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The period of revolution of planet A around the sun is 8 times that of B. The distance of A from the sun is how many times greater than that of B from the sun ? |
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Answer» Solution :Period of revolution of planet `A(T_(A))=8T_(B)`. According to Kepler.s 111 law of planetary motion `T^(2) PROP r^(3)`. Therefore `((r_(A))/(r_(B)))^(3)=((T_(A))/(T_(B))^(2))=((8T_(B))/(T_(B))^(2))=64` or `(r_(A))/(r_(B))=4" or " r_(A)=4r_(B)` |
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