(a) Assuming the earth to be a sphere of uniform density, calculate the value of acceleration due to gravity at a point (i) $1600$ km above the earth, (ii) $1600$ km below the earth, (b) Also find the rate of variation of acceleration due to gravity above and below the earth's surface. Radius of earth $= 6400$ km, $g = 9.8 m / s^{2}$ .

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Solution

A) (i) $g$ at altitude $h$
$g^{h} = g {[\frac{R}{R + h}]}^{2} = 9.8 {[\frac{6400}{6400 + 1600}]}^{2} = 6.272 {m / s}^{2}$
(ii) $g$ at depth $h$
$g^{d} = g [1 - \frac{d}{r}] = 9.8 [1 - \frac{1600}{6400}] = 7.35 m {/ s}^{2}$
B) Ratio of variation of $^{'} g^{'}$
i) $\Rightarrow \frac{g - g_{h}}{g} \times 100 \Rightarrow 36$
ii) $\Rightarrow \frac{g - g_{d}}{g} \times 100 \Rightarrow 25$

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A) (i)g at altitude hgh=g[RR+h]2=9.8[64006400+1600]2=6.272m/s2(ii)g at depth hgd=g[1−dr]=9.8[1−16006400]=7.35m/s2B) Ratio of variation of ′g′i)⇒g−ghg×100⇒36ii)⇒g−gdg×100⇒25