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A body weighs one Newton at the surface of the earth. What will be it’s weight at a height equal to half the radius of the earth ? |
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Answer» SOLUTION :`g^'/g=(GM)/(R+h)^2xxR^2/(GM)=R^2/(R+R/2)^2 THEREFORE=((R^2xx4)/(9R_2))g=4/9g "Force" `=72xx4/9=32 N` |
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