CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

Two particles of equal mass go round a circle of radius R under the action of their mutual gravitational attraction. Find the speed of each particle. 


Solution

$$\begin{array}{l}f = \dfrac{{Gm.m}}{{{{\left( {2R} \right)}^2}}} = \dfrac{{G{m^2}}}{{\left( {4R} \right)}}\\f = \dfrac{{m{v^2}}}{R}\\\dfrac{{G{m^2}}}{{\left( {4R} \right)}} = \dfrac{{m{v^2}}}{R}\\v = \dfrac{1}{2}\sqrt {\dfrac{{Gm}}{R}} \end{array}$$

Physics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image