what is the distance from earth's surface to a point above its surface where the acceleration due to fravity is 1/8th that of the surface
问题描述：
what is the distance from earth's surface to a point above its surface where the acceleration due to fravity is 1/8th that of the surface
答
在地球表面，GMm/R^2=mg，则g=GM/R^2
高度h处，GMm/(R+h)^2=mg'，则g'=GM/(R+h)^2
g'/g=(R/(R+h))^2=1/8，则(R+h)/R=2*sqrt(2)，则h=(2sqrt(2)-1)R
答
很简单啊,万有引力公式：F=GMm/(R^2),所以地球表面的重力加速度是FM/(R^2),这里M是地球质量,R是地球半径.
设1/8的地方离地心距离是r,重力加速度是FM/(r^2)=1/8*FM/(R^2),所以r=2根号2R,也就是距离地面大约1.8倍地球半径的位置