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Gravitational due to an infinitely long thin rod of linear density P from a point of mass M and perpendicular distance from rod to point is L
Gravitational due to an infinitely long thin rod of linear density P from a point of mass M and perpendicular distance from rod to point is L


3 years ago

Sai Ram Charan
31 Points
							Data provided is inconsistent; anyway, we have the following data-Length of rod=1/0 , magnitude of point mass= ‘m’​, distance of seperation= lTherefore, from Newton’s Law of universal Gravitation, we have F= $-G{m_{1}m_{2} \over {\vert \mathbf {r} _{12}\vert }^{2}}\,\mathbf {\hat {r}} _{12}}$(in vector calculuss form)Clearly, the force is  $\frac{G P \frac{1}{0} m}{l^{2}}$

3 years ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions