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Given that acceleration due to gravity v...

Given that acceleration due to gravity varies inversely as the square of the distance from the center of earth, find its value at a height of 64 km from the earth's surface , if the value at the surface be `9.81ms^(-2)`. Radis of earth`=6400km`.

Text Solution

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Given `g=(1)/(r^2)`
If g be the value required then
`(g)/(9.81)=((R )/(R+h))^2implies(g)/(9.81)[(R+h)/(R )]^-2`
`impliesg=9.81[1+(h)/(R )]^-2=9.81[1-2(h)/(R )]`
`=9.81[1-2((64)/(6400))]=9.81[1-((2)/(100))]`
`=9.81[1-((1)/(50))=9.81[(49)/(50)]=9.61(m)/(s^2)`
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