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At a point above the surface of the eart...

At a point above the surface of the earth, the gravitational potential is `-5.12 xx 10^(7)` J/kg and the acceleration due to gravity is `6.4 m//s^(2)`. Assuming the mean radius of the earth to be 6400 km, calculate the height of the point above the earth's surface.

Text Solution

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Let `r` be the distance of the given point from the centre of the earth. Then,
Gravitational potential `=-(GM)/(r )=-5.12xx10^(7)J//kg`………`(1)`
and acceleration due to gravity,
`g=(GM)/(r^(2))=6.4m//s^(2)` ...............`(2)`
Divding `(1)` by `(2)`, we get
`r=(5.12xxx10^(7))/(6.4)=8xx10^(6)m=8000km`
Height of the point from earth.s surface `=r-R=8000-6400=1600km`
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