CBSE Class 11-science Answered
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![question image](https://images.topperlearning.com/topper/new-ate/topr_3032981437219373Screenshot20220927150543.jpeg)
Asked by ankurgoyal616 | 27 Sep, 2022, 15:07: PM
![](https://images.topperlearning.com/topper/tinymce/imagemanager/files/73487d1be3bd311c739f39b786f4507363331491b5edf7.04312991f4.png)
Gravitational attraction force between the two masses provides the necessary centripetal force to go in a circular path of radius R .
Gravitational force of attraction = ![begin mathsize 14px style G space cross times m squared over left parenthesis 2 R right parenthesis squared space equals space G space cross times fraction numerator m squared over denominator 4 space R squared end fraction end style](https://images.topperlearning.com/topper/tinymce/cache/aa648f29e697fa7fd6edee19486e4ae4.png)
![begin mathsize 14px style G space cross times m squared over left parenthesis 2 R right parenthesis squared space equals space G space cross times fraction numerator m squared over denominator 4 space R squared end fraction end style](https://images.topperlearning.com/topper/tinymce/cache/aa648f29e697fa7fd6edee19486e4ae4.png)
where G is gravitation universal constant and m is mass of each particle.
Centripetal force = m ( v2 / R )
![begin mathsize 14px style m space cross times v squared over R space equals space G space cross times fraction numerator m squared over denominator 4 R squared end fraction end style](https://images.topperlearning.com/topper/tinymce/cache/e4ecb4441c43bd91e25385578a46c68a.png)
From above expression we get
![begin mathsize 14px style v space equals space 1 half square root of fraction numerator G space m over denominator R end fraction end root end style](https://images.topperlearning.com/topper/tinymce/cache/14f28445688ba9f8dad6635bd2cf282e.png)
Answered by Thiyagarajan K | 27 Sep, 2022, 20:58: PM
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