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The distance from the earth's surface at which the acceleration due to gravity equals to 1m/s^2 is about |
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Answer» Solution :MASS of the earth `VxxP=4//3pir^3rho g=GM//R^2=G/R^2(4/3piR^3rho)=4//3piRGrho` THUS, `galpharho` since R S G are constant . If `RHO` is DOUBLED, then g'=2g=2xx98=19.6m//s^2` |
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