Question

HOPE THAT SOMEBODY ANSWERS THIS:​

Answer 1

Gh = g / (1+h/r)^2

Gh = g (1+h/r)^(-2)

if n<<<R

Gh = g(1-2h/r)

∆g /g ×100 = -2h/r ×100

h= 32 ×10^2m

h= 32 km

Answer 2

We know that gravity at any height is given as;

$g' = g \huge( \small 1 - \frac{2h}{R} \huge)$

Decrease by 1% implies that g' = 99/100g

Substitute ,we get

$\frac{99}{100 \cancel{g}} = \cancel{g }\huge( \small 1 - \frac{2h}{R} \huge) \\ \\ \frac{99}{100} = 1 - \frac{2h}{R} \\ \\ \frac{2h}{R} = 1 - \frac{99}{100} \\ \\ \frac{2h}{R} = \frac{1}{100} \\ \\ h = \frac{R}{200} \\ \\ h = \frac{64 \cancel{00}}{2 \cancel{00} } \\ \\\underline{\underline{h = \: 32 \: km}}$