A particle of mass 'm' is raised to a height `h = R` from the surface of earth. Find increase in potential energy. `R =` radius of earth. `g =` acceleration due to gravity on the surface of earth.
Text Solution
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Let `M,R` be the mass and radius of the earth. Then `G=GM//R^(2)` or `GM=gR^(2)`…………i Potential energy of the object on the surface of earth `U_(1)=(-GMm)/R` The potential energy of the object at a height equal to radius of the earth is `U_(2)=(-GMm)/(2R)` Gain in `PE` is `U_(2)-U_(1)=-(GMm)/(2R)+(GMm)/R=(GMm)/(2R)` `=((gR^(2))m)/(2R)=1/2mgR`
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