Calculate the acceleration due to gravity at a point
a] $64 k m$ above and
b] $32 k m$ below the surface of earth.
Given Radius of Earth $= 6400 k m$ . Acceleration due to gravity at the surface of earth $= 9.8 m s^{- 2}$ .

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Solution

$g_{h} = \frac{g_{s}}{{(1 + \frac{h}{R})}^{2}}$
a) $g_{h} = \frac{9.8}{{(1 + \frac{64}{6400})}^{2}}$
$= \frac{9.8}{{(1 + \frac{1}{100})}^{2}}$
$g_{h} = 9.607 m / s^{2}$
b) $g_{d} = g_{s} (1 - \frac{d}{R})$
$= 9.8 (1 - \frac{d}{R})$
$= 9.8 (\frac{199}{200})$
$g_{d} = 9.751 m / s^{2}$

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Calculate the acceleration due to gravity at a point a] 64 km above and b] 32 km below the surface of earth.Given Radius of Earth =6400 km. Acceleration due to gravity at the surface of earth =9.8 ms−2.

gh=gs(1+hR)2a) gh=9.8(1+646400)2 =9.8(1+1100)2 gh=9.607m/s2b) gd=gs(1−dR) =9.8(1−dR) =9.8(199200) gd=9.751m/s2

Calculate the acceleration due to gravity at a point
a] $64 k m$ above and
b] $32 k m$ below the surface of earth.
Given Radius of Earth $= 6400 k m$ . Acceleration due to gravity at the surface of earth $= 9.8 m s^{- 2}$ .

$g_{h} = \frac{g_{s}}{{(1 + \frac{h}{R})}^{2}}$
a) $g_{h} = \frac{9.8}{{(1 + \frac{64}{6400})}^{2}}$
$= \frac{9.8}{{(1 + \frac{1}{100})}^{2}}$
$g_{h} = 9.607 m / s^{2}$
b) $g_{d} = g_{s} (1 - \frac{d}{R})$
$= 9.8 (1 - \frac{d}{R})$
$= 9.8 (\frac{199}{200})$
$g_{d} = 9.751 m / s^{2}$