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Concept:
- Kepler's Law of Planetry Motion: When a planet is revolving around its star (Sun for earth), then the square of the time period of revolution is directly proportional to the cube of the radius of revolution.
T2 ∝ R3
⇒ T2 = k R3
- The time period of the revolution of the earth across the sun is one year.
- One year of the earth consists of 365 days.
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.
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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