Be aware,stars are moving

Classical Mechanics Level 4

6 stars of equal masses m each are moving about the center of mass of the system such that they are always on the vertices of a regular hexagon of the side-length having length a each.Their common time period can be determined in the form of $w \pi \times \sqrt{ \frac {x \sqrt{3} a^{3} }{Gm(y+z \sqrt{3} ) } }$ where w,x,y,z are positive integers.Find w+x+y+z .

The answer is 15.

1 solution

Shashwat Mishra
Jan 23, 2014

Drawing a diagram it is easily seen

That the stars move in a circle of radius $a$
2 stars are at the distance $a$ and each making an angle $60^{\circ}$ with respect to the centre from the single star whose motion is being analysed; 2 stars are at distance $a\sqrt{3}$ subtending $30^{\circ}$ each and one star is directly opposite from star across the centre at distance $2a$ .

Total gravitation force providing the centripetal force to the star is given by

$F=m\omega^2a = GMm\left ( \frac{2cos60^{\circ}}{a^2}+\frac{2cos30^{\circ}}{(a\sqrt{3})^2}+\frac{1}{4a^2}\right ) \rightarrow \frac{GMm}{a^2}\left (\frac{5\sqrt{3}+4}{4\sqrt{3}} \right )$

$\omega= \sqrt{\frac{GM\left(5\sqrt{3}+4 \right )}{a^34\sqrt{3}}}$ $T= 2\pi\setminus\omega= \boxed {2\pi\sqrt{\frac{a^34\sqrt{3}}{GM\left(4+5\sqrt{3} \right )}}}$

There is some ambiguity in this problem. The 4 under the square root can be moved out. So we can have x = 1 or x = 4, and the responses can be 14 or 15.

Abdelhamid Saadi - 7 years, 4 months ago

it seen in the expression that x has been given its place inside the roots....so x is specificaly given for that

Chitres Guria - 7 years, 4 months ago

yes, There is an obvious ambiguity

Pranjit Handique - 7 years, 3 months ago

x can be either 1 or 4 or even 16 ...........

Pranjit Handique - 7 years, 3 months ago

Be aware,stars are moving

The answer is 15.

1 solution

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