Question

Four identical particles of mass $$M$$ are located at the corners of a square of side $$'a'$$. What should be their speed if each of them revolves under the influence of other's gravitational field in a circular orbit circumscribing the square?

Answer 1

Correct option is C. $$1.16 \sqrt {\dfrac {GM}{a}}$$

$$\textbf{Step1:Draw diagram and find radius of circular orbit}$$

$$(Hypotaneous)^2=a^2+a^2$$

$$Hypotaneous=\sqrt 2$$

$$Diameter=\sqrt 2$$

$$ \text{Radius} =\dfrac{\text{Diameter}}{\sqrt 2}$$

$$\textbf{Step2:Net force on particle towarda centre of circle}$$

Both $$F=\dfrac{GM^2}{a^2}$$ are at $$90^0$$ so their resultant is $$\dfrac{\sqrt 2GM^2}{a^2}$$ and act at angle bisector.

Net force on particle towards centre of circle is $$F_{C} = \dfrac {GM^{2}}{2a^{2}} + \dfrac {GM^{2}}{a^{2}} \sqrt {2}$$
$$= \dfrac {GM^{2}}{a^{2}} \left (\dfrac {1}{2} + \sqrt {2}\right )$$

$$\textbf{Step3:Calculation of speed}$$
This resultant force will act as centripetal force. Distance of particle from centre of circle is $$\dfrac {a}{\sqrt {2}}$$.
$$r = \dfrac {a}{\sqrt {2}}, F_{C} = \dfrac {mv^{2}}{r}$$
$$\dfrac {mv^{2}}{\dfrac {a}{\sqrt {2}}} = \dfrac {GM^{2}}{a^{2}} \left (\dfrac {1}{2} + \sqrt {2}\right )$$
$$v^{2} = \dfrac {GM}{a} \left (\dfrac {1}{2\sqrt {2}} + 1\right )$$
$$v^{2} = \dfrac {GM}{a} (1.35)$$
$$v = 1.16\sqrt {\dfrac {GM}{a}}$$.