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

Step-by-step 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=(GM_1M_2)/(r^2)`

Force
`F=mg'=(GMm)/(4R^2)-----1`

Force on earth surface
`F=mg=(GMm)/(R^2)------2`

Divide 1 and 2 we get

`(g')/(g)=(R^2)/(4R^2)`

`g'=(g)/(4)`