题目内容
同一遥感卫星离地面越近时,获取图象的分辨率也就越高.则当图象的分辨率越高时,卫星的( )A.向心加速度越小
B.角速度越小
C.线速度越小
D.周期越小
【答案】分析:分辨率越高,可知r越小,根据万有引力提供向心力,找出向心加速度、角速度、线速度、周期与轨道半径的关系.
解答:解:分辨率越高,可知r越小,根据万有引力提供向心力,G
=ma=mrω2=m
=mr
,得:a=
,v=
,ω=
,T=2π
.可知r越小,向心加速度越大,线速度越大,角速度越大,周期越小.故A、B、C错,D对.
故选D.
点评:解决本题的关键是熟练掌握根据万有引力提供向心力,G
=ma=mrω2=m
=mr
.
解答:解:分辨率越高,可知r越小,根据万有引力提供向心力,G
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028201620548267726/SYS201310282016205482677013_DA/0.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028201620548267726/SYS201310282016205482677013_DA/1.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028201620548267726/SYS201310282016205482677013_DA/2.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028201620548267726/SYS201310282016205482677013_DA/3.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028201620548267726/SYS201310282016205482677013_DA/4.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028201620548267726/SYS201310282016205482677013_DA/5.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028201620548267726/SYS201310282016205482677013_DA/6.png)
故选D.
点评:解决本题的关键是熟练掌握根据万有引力提供向心力,G
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028201620548267726/SYS201310282016205482677013_DA/7.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028201620548267726/SYS201310282016205482677013_DA/8.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028201620548267726/SYS201310282016205482677013_DA/9.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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