g at earths surface=GMe/re^2
1/20 g= GMe/r^2= g(1/(r/re)^2
(r/re )^2= 20
solve for r.
What is the distance in meters from the Earth's center to a point out-side the Earth where the gravitational acceleration due to the Earth is 1/20 of its value at the Earth's surface?
1 answer