ROTATIONAL MOTION: What Distance Separates Two Projectiles Revolving Around a Common Center of Mass?

Q: Two projectiles separated by distance ( d_t ) revolve around their center of mass ( cm = ½ d_t ). Each projectile has a mass ( m ) of 4.81 x 10²⁰ kg, and they have an instantaneous rotational speed ( ⍵ ) of 1.25 x 10^-10 rad/s relative to the center of mass. What is the value of the distance ( d_t ) that separates the projectiles?

A: The center of mass ( cm ) is a system coordinate that identifies where the average mass within a system is located. Consider a system on the earth’s surface within which two equivalent masses ( m1 = m2 ) are connected via a horizontal lever of length ( d_t ) at rest atop a pivot. The gravitational force of attraction pulls each mass downward, and thus, the center-of-mass location is above the pivot:

The center-of-mass equation pertinent to this system is as follows:

cm = [ ( m₁d₁ + m₂d₂ )/( m₁ + m₂ ) ]

As long as the lever’s length remains constant, the center of mass will maintain its location. The numerical value of a ( cm ) measurement is dependent on the vantage point from which an observer makes measurements. In the system below, two projectiles of equal mass rotate in a counterclockwise direction around ( cm = d₁ ):

An observer located on ( m₁ ) will derive the following center-of-mass value:

cm = [ ( m₁ )( d_o= 0 ) + ( m₂ )( d₁ + d₂ ) ]/( m₁ + m₂ )

cm = [ 0 + ( m₂ )( d₁ + d₂ ) ]/( m₁ + m₂ )

d₁ = d₂

m₁ = m₂

cm = [ ( m )( 2d₁ ) ]/( 2m )

cm = d₁                                                              

In the diagram that follows, measurements are made using ( cm = d_o = 0 ) as the point of reference. The x-axis is now divided into ( + x ) and ( – x ) values, with ( d₂ = + x ) and ( – d₁ = – x ), respectively. If the counterclockwise direction is designated as being positive, the forces acting on lever arms ( – d₁ ) and ( d₂ ) have values of +F₁ and +F₂:

As a consequence, the following value of ( cm ) is derived:

F_t = F₁ + F₂

( m₁ + m₂ )( a ) = m₁a + m₂a

The forces acting upon the levers extending from ( cm ) exert torques ( 𝛕 ) on the system. With lever measurements taken into account, the torques acting upon the system may be evaluated in the following manner:

𝛕 = Fdsinθ

 θ = 90⁰

𝛕 = Fd

F = ma

𝛕 = ( ma )( d )

( m₁ + m₂ )( a )( d_o ) = ( m₁a )( – d₁ ) + ( m₂a )( d₂ )

( m₁ + m₂ )( d_o ) = ( m₁ )( – d₁ ) + ( m₂ )( d₂ )

( m₁ + m₂ )( d_o ) = ( m₂ )( d₂ ) – ( m₁ )( d₁ )

d_o = [ ( m₂ )( d₂ ) – ( m₁ )( d₁ ) ]/( m₁ + m₂ )

d_o = cm

cm = [ ( m₂ )( d₂ ) – ( m₁ )( d₁ ) ]/( m₁ + m₂ )

m₁ = m₂

 d₁ = d₂

cm = 0

0 = [ ( m₂ )( d₂ ) – ( m₁ )( d₁ ) ]/( m₁ + m₂ )

0 = ( m₂ )( d₂ ) – ( m₁ )( d₁ )

( m₂ )( d₂ ) = ( m₁ )( d₁ )

The equation above has two unknown variables. In these circumstances, another equation consisting of the unknown values must be used to further simplify the problem:

d_t = d₁ + d₂

We now have an equation containing two unknown values, and we have a second equation containing three unknown variables. We must use substitution to simplify the latter equation. The force equations pertinent to this system will allow us to solve d₁ ( or d₂ ), so we simplify the equation above in the following manner:

d_t = d₁ + d₂

d_t = d₁ + ( m₁/m₂ )( d₁ )

d_t = ( d₁ )[ 1 + ( m₁/m₂ ) ]

d_t = ( d₁ )( m₂ + m₁ )/m₂ 

( d_t )[ m₂/( m₂ + m₁ ) ]  = d₁

The force that enables the projectiles to revolve around their center of mass is both gravitational and centripetal in nature; thus: 

F_g = F_c

Gm₁m₂/d² = m₁⍵²d₁ 

Notice that the value of ( d ) in F_g ≠ ( d₁ ) in F_c! Although the projectiles revolve around their center of mass, the total distance that separates them must be used in F_g, while the radial distance ( r₁ ) from ( cm ) to ( m₁ ) is used in F_c. The angular force ( F_c ) acting on the satellites is derived in the following manner:

v = r₁⍵

v² = r₁²⍵²

( v²/r₁ ) = r₁⍵²

F_c = mv²/r₁ = mr₁⍵²

r₁ = d₁

Gm₁m₂/d² = m₁⍵²d₁ 

We now have two equations with two unknowns. Since the value of ( ⍵ ) is known, we solve the equation for ( ⍵ ). We subsequently substitute ( d₁ ) with an equivalent value expressed in terms of ( d ):

Gm₁m₂/d²m₁d₁ = ⍵²

d₁ = ( d )[ m₂/( m₂ + m₁ ) ]

( Gm₁m₂ )/[ ( d²m₁m₂d )/( m₂ + m₁ ) ] = ⍵²

( G )/[ ( d³/( m₂ + m₁ ) ] = ⍵²

[ ( G )( m₂ + m₁ ) ]/d³ = ⍵²

d³ = [ ( G )( m₂ + m₁ ) ]/⍵²

m₂ = m₁

d³ = [ ( G )( 2m ) ]/⍵²

Simple substitution may now be used to determine the value of ( d ):

d³ = [ ( 6.673 x 10^-11 Nm²/kg² )( 2 )( 4.81 x 10²⁰ kg ) ]/( 1.25 x 10^-10 rad/s )²

d³ = ( 6.419426 x 10¹⁰ Nm²/kg )/( 1.5625 x 10^-20 rad²/s² )

d³ = 4.10843264 x 10³⁰ N*m²s²/kg*rad² 

d³ = 4.10843264 x 10³⁰ m³/rad²

d = 1.60 x 10¹⁰ m/rad^2/3