Part A Exact solutions for gravitational problems involving more than two bodies

Question

Part A Exact solutions for gravitational problems involving more than two bodies are notoriously difficult. One solvable problem involves a configuration of three equal-mass objects spaced in an equilateral triangle. Forces due to their mutual gravitation cause the configuration to rotate. Suppose three identical stars, each of mass M, form an equilateral triangle of side L. Find the expression for the speed of their orbital motion. Your expression should be in terms of the given variables and any other known constants such as the gravitational constant G. Hint: The center of the orbits of the stars is at the center of the equilateral triangle. Submit Answer Tries 0/6 Part B Find an expression for the period of their orbital motion.D T= Submit Answer Tries 0/6

Explanation / Answer

M = mass of each star
L = distance between two stars
R = orbital radius

Lets first calculate force on one star, by two other other stars.

Accodfing to law of Gravitation, force on one star by other one star.

F = (G*M2 )/L2

it can be infered easily that

Cos 30 =(L/2)/R

L= 2*( R*cos 30 )

So, force

F = ( G *M2) / ( 2*R*cos 30 ) 2

Net force on one star by two stars,

Fn= F*cos30 + F *cos30 = F * 1.73
So :
Fn= [ ( G * M2) / ( 2*R*cos30)2] * 1.73

Accleration,

a = Fn / M

a = [ {(G*M2) / (2*R*cos30)2} * 1.73205 ) /M]

Now, for centrigugal accleration,

a = v2 / R

v = sqrt ( a * R ) )

v = sqrt (((G*M2 /( 2*R*cos30)2)*1.73)/M )*R) ............eq1

Then :

time period
T = distance / time = ( 2 * pi * R ) / v

So the final equation for t in terms of R and m :
T = ( 2 * pi * R ) / ( sqrt (((G*M2 /( 2*R*cos30)2)*1.73)/M )*R) )

The time period,

T = sqrt [ ( 4 *pi2*R ) / ( ( (G*M2 /( 2*R*cos30)2)*1.73)/M )*R)]

put R = L/(2*cos30) and solving

T = sqrt (4*pi2*L3/3*G*M)

by solving eq1

v = sqrt(G*M/L)

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