0
You visited us 0 times! Enjoying our articles? Unlock Full Access!
Question

# A large number of identical point masses each of mass m are placed along x-axis, at x=0,1,2,4,... The magnitude of gravitational force on the mass at origin (x=0), will be

A
Gm2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
43Gm2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
23Gm2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
54Gm2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is B 43Gm2Let, F1, F2, F4, F8,.... be the forces of gravitation on mass m placed at x=0 due masses placed at x=1, 2, 4, 8... respectively. The gravitational force between any two point masses is given by F=Gm1m2r2 According to principle of superposition, the net gravitational force on mass m placed at x=0 can be written as →F=→F1+→F2+→F4+→F8... ⇒F=Gm212+Gm222+Gm242+Gm282+... ⇒F=Gm2(11+14+116+164...)...(1) The above expression consists a infinite G.P with common ratio, r=14 So, sum of the G.P, S=a1−r where a is the first term r is the common ratio. ⇒S=11−14=43 From equation (1), ⇒F=F1+F2+F4+F8......=Gm2S ∴F=43Gm2 Hence, option (b) is the correct answer. Why this question: To make students understand that gravitational force follows the principle of superposition " The net force on a particle is equal to the vector sum of all the forces acting on that particle".

Suggest Corrections
1
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program