Q: A satellite circles a planet with a tangential velocity of 1.70 x 104 m/s. The orbital radius ( r1 ) is 5.25 x 106 m. A second satellite of equal mass revolves around the same planet with an orbital radius ( r2 ) of 8.60 x 106 m. What is the orbital speed of the second satellite?
A: Newton’s Third Law of Motion states that forces occur in pairs. The planet exerts a gravitational force of attraction ( Fg = GMm/r2 ) on each satellite. If the satellites were at rest with respect to the planet, the force and acceleration exerted upon them would be linear in nature; however, since the satellites revolve around the planet, the planet’s force of attraction is centripetal ( Fc ). The magnitude of the centripetal forces acting on the satellites is determined by their respective distances from the planet. Additionally, the acceleration of the objects is due to a change in direction of their linear velocity ( Δv ) as opposed to a change in magnitude. As a consequence, expectations regarding each satellite’s velocity can be confirmed with a brief review of angular momentum ( L ) and the moment of inertia of revolving systems.
It is very difficult to sit on a stationary bicycle without having it tip over; conversely, rolling bicycle wheels resist being tilted outside of the vertical plane within which they reside. This resistance to a change in orientation resides in an imaginary moment of inertia ( I ) vector that is perpendicular to the central axis of rotation of the wheels, and its magnitude is determined by each wheel’s mass ( m ) and radius ( r ). Thus, when a moving bicycle’s front wheel is slightly turned, the bicycle resists tilting over sideways, because the wheel’s moment of inertia resists being rotated off its axis:
When a system is isolated from outside forces, its momentum will remain unchanged. The linear momentum ( p = mv ) of an object is a product of its mass and velocity. If such a body is attached to the end of a lever, the lever’s origin will experience a torque if the object moves in a direction that is perpendicular to the lever. Since the motion of the body will now be angular as opposed to linear, the resulting angular momentum is quantified as follows:
L = ( mv)( r )( sin θ )
The lever in question ( r ) need not be physical; anything that restricts a moving object’s motion from a distance has the potential to be a lever, and this includes forces that influence the motion of objects from a distance. The angular momentum of a system is maximized when momentum is imparted on a lever at a 900 angle. Since the sin 900 = 1, the angular momenta of the satellites may be evaluated with the following equation:
L = mvr
The angular velocity of an object acted upon by a centripetal force is v = r⍵. Inserting this value into the equation above yields the following result:
L = ( m )( r⍵ )( r ) = ( mr2 )( ⍵ )
The ( mr2 ) term is the moment of inertia ( I ), and thus:
L = I⍵
The moment of inertia is the rotational equivalent of mass. If the lever arm of the system decreases, the system’s ability to resist a change in motion decreases as well. This is why a spinning figure skater’s angular velocity ( ⍵ ) increases as she pulls her arms inward; as her moment of inertia decreases while her rotational velocity and kinetic energy increase, and the momentum within the system is preserved. For this reason, we can predict that the angular and linear velocity component of a satellite in orbit will lessen as its distance from a planet increases.
Let’s now consider a circumstance in which two satellites of equal mass travel side by side around a planet. If satellite ( a ) moves inward to a smaller orbital radius ( r1 ), its angular and linear velocity will increase. In order for satellite ( b ) located at ( r2 ) to complete revolutions around the planet in the same amount of time, its angular velocity ( ⍵2 ) must increase so that it is equal in magnitude to ( ⍵1 ):
⍵1 = ⍵2
r2 > r1
ac = r⍵2
r2⍵2 > r1⍵2
Furthermore,
v = r⍵
v2 = r2⍵2
v2/r2 = ⍵2
⍵12 = ⍵22
r2⍵2 > r1⍵2
( r2 )( v22/r22 ) > ( r1 )( v12/r12 )
( r2/r22 )( v22 ) > ( r1/r12 )( v12 )
( v22/r2 ) > ( v12/r1 )
v2 > v1
Since v = r⍵
v2 = r2⍵2
v2/r = r⍵2
ac = v2/r
ac2 > ac1
Fc2 > Fc1
In order for satellite ( b ) to have the same angular velocity ( ⍵ = v/r ) as satellite ( a ), where rb > ra , we reach the inevitable conclusion that vb > va must hold true. Additionally, since the orbital circumference of 2πrb > 2πra , and d = vt, satellite ( b ) must travel with a greater linear velocity component to complete a revolution in the same unit of time as satellite ( a ). Furthermore, since F = ma, and acb > aca , the centripetal force ( Fc ) exerted upon satellite ( b ) by the planet would have to be greater than the centripetal force exerted upon satellite ( a ) to establish the same angular speed from a greater distance; however, there is only one planet within our system! A force exerted by a planet upon two satellites with different radial orbits will cause the outermost satellite to accelerate centripetally with less magnitude than the innermost satellite. Thus, if ac1 > ac2, and r1 < r2, there is an inverse relationship between the centripetal force exerted upon each satellite and their relative orbital distances from the planet:
F 1/∝ r
Additionally, if Fc1 > Fc2 , the angular velocity ( ⍵ ) of satellite ( b ) will be less than that of satellite ( a ):
Fc = mac = mr⍵2
Fc1 > Fc2
m1ac1 > m2ac2
m1 = m2
ac1 > ac2
ac = r⍵2
r1⍵12 > r2⍵22
r1 < r2
⍵12 > ⍵22
⍵1 > ⍵2
Since ac = v2/r
v12/r1 > v22/r2
v12/v22 > r1/r2
v1/v2 > √ r1/r2
v1 > ( √ r1/r2 )( v2 )
r1 < r2
v1 > v2
The linear velocity component of the outermost satellite is less than the linear velocity component of the innermost satellite traveling within the gravitational field of the planet. The planetary system’s forces of interest are the gravitational force of attraction exerted upon each satellite and the consequential centripetal forces exerted upon each satellite expressed in terms of their linear velocity components:
Fg = Gm1m2/r2
Fc = mv2/r
Since two satellites are within the system, some method of comparing the effect of Fg on satellite ( a ) and satellite ( b ) must be devised. If satellite ( a ) is some distance ( r1 ) from the planet, and satellite ( b ) is a distance ( r2 ) from the planet, the centripetal forces experienced by each satellite are as follows:
F1 = m1v12/r1
F2 = m2v22/r2
The value of F2 is some fraction of F1, and this relationship is proportional to some ratio of the radii of the satellites:
F 1/∝ r
F1/F2 ∝ r2/r1
Since mv2/r = GMm/r2 applies to both satellites, we solve the equation for ( r ) and form a ratio between r2 and r1:
mv2/r = GMm/r2
mv2r = GMm
r = GM/v2
r2/r1 = [ ( GM/v22 )/( GM/v12 ) ]
r2/r1 = v12/v22
√ r2/r1 = v1/v2
( v2 )( √ r2/r1 ) = v1
If r2 > r1, then v1 > v2
We now have all of the information needed to solve the problem:
v1 = 1.70 x 104 m/s
v2 = ?
r1 = 5.25 x 106 m
r2 = 8.60 x 106 m
v2 = [ ( v1 )/( √ r2/r1 ) ]
v2 = ( 1.70 x 104 m/s ) / [ √ ( 8.60 x 106 m / 5.25 x 106 m ) ]
v2 = ( 1.70 x 104 m/s ) / ( √ 1.64 )
v2 = ( 1.70 x 104 m/s ) / ( 1.28 )
v2 = 1.33 x 104 m/s
RENAISSANCE PHYSICS 