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Four particles, each of mass M and equidistant "from each other, move along a circle of radius R under the action of the mutual gravitational attraction. The speed of each particle is ......... (G = universal gravitational constant)

Answer» <html><body><p>`sqrt((<a href="https://interviewquestions.tuteehub.com/tag/gm-1008640" style="font-weight:bold;" target="_blank" title="Click to know more about GM">GM</a>)/R (1+2sqrt2))` <br/>`1/2sqrt((GM)/R (1+2sqrt2))` <br/>`sqrt((GM)/R)` <br/>`sqrt(2sqrt2 (GM)/R)`</p>Solution :`implies` The force of attraction due to to <a href="https://interviewquestions.tuteehub.com/tag/particles-1147533" style="font-weight:bold;" target="_blank" title="Click to know more about PARTICLES">PARTICLES</a> B and D on <a href="https://interviewquestions.tuteehub.com/tag/particle-1147478" style="font-weight:bold;" target="_blank" title="Click to know more about PARTICLE">PARTICLE</a> A is <br/> `F_1 = F_2 = (GM^2)/(2R^2)` <br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/KPK_AIO_PHY_XI_P1_C08_E05_058_S01.png" width="80%"/> <br/>and force attraction exerted by particle at C on particle at A <br/> `F_3 = (GM^2)/(4R^2)` <br/> `:.` The <a href="https://interviewquestions.tuteehub.com/tag/total-711110" style="font-weight:bold;" target="_blank" title="Click to know more about TOTAL">TOTAL</a> force on particle A <br/> `F= F_1 + F_2 + F_3` <br/> but resultant force of `F_1 +F_2` <br/> `F.=sqrt(F_1^2 +F_2^2+2F_1^2 cos 90^@) `<br/> `F. = sqrt(F_1^2 +F_1^2) = sqrt(2F_1^2)` <br/> `:. F. = sqrt2F_1` <br/> `:. F. = sqrt2 F_1 +F_3` <br/> `=sqrt2 (GM^2)/(2R^2) + (GM^2)/(4R^2)`<br/> `:. (Mv^2)/R = (GM^2)/(2R^2) [sqrt2 +1/2]` <br/>`v^2 =(GM)/(2R) [(2sqrt2 +1)/2]` <br/> `:. v = 1/2 sqrt((GM)/R (1+ 2sqrt2)`</body></html>


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