Question:
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?
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?
Updated On: Aug 16, 2024
- 4
- 5
- 2
- 3
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The Correct Option is A
Solution and Explanation
Period of revolution of planet $ A (T_A) = 8T_B $
According to Kepler's III law of planetary motion
$T^2 \propto R^3 $.
Therefore $\bigg( \frac{ r_A}{r_B } \bigg)^3 = \bigg( \frac{ T_A}{T_B } \bigg)^2= \bigg( \frac{ 8 T_B}{T_B } \bigg)^2 = 64 $
or $ \frac{ r_A}{r_B} = 4 \, \, \, $ or $ \, \, \, \, r_A = 4 r_B $.
According to Kepler's III law of planetary motion
$T^2 \propto R^3 $.
Therefore $\bigg( \frac{ r_A}{r_B } \bigg)^3 = \bigg( \frac{ T_A}{T_B } \bigg)^2= \bigg( \frac{ 8 T_B}{T_B } \bigg)^2 = 64 $
or $ \frac{ r_A}{r_B} = 4 \, \, \, $ or $ \, \, \, \, r_A = 4 r_B $.
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Concepts Used:
Gravitation
In mechanics, the universal force of attraction acting between all matter is known as Gravity, also called gravitation, . It is the weakest known force in nature.
Newton’s Law of Gravitation
According to Newton’s law of gravitation, “Every particle in the universe attracts every other particle with a force whose magnitude is,
- F ∝ (M1M2) . . . . (1)
- (F ∝ 1/r2) . . . . (2)
On combining equations (1) and (2) we get,
F ∝ M1M2/r2
F = G × [M1M2]/r2 . . . . (7)
Or, f(r) = GM1M2/r2
The dimension formula of G is [M-1L3T-2].