JEE MAIN - Physics (2021 - 31st August Evening Shift - No. 20)
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)
$$1 - {r \over {{R_E}}} - {{{r^2}} \over {R_E^2}} - {{{r^3}} \over {R_E^3}}$$
$$1 + {r \over {{R_E}}} + {{{r^2}} \over {R_E^2}} + {{{r^3}} \over {R_E^3}}$$
$$1 + {r \over {{R_E}}} - {{{r^2}} \over {R_E^2}} + {{{r^3}} \over {R_E^3}}$$
$$1 + {r \over {{R_E}}} - {{{r^2}} \over {R_E^2}} - {{{r^3}} \over {R_E^3}}$$
Explanation
$${g_{up}} = {g \over {{{\left( {1 + {r \over R}} \right)}^2}}}$$
$${g_{down}} = g\left( {1 - {r \over R}} \right)$$
$${{{g_{down}}} \over {{g_{up}}}} = \left( {1 - {r \over R}} \right){\left( {1 + {r \over R}} \right)^2}$$
$$ = \left( {1 - {r \over R}} \right)\left( {1 + {{2r} \over R} + {{{r^2}} \over {{R^2}}}} \right)$$
$$ = 1 + {r \over R} - {{{r^2}} \over {{R^2}}} - {{{r^3}} \over {{R^3}}}$$
$${g_{down}} = g\left( {1 - {r \over R}} \right)$$
$${{{g_{down}}} \over {{g_{up}}}} = \left( {1 - {r \over R}} \right){\left( {1 + {r \over R}} \right)^2}$$
$$ = \left( {1 - {r \over R}} \right)\left( {1 + {{2r} \over R} + {{{r^2}} \over {{R^2}}}} \right)$$
$$ = 1 + {r \over R} - {{{r^2}} \over {{R^2}}} - {{{r^3}} \over {{R^3}}}$$
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