Question 145: Gravitation

Question

If R
E
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 < R
E
)

Accepted Answer

$${g_{up}} = {g \over {{{\left( {1 + {r \over R}} \right)}^2}}}$$
$${g_{down}} = g\left( {1 - {r \over R}} \right)$$
$${{{g_{down}}} \over {{g_{up}}}} = \left( {1 - {r \over R}} \right){\left( {1 + {r \over R}} \right)^2}$$
$$ = \left( {1 - {r \over R}} \right)\left( {1 + {{2r} \over R} + {{{r^2}} \over {{R^2}}}} \right)$$
$$ = 1 + {r \over R} - {{{r^2}} \over {{R^2}}} - {{{r^3}} \over {{R^3}}}$$

Answer

$$1 - {r \over {{R_E}}} - {{{r^2}} \over {R_E^2}} - {{{r^3}} \over {R_E^3}}$$

Answer

$$1 + {r \over {{R_E}}} + {{{r^2}} \over {R_E^2}} + {{{r^3}} \over {R_E^3}}$$

Answer

$$1 + {r \over {{R_E}}} - {{{r^2}} \over {R_E^2}} + {{{r^3}} \over {R_E^3}}$$

Answer

$$1 + {r \over {{R_E}}} - {{{r^2}} \over {R_E^2}} - {{{r^3}} \over {R_E^3}}$$

Gravitation

If R E 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 < R E )

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