Questions: A distant star has a single planet circling it in a circular orbit of radius 7.51 x 10^11 m. The period of the planet's motion about the star is 726 days. What is the mass of the star? The value of the universal gravitational constant is 6.67259 x 10^-11 N · m^2 / kg^2. Answer in units of kg.

Solution Steps

We are given:

• Radius of the planet's orbit: \( r = 7.51 \times 10^{11} \, \text{m} \)
• Period of the planet's motion: \( T = 726 \, \text{days} \)
• Universal gravitational constant: \( G = 6.67259 \times 10^{-11} \, \text{N} \cdot \text{m}^2 / \text{kg}^2 \)

To use the period in our calculations, we need to convert it to seconds: \[ T = 726 \, \text{days} \times 24 \, \text{hours/day} \times 3600 \, \text{seconds/hour} \] \[ T = 726 \times 24 \times 3600 \] \[ T = 62,726,400 \, \text{s} \]

Kepler's third law in terms of the gravitational constant is: \[ T^2 = \frac{4 \pi^2 r^3}{G M} \] Solving for the mass \( M \) of the star: \[ M = \frac{4 \pi^2 r^3}{G T^2} \]

\[ M = \frac{4 \pi^2 (7.51 \times 10^{11} \, \text{m})^3}{(6.67259 \times 10^{-11} \, \text{N} \cdot \text{m}^2 / \text{kg}^2) (62,726,400 \, \text{s})^2} \]

First, calculate \( r^3 \): \[ r^3 = (7.51 \times 10^{11} \, \text{m})^3 = 4.2351 \times 10^{35} \, \text{m}^3 \]

Next, calculate \( T^2 \): \[ T^2 = (62,726,400 \, \text{s})^2 = 3.9347 \times 10^{15} \, \text{s}^2 \]

Now, substitute these values into the equation for \( M \): \[ M = \frac{4 \pi^2 (4.2351 \times 10^{35} \, \text{m}^3)}{(6.67259 \times 10^{-11} \, \text{N} \cdot \text{m}^2 / \text{kg}^2) (3.9347 \times 10^{15} \, \text{s}^2)} \]

\[ M = \frac{4 \pi^2 \times 4.2351 \times 10^{35}}{6.67259 \times 10^{-11} \times 3.9347 \times 10^{15}} \]

\[ M = \frac{1.6755 \times 10^{37}}{2.6268 \times 10^5} \]

\[ M = 6.3804 \times 10^{31} \, \text{kg} \]

Final Answer