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Kepler’s 3rd Law Examples

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1 Kepler’s 3rd Law Examples

2 𝑇 2 = π‘Ÿ 3 ;𝑇=π‘‚π‘Ÿπ‘π‘–π‘‘π‘Žπ‘™ π‘π‘’π‘Ÿπ‘–π‘œπ‘‘, π‘Ÿ=π‘šπ‘’π‘Žπ‘› π‘‘π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’ π‘“π‘Ÿπ‘œπ‘š π‘‘β„Žπ‘’ 𝑠𝑒𝑛
Example 1 A dwarf planet discovered out beyond the orbit of Pluto is known to have an orbital period of years. What is its average distance from the sun? 𝑇 2 = π‘Ÿ 3 ;𝑇=π‘‚π‘Ÿπ‘π‘–π‘‘π‘Žπ‘™ π‘π‘’π‘Ÿπ‘–π‘œπ‘‘, π‘Ÿ=π‘šπ‘’π‘Žπ‘› π‘‘π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’ π‘“π‘Ÿπ‘œπ‘š π‘‘β„Žπ‘’ 𝑠𝑒𝑛 = π‘Ÿ 3 = π‘Ÿ 3 π‘Ÿ=72.35π΄π‘ˆ

3 Example 2 From a telecommunications point of view, it’s advantageous for satellites to remain at the same location relative to a location on Earth. This can only occur if the satellite’s orbital period is the same as Earth’s period of rotation, approximately 24 hours. At what distance from the center of can geosynchronous orbit be found? 𝑇 2 = 4 πœ‹ 2 𝐺 𝑀 𝐸 π‘Ÿ 3 = 4 πœ‹ π‘₯ 10 βˆ’11 π‘˜π‘” βˆ’1 π‘š 3 𝑠 2 Γ—5.98π‘₯ π‘˜π‘” π‘Ÿ 3 =9.898π‘₯ 10 βˆ’14 π‘Ÿ 3 7.54π‘₯ = π‘Ÿ 3 π‘Ÿ=βˆ›7.54π‘₯ β‰ˆ4.23π‘₯ 10 7 π‘š 𝑇=86400𝑠 𝐺=6.67π‘₯ 10 βˆ’11 π‘˜π‘” βˆ’1 π‘š 3 𝑠 2 𝑀 𝐸 =5.98π‘₯ π‘˜π‘”


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