Question

Four particle each of mass M and equidistant from each other , moves along a Circle of radius R under the action of their mutual gravitational attraction , the speed of each particle is [JEE main 2014] ​

Answer 1

ANSWER

Net force on any one particle

(2R)  

2

 

GM  

2

 

​  

+  

(R  

2

​  

)  

2

 

GM  

2

 

​  

cos45  

o

+  

(R  

2

​  

)  

2

 

GM  

2

 

​  

cos45  

o

 

=  

R  

2

 

GM  

2

 

​  

[  

4

1

​  

+  

2

​  

 

1

​  

]

This force will be equal to centripetal force so

R

Mu  

2

 

​  

=  

R  

2

 

GM  

2

 

​  

[  

4

1+2  

2

​  

 

​  

]

u=  

4R

GM

​  

[1+2  

2

​  

]

​  

=  

2

1

​  

 

R

GM

​  

(2  

2

​  

+1)

​

Answer 2

• Explanation:- Net force acting on any one particle M

• GM²/(2R)² + GM² /(R√2)²cos45+GM /(R√2)²cos45°
• ⟹ GM²/4R² + 2 GM²/2R* 1/√2
• ⟹ GM²/R²(1/4+1/√2)

This force will equal to centripetal force

So, Mu²/R = GM²/R ( 1/4+ 1/√2)

$⟹u \: = \sqrt{ \frac{GM {}^{2} }{4R {}^{2} \:(1 + 2 \sqrt{2)} } } \: \\ \\ = \frac{1}{2} \sqrt{ \frac{GM}{R}(2 \sqrt{2} + 1) }$

Hence, speed of each particle in a circular motion is

$= \frac{1}{2} \sqrt{ \frac{GM}{R} (2 \sqrt{2} + 1 }$ Answer