Calculate the mass of the Sun based on data for Earth’s orbit and compare the value obtained with the Sun’s actual mass.

A centripetal force is a force exerted toward the center of a circular road that is always perpendicular to the velocity of the body.

The gravitational force between the two objects is given by,

$F_g=\frac{G{m}_1{m}_2}{r^2}$

Where G is the universal gravitation constant, r is the distance between the two objects, and ${\text{m}}_{\text{1}}{\text{,m}}_{\text{2}}$are the masses of two bodies, respectively.

In angular motion, the centripetal acceleration is given,

localid="1655454943790" $\begin{array}{c}{F}_c=m{a}_c\\ =m{\omega }^{2}r\end{array}$

As the earth revolves around the sun, so there is the only force that is the gravitational force which provides centripetal force to the earth for revolution. Therefore, the centripetal force on the earth is equal to the gravitational force between the earth and the sun.

Mathematically, it can be written as,

$\begin{array}{c}{F}_g=F_c\\ \frac{G{m}_{earth}{m}_{sun}}{r^2}=m_{earth}{\omega }^{2}r\\ m_{sun}=\frac{{\omega }^2{r}^3}{G}\end{array}$

The earth completes one revolution of the sun in one year. So, the angular velocity of the earth is,

$\begin{array}{l}\omega =1\frac{\text{rev}}{\text{y}}\\ \omega =1\frac{\text{rev}}{\text{y}}\times \frac{\text{2}\pi \text{\hspace{0.33em}rad}}{\text{1\hspace{0.33em}rev}}\times \frac{1\text{\hspace{0.33em}y}}{365\text{\hspace{0.33em}day}}\times \frac{1\text{\hspace{0.33em}day}}{24\text{\hspace{0.33em}hr}}\times \frac{1\text{\hspace{0.33em}hr}}{3600\text{\hspace{0.33em}s}}\\ \omega =1.99\times {10}^{-7}\frac{\text{rad}}{\text{s}}\end{array}$

Distance of the sun from the earth in Table 6.2 $r=1.496\times {10}^{11}\text{\hspace{0.33em}m}$.

So, the mass of the sun,

$\begin{array}{l}{m}_{sun}=\frac{{\left(1.99\times {10}^{-7}\right)}^2\times {\left(1.496\times {10}^{11}\right)}^3}{6.673\times {10}^{-11}}\\ m_{sun}=1.9869\times {10}^{30}\text{\hspace{0.33em}kg}\end{array}$

Hence, the mass of the Sun based on the data from Earth’s orbit is ${\text{1.9869×10}}^{\text{30}}\text{\hspace{0.33em}kg}$, which is very close to the Sun’s actual mass that is ${\text{1.9889×10}}^{\text{30}}\text{\hspace{0.33em}kg}$. As a result, the computed number for the Sun's mass agrees with the real value.