15 Questions 60 Marks 15 Mins
Concept:
T2 ∝ R3 ⇒ T2 = k R3
Calculation: Let a present, time period of revolution is T = 365 days -- (1) The radius of earth R By Kepler's Law T2 = k R3 -- (2) If the radius is halved then the new radius is R', and the new time period is T' Radius is halved \(R'=\frac{R}{2}\)---- (3) T'2 = k R'3 ⇒ \(T'^2 = k(\frac{R}{2})^3\) ⇒ \(T'^2 = k(\frac{R^3}{8})\) --- (4) Equation (3) in (4) \(T'^2 = k(\frac{R^2}{8})\) ⇒ \(T' = (\frac{T}{\sqrt{8}})\) (using eq 2 ) ⇒ \(T' = (\frac{365}{\sqrt{8}})\) ⇒ T' = 129 So, 129 days is the correct answer. India’s #1 Learning Platform Start Complete Exam Preparation
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Practice Question Bank
Mock Tests & Quizzes Trusted by 3.4 Crore+ Students Text Solution 1032 days 365 days 129 days 556 days Answer : A Solution : By Kepler's second law, `T^(2)prop r^(3)` <br> `therefore ((T_(2))/(T_(1)))^(2)=((r_(2))/(r_(1)))^(3)=((2)/(1))^(3) = 8` <br> `therefore (T_(2))/(T_(1))=sqrt(8)=2sqrt(2) " " therefore T_(2)=2sqrt(2)xx T_(1)` <br> `therefore T_(2)=365xx2xx1.414=1032` days |
What would be the duration of the year if the distance between the Earth and sun has been increased by two times its present distance?
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