题目内容
(10分)如图所示,在距离一质量为M、半径为R、密度均匀的球体R远处有一质量为m的质点。此时M对m的万有引力为F1,当从M中挖去一半径为R/2的球体时,剩余部分对m的万有引力为F2,则F1与F2的比值为多少?![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241620431733806.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241620431733806.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824162043345654.png)
试题分析:若质点与大球球心相距为2R,其万有引力为F1,则有
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241620437351317.png)
大球质量
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824162044047892.png)
小球质量
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241620444841746.png)
小球球心与质点相距
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824162044655496.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241620448271643.png)
剩余部分对质点m的万有引力
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241620450452043.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824162043345654.png)
点评:本题难度中等,解决本题的技巧是利用打补丁法,由于小球为质量分布不均匀的球体,万有引力定律不能适用,当补齐之后r为球心间的距离,根据力的矢量叠加求此问题
![](http://thumb2018.1010pic.com/images/loading.gif)
练习册系列答案
相关题目