题目内容
地球半径为R,在距球心r处(r>R)有一同步卫星. 另有一半径为2R的星球A,在距球心3r处也有一同步卫星,它的周期是72h,那么A星球平均密度与地球平均密度的比值为
- A.1∶9
- B.3∶8
- C.27∶8
- D.1∶8
B
分析:根据万有引力提供向心力G
=mr(
)2,求出天体的质量,再求出密度,看与什么因素有关.
解答:解:万有引力提供向心力G
=mr(
)2,M=
.密度ρ=
=
.
因为地球的同步卫星和星球A的同步卫星的轨道半径比为1:3,地球和星球A的半径比为1:2,两同步卫星的周期比1:3.所以地球和A星球的密度比为8:3.故B正确,A、C、D错误.
故选B.
点评:解决本题的关键掌握万有引力提供向心力G
=mr(
)2.
分析:根据万有引力提供向心力G
![](http://thumb.1010pic.com/pic5/doc2html/Upload/2013-01/22/8f067c68-a6c3-4bbf-b44c-e5c2c14451d2/paper.files/image001.png)
![](http://thumb.1010pic.com/pic5/doc2html/Upload/2013-01/22/8f067c68-a6c3-4bbf-b44c-e5c2c14451d2/paper.files/image002.png)
解答:解:万有引力提供向心力G
![](http://thumb.1010pic.com/pic5/doc2html/Upload/2013-01/22/8f067c68-a6c3-4bbf-b44c-e5c2c14451d2/paper.files/image001.png)
![](http://thumb.1010pic.com/pic5/doc2html/Upload/2013-01/22/8f067c68-a6c3-4bbf-b44c-e5c2c14451d2/paper.files/image002.png)
![](http://thumb.1010pic.com/pic5/doc2html/Upload/2013-01/22/8f067c68-a6c3-4bbf-b44c-e5c2c14451d2/paper.files/image003.png)
![](http://thumb.1010pic.com/pic5/doc2html/Upload/2013-01/22/8f067c68-a6c3-4bbf-b44c-e5c2c14451d2/paper.files/image004.png)
![](http://thumb.1010pic.com/pic5/doc2html/Upload/2013-01/22/8f067c68-a6c3-4bbf-b44c-e5c2c14451d2/paper.files/image005.png)
因为地球的同步卫星和星球A的同步卫星的轨道半径比为1:3,地球和星球A的半径比为1:2,两同步卫星的周期比1:3.所以地球和A星球的密度比为8:3.故B正确,A、C、D错误.
故选B.
点评:解决本题的关键掌握万有引力提供向心力G
![](http://thumb.1010pic.com/pic5/doc2html/Upload/2013-01/22/8f067c68-a6c3-4bbf-b44c-e5c2c14451d2/paper.files/image001.png)
![](http://thumb.1010pic.com/pic5/doc2html/Upload/2013-01/22/8f067c68-a6c3-4bbf-b44c-e5c2c14451d2/paper.files/image002.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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