First slide

Gravitational force and laws

Question

Particles of masses 2M, m and M are respectively at points A, Band C with $AB = \frac{1}{2} (BC)$ . m is much-much smaller than M and, at time t = 0, they are all at rest as given in figure. At subsequent times before any collision takes place.

Moderate

A

m will remain at rest
B

m will move towards M
C

m will move towards 2M
D

m will have oscillatory motion

Solution

Force on B due to $A = F_{BA} = \frac{G (2 Mm)}{(AB)^{2}}$ towards BA
Force on B due to
$C = F_{BC} = \frac{GMm}{(BC)^{2}}$ towards BC
As (BC) = 2AB
$\Rightarrow F_{BC} = \frac{GMm}{(2 AB)^{2}} = \frac{GMm}{4 (AB)^{2}} < F_{BA}$
Hence, m will move towards BA (i.e., 2M)

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