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Grade 11IIT JEE Entrance Exam

the gravitational force between a point like mass M and an infinitely long,thin rod of linear mass density d at a perpandicular distance L from M is ??

Profile image of prachi agrawal
15 Years agoGrade 11
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3 Answers

Profile image of AKASH GOYAL AskiitiansExpert-IITD
15 Years ago

Dear Prachi

Gravitational field at a distance r from the infinitely long rod with liner mass density d = 2Gd/r

here r=L and force will be field x mass

force= 2GMd/L

 

All the best.                                                           

AKASH GOYAL

AskiitiansExpert-IIT Delhi

 

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Profile image of Sudheesh Singanamalla
15 Years ago

Gravitational Force is -Gm1m2 / r^2

since r is the radius and it is the prependicular distance between the rod and the point of mass M

the mass of the rod is d*V , since a rod is cylindrical the volume is pi*r*r*h and h = infinity

so mass of rod = d*infinity = infinity

Gravitational Force =  - G * infinity * M / L^2

                           =  -  infinity / l^2

since infinity divided by any positive number is also infinity

                           =  - infinity !

 

therefore gravitational force is - infinity

please approve if the answer is correct

Profile image of Shaikh Jinnatul
4 Years ago
There will be no change in the force onM due to another mass ,
Therefore , total mass = F+2F
                                        =3F
Therefore , the forces are F,2F.