Question

The gravitational force acting on a particle, due to a solid sphere of uniform density and radius `R`, at a distance of `3R` from the centre of the sphere is `F_1.` A spherical hole of radius `(R//2)` is now made in the sphere as shown in diagram. The sphere with hole now exerts a force `F_2` on the same particle. ratio of `F_1` to `F_2` is

Answer 1

Let mass of the removed sphere `=M`
Then mass of the original sphere `=8M`(since mass `propR^(3)`)
`F_(1)=(8GMm)/(9R^(2))` and `F_(2)=(8GMm)/(9R^(2))-(GMm)/(((5R)/2)^(2))`
Therefore, `(F_(1))/(F_(2))=50/41` (on simplifying)