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The height at which the acceleration due to gravity becomes g/9. (Where g = acceleration due to gravity) in terms of R. Where R is the radius of earth |
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Answer» SOLUTION :`G=(GM)/R^2` and g'=(GM)/(R+h)^2 THEREFORE g'//g=^1//9=GM/((R+h)^2)xxR^2/(GM)=R^2/((R+h)^2) thereforeR^2=(R+h)^2 therefore3R=R+h=2R` |
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