Chapter 7, Problem 90A

To determine

The mass of the Sun using Newton’s version of Kepler’s third law and the data of Earth.

Answer to Problem 90A

  $2.00\times {10}^{30}\text{kg}$

Explanation of Solution

Given:

Mass of Earth is, $M_E=5.98\times {10}^{24}\text{kg}$

Mean distance of Sun from the Earth is $r=1.50\times {10}^{11}\text{}\text{m}$

Time taken by Earth to circle the Sun is, $T=365\text{days}$

Formula used:

Newton’s version of Kepler’s third law: Period of a Planet orbiting the Sun is

  $T=2\pi \sqrt{\frac{r^3}{G{m}_S}}$

Where, $G$ is the gravitational constant, $r$ is the orbital radius, and $m_S$ is the mass of the Sun.

Calculation:

The value of gravitational constant is

  $G=6.67\times {10}^{-11}\text{N}{\text{.m}}^{\text{2}}{\text{/kg}}^{\text{2}}$

Using all the known values in above formula of Time period, the mass of Sun can be calculated as,

  $\begin{array}{c}T=2\pi \sqrt{\frac{r^3}{G{m}_S}}\\ 365\text{days}\times \frac{24\text{}\text{h}}{1\text{day}}\times \frac{3600\text{}\text{s}}{1\text{h}}=2\pi \sqrt{\frac{{\left(1.50\times {10}^{11}\right)}^3}{6.67\times {10}^{-11}\times m_S}}\\ 5.02\times {10}^6=\sqrt{\frac{0.506\times {10}^{44}}{m_S}}\\ m_s=2.00\times {10}^{30}\text{kg}\end{array}$

Conclusion:

Thus, the mass of the Sun is $2.00\times {10}^{30}\text{kg}$ .