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If g is acceleration due to gravity on the surface of the earth ,having radius R , the height at which the acceleration due to gravity reduces tog2 is

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a

R/2

b

2R

c

R/2

d

2-1R

answer is D.

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Detailed Solution

gh=gRR+h2 ⇒g2=gRR+h2 ∵ gh=g2 ⇒12=RR+h2 ⇒12=RR+h ⇒R+h=2R ⇒h=2-1R

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