**Kepler's Law**![Image of Mars]**What does Kepler's third law predict for the period of Mars' orbit around the Sun? Give your answer in Days.**- \( G = 6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \)- \( M_{\text{Mars}} = 6.4 \times 10^{23} \, \text{kg} \)- \( M_{\text{Earth}} = 5.97 \times 10^{24} \, \text{kg} \)- \( M_{\text{Sun}} = 2.00 \times 10^{30} \, \text{kg} \)Distance from Sun to Mars \( = 2.28 \times 10^{11} \, \text{m} \)There are 86400 seconds in 1 day.\[ \boxed{\phantom{0}} \] Days

Given: G=6.67×10-11Nm2/kg2 Mmars=6.4×1023kg Mearth=5.97×1024kg Msun=2.00×1030kg Distance from sun to…

What does Kepler's third law predict for the period of Mars' orbit around the Sun? Give your answer in Days. G=6.67x10-11 Nm2/kg? MMars=6.4x1023 kg MEarth=5.97x1024 kg Msun= 2.00x1030 kg Distance from Sun to Mars = 2.28x1011 m There are 86400 seconds in 1 day. Days

What does Kepler's third law predict for the period of Mars' orbit around the Sun? Give your answer in Days. G=6.67x10-11 Nm2/kg? MMars=6.4x1023 kg MEarth=5.97x1024 kg Msun= 2.00x1030 kg Distance from Sun to Mars = 2.28x1011 m There are 86400 seconds in 1 day. Days

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

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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Question

**Kepler's Law**

![Image of Mars]

**What does Kepler's third law predict for the period of Mars' orbit around the Sun? Give your answer in Days.**

- \( G = 6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \)
- \( M_{\text{Mars}} = 6.4 \times 10^{23} \, \text{kg} \)
- \( M_{\text{Earth}} = 5.97 \times 10^{24} \, \text{kg} \)
- \( M_{\text{Sun}} = 2.00 \times 10^{30} \, \text{kg} \)

Distance from Sun to Mars \( = 2.28 \times 10^{11} \, \text{m} \)

There are 86400 seconds in 1 day.

\[ \boxed{\phantom{0}} \] Days

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