Three uniform spheres of mass $m$ and radius $r$ are placed in contact. What is the net force of gravitation on the second sphere?

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

\frac{5 G m^{2}}{16 r^{2}}

\frac{G m^{2}}{16 r^{2}}

\frac{G m^{2}}{r^{2}}

Zero

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

The net force of gravitation on the second sphere is zero.
Universal law of gravity states, “Every particle attracts every other particle in the universe with a force that is directly proportional to the product of the masses and inversely proportional to the square of the distance between them.” Mathematically,

F = \frac{G m_{1} m_{2}}{r^{2}}

where

G

is the universal gravitation constant,

m_{1}

is the mass of the first object,

m_{2}

is the mass of second object and

r

is the distance between them.
It is given that the mass of each sphere is

m

and the radius of each sphere is

r

.
Three spheres of the same mass and radius are spaced apart by the same distance. For the middle sphere, the distance would be the same as the other two spheres, and the force would also be the same, but the direction would be opposite. Therefore, for the second sphere, two opposite directions of the same magnitude force cancel each other out, and the net force becomes zero.

Watch 3-min video & get full concept clarity