Gravitation Question 303
Question: elliptical orbit of a planet m about the sun S. The shaded area SCD is twice the shaded area SAB. If $ [t_{1}] $ is the time for the planet to move from C to D and $ [t_{2}] $ is the time to move from A to B then
Options:
A) $ [t_{i}=4t_{2}] $
B) $ [t_{i}=2t_{2}] $
C) $ [t_{i}=t_{2}] $
D) $ [t_{i}>t_{2}] $
Show Answer
Answer:
Correct Answer: B
Solution:
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According to Kepler’s law, the areal velocity of a planet around the sun always remains constant. $ [SCD : A_{1}-t_{1}] $ (areal velocity constant) $ [SAB : A_{2}-t_{2}] $
$ \frac{A_{1}}{t_{1}}=\frac{A_{2}}{t_{2}},t_{1}=t_{2}.\frac{A_{1}}{A_{2}}$, given,$ A_{1}=2{A}_{2} $
$ [=t _{2}.\frac{2A _{2}}{A _{2}}\therefore t _{1}=2t _{2}] $
