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The value of acceleration due to gravity...

The value of acceleration due to gravity at Earth's surface is `9.8ms^(-2)`. The altitude (in metre) above its surface at which the acceleration due to gravity decreases to `4.9ms^(-2)`, is close to :(Radius of earth `=6.4xx10^(6)m`)

A

`2.6xx10^(6)m`

B

`6.4xx10^(6)m`

C

`9.0xx10^(6)m`

D

`1.6xx10^(6)m`

Text Solution

Verified by Experts

The correct Answer is:
A
`(g)/(2)=(Gm)/((R+h)^(2))`
`g=(Gm)/(R^(2))`
`(1)/(2)=(R^(2))/((R+h)^(2))impliesR+h=sqrt(2)R`
`R=0.41R`
`=0.41xx6.4xx10^(6)m=2.6xx10^(6)m`
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