Question

An Earth satellite is orbiting at a distance from the Earth's surface equal to one Earth radius (4 000 miles). At this location, the acceleration due to gravity is what factor times the value of g at the Earth's surface?

Answer 1

Answer:0.25 times

Explanation:

Given

Distance of satellite from earth surface=Radius of earth

Force on the satellite is F=mg'

where g'=acceleration due to gravity at that point

Distance from center of Earth=R+R=2R

Gravitational Force is given by

$F=\frac{GM_1M_2}{r^2}$

Force $F=mg'=\frac{GMm}{4R^2}-----1$

Force on earth surface $F=mg=\frac{GMm}{R^2}------2$

Divide 1 and 2 we get

$\frac{g'}{g}=\frac{R^2}{4R^2}$

$g'=\frac{g}{4}$