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
人气：307 ℃ 时间：2020-04-20 05:07:07
解答
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
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