If RE be the radius of Earth, then the ratio between the acceleration due to gravity at a depth 'r' below and a height 'r' above the earth surface is : (Given : r < RE)
Solution
${g_{up}} = {g \over {{{\left( {1 + {r \over R}} \right)}^2}}}$<br><br>${g_{down}} = g\left( {1 - {r \over R}} \right)$<br><br>$${{{g_{down}}} \over {{g_{up}}}} = \left( {1 - {r \over R}} \right){\left( {1 + {r \over R}} \right)^2}$$<br><br>$$ = \left( {1 - {r \over R}} \right)\left( {1 + {{2r} \over R} + {{{r^2}} \over {{R^2}}}} \right)$$<br><br>$= 1 + {r \over R} - {{{r^2}} \over {{R^2}}} - {{{r^3}} \over {{R^3}}}$
About this question
Subject: Physics · Chapter: Gravitation · Topic: Gravitational Field and Potential
This question is part of PrepWiser's free JEE Main question bank. 129 more solved questions on Gravitation are available — start with the harder ones if your accuracy is >70%.