"Men often
become what they believe themselves to be. If I believe I cannot do something,
it makes me incapable of doing it. But when I believe I can, then I acquire the
ability to do it even if I didn't have it in the beginning."

– Mahatma Gandhi

Today we shall discuss a few multiple choice questions involving gravitation. The first question pertains to a simple binary star system.

In a simple binary star system two stars orbit around
their centre of mass under their mutual gravitational attractive force. The two
stars constituting the binary star system need not be of the same mass. Note
that Kepler’s laws are applicable to binary stars also. Click to see this.

Star systems with more than two stars (multiple
star systems) are also present. About half of the stars in our galaxy, the
milky way, are part of binary star systems or multiple star systems. Here is
the first question:

(1) Consider a simple binary star system (Fig.)
in which the member stars are identical, orbiting in a circle of radius

*r*. If each star has mass*m*, the orbital speed of each star is
(a) √(

*Gm/r*)
(b) √(

*Gm/*2*r*)
(c) √(

*Gm/*4*r*)
(d) √(2

*Gm/r*)
(e) √(4

*Gm/r*)
The gravitational force between the stars
supplies the centripetal force required for the circular motion. Therefore we
have

*Gmm*/(2

*r*)

^{2}=

*mv*

^{2}

*/r*where

*G*is the gravitational constant and

*v*is the orbital speed.

The above equation gives

*v =*√(

*Gm/*4

*r*)

The following question is related to a simple
multiple star system:

(2) Consider a multiple star system containing

*four*stars of the same mass*m*arranged symmetrically and orbiting around their centre of mass C (Fig.) in the clockwise direction in a circle of radius*r*. What is the orbital speed of each star?
(a) 4[(

*Gm/r*)(1+2√2)]^{1/2}
(b) 2[(

*Gm/*2*r*)(1+2√2)]^{1/2}
(c) [(

*Gm/r*)(1+2√2)]^{1/2}
(d) (½)[(

*Gm/r*)(1 + 2√2)]^{1/2}
(e) (¼)[(

*Gm/r*)(1+2√2)]^{1/2}
The gravitational force F

_{1}on a star due to the diametrically opposite star is given by*F*

_{1}=

*Gmm/*(

*2r*)

^{2}=

*Gm*

^{2}

*/*4

*r*

^{2}

[Note that the distance between the diametrically
opposite stars is 2

*r*]
Evidently

*F*_{1}is directed towards the centre of mass C of the star system.
The other two stars exert gravitational forces

*F*_{2}*and**F*_{3 }as shown in the figure. These forces are directed at 45º with respect to*F*_{1}and they have the*same*magnitude given by*F*

_{2}

*=*

*F*

_{3}=

*Gmm/*(

*r*√2)

^{2}=

*Gm*

^{2}/2

*r*

^{2}

[Note that the distance between adjacent stars is
(√2)

*r*]*F*

_{2}

*and*

*F*

_{3}have equal components (

*Gm*

^{2}/2

*r*

^{2})cos 45º directed towards the centre of mass C of the star system. The net

*radial*force acting on a star is therefore equal to

*F*

_{1 }+

*F*

_{2}cos 45º +

*F*

_{3}cos 45º =

*Gm*

^{2}

*/*4

*r*

^{2 }+

*(*

*Gm*

^{2}/2

*r*

^{2})cos 45º +

*(*

*Gm*

^{2}/2

*r*

^{2})cos 45º.

The radial force thus works out to (

*Gm*^{2}*/r*^{2})[(1/4) + (1/√2)] = (*Gm*^{2}*/r*^{2})[(1 + 2√2)/4]
Since
the above radial force supplies the centripetal force required for circular
motion, we have

*mv*

^{2}/

*r*=

^{ }(

*Gm*

^{2}

*/r*

^{2})[(1 + 2√2)/4]

This gives the orbital speed,

*v =*(½)[(*Gm/r*)(1 + 2√2)]^{1/2}, as given in option (d).
(3) A particle of mass 4

*m*is kept fixed at point P (Fig.) in the xy-plane. Another*free*particle of mass*m*at the origin is found to be unperturbed when a third*fixed*particle of mass 6*m*also is present in the xy-plane. If the x-coordinate and y-coordinate of the particle of mass 4*m*are –√2 and +√2 respectively, what are the x-coordinate and y-coordinate respectively of the particle of mass 6*m*?
(a) +√3 and –√3

(b) –√3 and +√3

(c) –√6 and +√6

(d) +√2 and –√2

(e) +3 and –3

The free particle of mass

*m*at the origin is unperturbed since the gravitational forces on it due to the fixed particles are equal in magnitude and opposite in direction. This can happen only if the three particles are on the same straight line, with the free particle of mass*m*in between the fixed particles.
Equating the magnitudes of the gravitational
forces, we have

*G*(4

*m*)

*m*/2

^{2}=

^{ }

*G*(6

*m*)

*m*/r

^{2}where

*G*is the constant of gravitation and

*r*is the distance of the particle of mass 6

*m*from the origin.

[Note that the distance of the particle of mass 6

*m*from thr origin is {(√2)^{2}+ (√2)^{2}}^{1/2 }= 2
The above equation gives

*Gm*

^{2 }= 6

*Gm*

^{2}/

*r*

^{2}from which

*r =*√6

Since the line joining the particles evidently
bisects the angle XOY’ the x-coordinate and y-coordinate of the particle of
mass 6

*m*must be of equal value*d*such that*√(*

*d*

^{2}+

*d*

^{2}) = √6

Therefore

*d =*√3
Since the
particle of mass 6

*m*is located in the 4^{th}quadrant, its x-coordinate and y-coordinate are +√3 and –√3 respectively [Option (a)].
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