KCET · Physics · Gravitation
A planet moving around sun sweeps area \(\mathrm{A}_{1}\) in 2 days, \(\mathrm{A}_{2}\) in 3 days and \(\mathrm{A}_{3}\) in 6 days. Then, the relation between \(\mathrm{A}_{1}, \mathrm{~A}_{2}\) and \(\mathrm{A}_{3}\) is
- A \(6 \mathrm{~A}_{1}=3 \mathrm{~A}_{2}=2 \mathrm{~A}_{3}\)
- B \(3 A_{1}=2 A_{2}=A_{3}\)
- C \(2 \mathrm{~A}_{1}=3 \mathrm{~A}_{2}=6 \mathrm{~A}_{3}\)
- D \(3 A_{1}=2 A_{2}=6 A_{3}\)
Answer & Solution
Correct Answer
(B) \(3 A_{1}=2 A_{2}=A_{3}\)
Step-by-step Solution
Detailed explanation
By Kepler's second law of motion,
\(\frac{A_{1}}{t_{1}}=\frac{A_{2}}{t_{2}}=\frac{A_{3}}{t_{3}}\)
\(\frac{A_{1}}{2}=\frac{A_{2}}{3}=\frac{A_{3}}{6}\)
\(\frac{3 A_{1}}{6}=\frac{2 A_{2}}{6}=\frac{A_{3}}{6}\)
\(3 A_{1}=2 A_{2}=A_{3}\)
\(\frac{A_{1}}{t_{1}}=\frac{A_{2}}{t_{2}}=\frac{A_{3}}{t_{3}}\)
\(\frac{A_{1}}{2}=\frac{A_{2}}{3}=\frac{A_{3}}{6}\)
\(\frac{3 A_{1}}{6}=\frac{2 A_{2}}{6}=\frac{A_{3}}{6}\)
\(3 A_{1}=2 A_{2}=A_{3}\)
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