A planet has radius equal to 2RE and density is equal to 4ρE. If the value of escape speed at surface of planet is α times vE(vE is escape speed on earth, RE is radius of earth and ρE is the density of earth) then -
A
α=2
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B
α=9
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C
α=6
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D
α=4
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Solution
The correct option is Dα=4 vE=√2GMERE=√2G×ρE×4πR3E3RE ⇒vE=√8πGρER2E3
If radius is 2RE and density is 4ρE then, v′E=√8πG(4ρE)(2RE)23 ⇒v′E=√16×8πGρER2E3 ⇒v′E=4vE