What does Kepler's third law predict for the period of an Asteroid's orbit around the Sun? Give your answer in Years G=6.67x10 11 Nm2/kg2 MAsteroid=8.8x1020 kg MEarth=5.97x1024 kg Msun= 2.00x1030 kg Distance from Sun to Asteroid = 5.50x1012 m There are 3.15x107 seconds in 1 earth year. Years
What does Kepler's third law predict for the period of an Asteroid's orbit around the Sun? Give your answer in Years G=6.67x10 11 Nm2/kg2 MAsteroid=8.8x1020 kg MEarth=5.97x1024 kg Msun= 2.00x1030 kg Distance from Sun to Asteroid = 5.50x1012 m There are 3.15x107 seconds in 1 earth year. Years
College Physics
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Chapter1: Units, Trigonometry. And Vectors
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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![**Kepler's Law**
**Image Description:** The image shows a detailed photo of an asteroid, highlighting its cratered surface.
**Problem Statement:**
What does Kepler's third law predict for the period of an Asteroid’s orbit around the Sun? Give your answer in years.
**Given Values:**
- Gravitational constant, \( G = 6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \)
- Mass of the Asteroid, \( M_{\text{Asteroid}} = 8.8 \times 10^{20} \, \text{kg} \)
- Mass of the Earth, \( M_{\text{Earth}} = 5.97 \times 10^{24} \, \text{kg} \)
- Mass of the Sun, \( M_{\text{Sun}} = 2.00 \times 10^{30} \, \text{kg} \)
- Distance from Sun to Asteroid, \( 5.50 \times 10^{12} \, \text{m} \)
- There are \( 3.15 \times 10^7 \) seconds in 1 Earth year.
**Objective:**
Calculate the orbital period of the asteroid in Earth years using Kepler's third law.
**Solution:**
Insert the corrected formula and steps to find the period in years, if required, following the data given.
(Note: This educational content would further include a detailed explanation or example calculations using Kepler's third law of planetary motion, though they are not included in the image text itself.)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa50e57d1-2bd6-427e-801e-187552a1e482%2F5992f5be-d2fe-413b-aadf-64b2e57b7738%2Flucav2q_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Kepler's Law**
**Image Description:** The image shows a detailed photo of an asteroid, highlighting its cratered surface.
**Problem Statement:**
What does Kepler's third law predict for the period of an Asteroid’s orbit around the Sun? Give your answer in years.
**Given Values:**
- Gravitational constant, \( G = 6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \)
- Mass of the Asteroid, \( M_{\text{Asteroid}} = 8.8 \times 10^{20} \, \text{kg} \)
- Mass of the Earth, \( M_{\text{Earth}} = 5.97 \times 10^{24} \, \text{kg} \)
- Mass of the Sun, \( M_{\text{Sun}} = 2.00 \times 10^{30} \, \text{kg} \)
- Distance from Sun to Asteroid, \( 5.50 \times 10^{12} \, \text{m} \)
- There are \( 3.15 \times 10^7 \) seconds in 1 Earth year.
**Objective:**
Calculate the orbital period of the asteroid in Earth years using Kepler's third law.
**Solution:**
Insert the corrected formula and steps to find the period in years, if required, following the data given.
(Note: This educational content would further include a detailed explanation or example calculations using Kepler's third law of planetary motion, though they are not included in the image text itself.)
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