Question

At what height from the surface of the earth, the
value of 'g' will be reduced by 40% of the value at the
surface ? Radius of earth is 6400 km.
[Hint. 'g' is reduced by 40%, i.e., it is 60% of its value
on the surface of the earth.]​

Answer 1

Answer:

try to solve it by g = GMe/(Re+h)²

Explanation:

then take error..

i think it is nearly to 6400/4

Answer 2

$at \: height \: h \: the \: value \: of \: g \: is \: reduced \: by \: 40\% \\ i.e. \: it \: becomes \: 60\% \: of \: that \: the \: surface \\ then \\ g_{h} = 64\% \: of \: g = \frac{64}{100}g \\ g_{h} = g {( \frac{r}{r + h}) }^{2} \\ \: \: \: \: \frac{64}{100}g = g( { \frac{r}{r + h} )}^{2} \\ \: \: \: \: \: \frac{8}{10} = \frac{r}{r + h} \\ \: \: \: \: 10r = 8r + 8h \\ \: \: \: \: h = \frac{r}{4} \\ \: \: \: \: h = \frac{6400}{4} = 1600km$