A uniform ring of mass m is lying at a distance $$3a$$ form the centre of a sphere of mass M just over the sphere (where$$ a is the radius of the ring as well as that of the $$sphere). Then magnitude of gravitational force between them is

Question

A uniform ring of mass m is lying at a distance $$3a$$ form the centre of a sphere of mass M just over the sphere (where$$ a is the radius of the ring as well as that of the $$sphere). Then magnitude of gravitational force between them is

Infinity Learn · Accepted Answer

For a small mass dm, distance from the center of sphere is $$2a$$. force along ‘dm’  G,M,$$dm4a2=FComponent$$ of force F $$sin$$   get cancelled net force $$2Fcos$$ θ . For the entire $$ring-$$ $$GM4a2m2$$×$$2cos$$θ$$GMm4a23a2a=3GMm8a2Alternative$$ solution:$$-field$$ intensity due the ring at the centre of the sphere is $$g=GMx(R2+x2)32=GMa(a2+3a2)32=3GM8a2$$ $$F=mg=3GMm8a2$$

A uniform ring of mass m is lying at a distance $\sqrt{3}$ a form the centre of a sphere of mass M just over the sphere (where a is the radius of the ring as well as that of the sphere). Then magnitude of gravitational force between them is

Unlock the full solution & master the concept.

Ready to Test Your Skills?

free study material

A uniform ring of mass m is lying at a distance 3a form the centre of a sphere of mass M just over the sphere (where a is the radius of the ring as well as that of the sphere). Then magnitude of gravitational force between them is

Unlock the full solution & master the concept.

Ready to Test Your Skills?

free study material

A uniform ring of mass m is lying at a distance $\sqrt{3}$ a form the centre of a sphere of mass M just over the sphere (where a is the radius of the ring as well as that of the sphere). Then magnitude of gravitational force between them is