Question 13.4: The Mass of the Sun In the opening storyline, you were wonde...

The Mass of the Sun

In the opening storyline, you were wondering how to determine the mass of the Sun. Now that we have discussed Kepler’s third law, calculate the mass of the Sun.

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Conceptualize Based on the mathematical representation of Kepler’s third law expressed in Equation 13.11, we realize that the mass of the central object in a gravitational system is related to the orbital size and period of objects in orbit around the central object.

$T^2=\left(\frac{4 \pi^2}{G M_S}\right) a^3=K_S a^3$ (13.11)

Categorize This example is a relatively simple substitution problem.

Solve Equation 13.11 for the mass of the Sun:

M_S =\frac{4 \pi^2 r^3}{G T^2}

Substitute numerical values, using data from Table 13.2:

M_S =\frac{4 \pi^2\left(1.496 \times 10^{11}  m\right)^3}{\left(6.674 \times 10^{-11}  N \cdot m^2 / kg^2\right)\left(3.156 \times 10^7  s\right)^2}=1.99 \times 10^{30}  kg

In Example 13.2, an understanding of gravitational forces enabled us to find out something about the density of the Earth’s core, and now we have used this understanding to answer your question about the mass of the Sun! To answer your question about the masses of planets, we can perform the same calculation using the orbital size and period of a moon of a planet to find the planet mass. Neither Kepler’s third law or Newton’s law of universal gravitation can be used to determine the mass of the orbiting object, and the planets and KBOs for which we have precise mass data are the ones with moons or ones about which we have placed a spacecraft in orbit.

Table 13.2 Useful Planetary Data
Body	Mass (kg)	Mean Radius (m)	Period of Revolution (s)	Mean Distance from the Sun (m)	$\frac{T^2}{r^3}(s^2/m^3)$
Mercury	$3.30 \times 10^{23}$	$2.44 \times 10^{6}$	$7.60 \times 10^{6}$	$5.79 \times 10^{10}$	$2.98 \times 10^{-19}$
Venus	$4.87 \times 10^{24}$	$6.05 \times 10^{6}$	$1.94 \times 10^{7}$	$1.08 \times 10^{11}$	$2.99 \times 10^{-19}$
Earth	$5.97 \times 10^{24}$	$6.37 \times 10^{6}$	$3.156 \times 10^{7}$	$1.496 \times 10^{11}$	$2.97 \times 10^{-19}$
Mars	$6.42 \times 10^{23}$	$3.39 \times 10^{6}$	$5.94 \times 10^{7}$	$2.28 \times 10^{11}$	$2.98 \times 10^{-19}$
Jupiter	$1.90 \times 10^{27}$	$6.99 \times 10^{7}$	$3.74 \times 10^{8}$	$7.78 \times 10^{11}$	$2.97 \times 10^{-19}$
Saturn	$5.68 \times 10^{26}$	$5.82 \times 10^{7}$	$9.29 \times 10^{8}$	$1.43 \times 10^{12}$	$2.95 \times 10^{-19}$
Uranus	$8.68 \times 10^{25}$	$2.54 \times 10^{7}$	$2.65 \times 10^{9}$	$2.87 \times 10^{12}$	$2.97 \times 10^{-19}$
Neptune	$1.02 \times 10^{26}$	$2.46 \times 10^{7}$	$5.18 \times 10^{9}$	$4.50 \times 10^{12}$	$2.94 \times 10^{-19}$
$\text{Pluto}^a$	$1.25 \times 10^{22}$	$1.20 \times 10^{6}$	$7.82 \times 10^{9}$	$5.91 \times 10^{12}$	$2.96 \times 10^{-19}$
Moon	$7.35 \times 10^{22}$	$1.74 \times 10^{6}$	—	—	—
Sun	$1.989 \times 10^{30}$	$6.96 \times 10^{8}$	—	—	—
${}^a$ In August 2006, the International Astronomical Union adopted a definition of a planet that separates Pluto from the other eight planets. Pluto is now defined as a “dwarf planet” like the asteroid Ceres.