technique can be employed to find the mass of other stars like Cygnus X-1.
Now we can say that,
${F_{gravitational}} = {F_{{\text{centripetal}}}}$
${F_{gravitational}} = \dfrac{{GMm}}{{{r^2}}} - - - - - - \left( 1 \right)$ where $G = $ Universal Gravitational constant, $M = $ mass of the Sun, $m = $ mass of the Earth and $r = $ orbital radius of the earth$ = $ distance between the Sun and the Earth.
${F_{{\text{centripetal}}}} = \dfrac{{m{v^2}}}{r} - - - - - - \left( 2 \right)$ where $m = $ mass of the Earth, $v = $ velocity of the Earth around the Sun and $r = $ orbital radius of the earth$ = $ distance between the Sun and the Earth.
Comparing both equation $\left( 1 \right)$ and $\left( 2 \right)$ we get,
$\dfrac{{GMm}}{{{r^2}}} = \dfrac{{m{v^2}}}{r}$
Eliminating $m$ from both sides we get,
$\dfrac{{GM}}{r} = {v^2}$
Arranging the equation we get,
$M = \dfrac{{r{v^2}}}{G} - - - - - - \left( 3 \right)$
The known values are,
$G = 6.67 \times {10^{ - 11}}{\text{ }}\dfrac{{N.{m^2}}}{{k{g^{}}}}$
$r = 1.5 \times {10^{11}}{\text{ }}m$
$v = \dfrac{{2\pi r}}{T}$ where $T$ is the time required for the Earth to complete one revolution around the Sun.
$\therefore T = 365.25 \times 24 \times 60 \times 60 = 3.15 \times {10^7}{\text{ }}s$
Now, $v = \dfrac{{2 \times 3.14 \times 1.5 \times {{10}^{11}}}}{{3.15 \times {{10}^7}}} = 3 \times {10^4}{\text{ }}\dfrac{m}{s}$
Substituting all the values in equation $\left( 3 \right)$ we get,
$M = \dfrac{{1.5 \times {{10}^{11}} \times {{\left( {3 \times {{10}^4}} \right)}^2}}}{{6.67 \times {{10}^{ - 11}}}} = 1.989 \times {10^{30}}{\text{ }}kg$
Thus, the mass of the Sun is $1.989 \times {10^{30}}{\text{ }}kg$.


Note:The orbital radius between the Sun and the Earth is a mean radius. The path of orbit between the Sun and the Earth is not circular. It is elliptical, so we have to take the mean radius. It must be made sure that the distance measurement must be in meters, time in seconds in order to find the proper result.