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QJEEM21Phc994: JEE Main 2021 Physics - Class 11 JEEM21Phc994 Solution: If R E be the radius of Earth, then the ratio between the…
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 )
Multiple Choice Options
A
1 − r R E − r 2 R E 2 − r 3 R E 3 1 - {r \over {{R_E}}} - {{{r^2}} \over {R_E^2}} - {{{r^3}} \over {R_E^3}} 1 − R E r − R E 2 r 2 − R E 3 r 3
B
1 + r R E + r 2 R E 2 + r 3 R E 3 1 + {r \over {{R_E}}} + {{{r^2}} \over {R_E^2}} + {{{r^3}} \over {R_E^3}} 1 + R E r + R E 2 r 2 + R E 3 r 3
C
1 + r R E − r 2 R E 2 + r 3 R E 3 1 + {r \over {{R_E}}} - {{{r^2}} \over {R_E^2}} + {{{r^3}} \over {R_E^3}} 1 + R E r − R E 2 r 2 + R E 3 r 3
D
1 + r R E − r 2 R E 2 − r 3 R E 3 1 + {r \over {{R_E}}} - {{{r^2}} \over {R_E^2}} - {{{r^3}} \over {R_E^3}} 1 + R E r − R E 2 r 2 − R E 3 r 3
Step-by-Step Verified Solution
g u p = g ( 1 + r R ) 2 {g_{up}} = {g \over {{{\left( {1 + {r \over R}} \right)}^2}}} g u p = ( 1 + R r ) 2 g g d o w n = g ( 1 − r R ) {g_{down}} = g\left( {1 - {r \over R}} \right) g d o w n = g ( 1 − R r ) g d o w n g u p = ( 1 − r R ) ( 1 + r R ) 2 {{{g_{down}}} \over {{g_{up}}}} = \left( {1 - {r \over R}} \right){\left( {1 + {r \over R}} \right)^2} g u p g d o w n = ( 1 − R r ) ( 1 + R r ) 2 = ( 1 − r R ) ( 1 + 2 r R + r 2 R 2 ) = \left( {1 - {r \over R}} \right)\left( {1 + {{2r} \over R} + {{{r^2}} \over {{R^2}}}} \right) = ( 1 − R r ) ( 1 + R 2 r + R 2 r 2 ) = 1 + r R − r 2 R 2 − r 3 R 3 = 1 + {r \over R} - {{{r^2}} \over {{R^2}}} - {{{r^3}} \over {{R^3}}} = 1 + R r − R 2 r 2 − R 3 r 3
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