The acceleration due to gravity at a height of 6400 km from the surface of the earth is $2.5…
The acceleration due to gravity at a height of 6400 km from the surface of the earth is $2.5 \mathrm{~ms}^{-2}$. The acceleration due to gravity at a height of 12800 km from the surface of the earth is (Radius of the earth $=6400 \mathrm{~km}$ )
$1.11 \mathrm{~ms}^{-2}$
$1.5 \mathrm{~ms}^{-2}$
$2.22 \mathrm{~ms}^{-2}$
$1.25 \mathrm{~ms}^{-2}$
Solution
$\mathrm{g}_1=2.5 \mathrm{~m} / \mathrm{s}^2, \mathrm{~h}_1=6400 \mathrm{~km}, \mathrm{~h}_2=12800 \mathrm{~km}$ The acceleration due to gravity at height $h$ is
$g=\frac{G M}{(R+h)^2} \Rightarrow g \propto \frac{1}{(R+h)^2}$
$\therefore \frac{\mathrm{g}_2}{\mathrm{~g}_1}=\frac{\left(\mathrm{R}+\mathrm{h}_1\right)^2}{\left(\mathrm{R}+\mathrm{h}_2\right)^2}=\frac{(6400+6400)^2}{(6400+12800)^2}=\frac{4}{9}$
$\therefore g_2=\frac{4}{9} g_1=\frac{4}{9} \times 2.5=1.11 \mathrm{~m} / \mathrm{s}^2$