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Question
- neptunes orbital period neptune orbits the sun at an average distance given in figure 9, which allows gases, such as methane, to condense and form an atmosphere. if the mass of the sun is 1.99 × 10³⁰ kg, calculate the period of neptunes orbit.
(image: sun at center, dashed circular orbit, neptune at bottom, arrow labeled r = 4.496×10¹² m, caption figure 9 (not to scale))
Step1: Recall Kepler's Third Law (modified for circular orbits, using Newton's law of gravitation)
The formula for the orbital period \( T \) of a planet orbiting the Sun is derived from equating the gravitational force to the centripetal force. The formula is:
where:
- \( r = 4.496 \times 10^{12} \, \text{m} \) (orbital radius of Neptune),
- \( G = 6.674 \times 10^{-11} \, \text{N·m}^2/\text{kg}^2 \) (gravitational constant),
- \( M = 1.99 \times 10^{30} \, \text{kg} \) (mass of the Sun).
Step2: Substitute the values into the formula
First, calculate \( r^3 \):
Next, calculate the denominator \( G M \):
Now, compute \( \frac{r^3}{G M} \):
Take the square root:
Finally, multiply by \( 2\pi \):
(Optional: Convert to years for better intuition. There are \( 3.154 \times 10^7 \, \text{s/year} \), so \( T \approx \frac{5.17 \times 10^9}{3.154 \times 10^7} \approx 164 \, \text{years} \), which matches known values.)
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The orbital period of Neptune is approximately \( \boldsymbol{5.17 \times 10^9 \, \text{seconds}} \) (or about 164 years).