Derive Gravitational Force and Field between two identical rods

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The discussion explores the derivation of gravitational force and field between two identical rods by treating them as a collection of small elemental masses. Participants suggest integrating the forces exerted by these elemental masses to calculate the overall gravitational interaction. The approach emphasizes the importance of considering the distribution of mass along the rods for accurate results. This method could provide a clearer understanding of the gravitational effects in such configurations. Overall, the integration of elemental masses is seen as a viable strategy for this problem.
RIJU101
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Homework Statement
What is the Gravitational force and field between two identical rods(radius of cross-section is R) of uniform density, which are placed parallel to each other? Derive the formula for two cases:

1.When two rods are finite in length
2.When two rods are of infinite length
Relevant Equations
Gravitational Force: F=GMm/r^2
Gravitational Field: I=GM/r^2
Maybe it can be done by considering both rods as many small elemental masses then deriving the force we can integrate it
 
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RIJU101 said:
Maybe it can be done by considering both rods as many small elemental masses then deriving the force we can integrate it
So try that and post your attempt.
 
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Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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