Explanation Given: The table of values. Planet Mercury Venus Earth Mars a 5.79 10.8 15 22.8 T 0.241 0.615 1.00 1.88 Formula used: Kepler’s Third Law, T 2 a 3 = k Calculation: Let us evaluate the ratio T 2 a 3 for all the planets in the table. T 2 a 3 = 0.241 2 ( 5.79 × 10 10 ) 3 = 2.99 × 10 − 34 T 2 a 3 = 0.615 2 ( 10.8 × 10 10 ) 3 = 3.0 × 10 − 34 T 2 a 3 = 1 2 ( 15 × 10 10 ) 3 = 2.96 × 10 − 34 T 2 a 3 = 1...

4th Edition

ISBN: 9781319434526

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 14.6, Problem 1E

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Whether the data in the table supports conclusions of Kepler’s Third Law.

The length of the Jupiter’s period for $a = 77.8 \times 10^{10}$

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Chapter 14.6, Problem 3PQ

Chapter 14.6, Problem 2E

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Whether the data in the table supports conclusions of Kepler’s Third Law. The length of the Jupiter’s period for a = 77.8 × 10 10

Solution Summary: The author analyzes how the data in the table supports Kepler's Third Law. The length of period for Jupiter is 11.89.

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Whether the data in the table supports conclusions of Kepler’s Third Law.

The length of the Jupiter’s period for $a = 77.8 \times 10^{10}$

Want to see the full answer?

Chapter 14 Solutions

Whether the data in the table supports conclusions of Kepler’s Third Law. The length of the Jupiter’s period for a = 77.8 × 10 10

Solution Summary: The author analyzes how the data in the table supports Kepler's Third Law. The length of period for Jupiter is 11.89.

Concept explainers

Videos

Whether the data in the table supports conclusions of Kepler’s Third Law. The length of the Jupiter’s period for a=77.8×1010

Want to see the full answer?

Chapter 14 Solutions

Whether the data in the table supports conclusions of Kepler’s Third Law.

The length of the Jupiter’s period for $a = 77.8 \times 10^{10}$