✔ 最佳答案
Assume Newtonian Gravity apply
Case (1 )- The ring is a rigid body
The orbit should be circular, therefore v = Circumference / Period
V = 2πr /T
v is directly proportional to r
Case (2) – The ring is a group of satellites
Consider the group is a lump mass system of mass = m
F = ma = m dv/dt = - GMm/r^2
a = dv/dt = -GM/r^2
dv/dt = dv/dr․dr/dt = -GM/r^2
dv/dr ․ v = -GM/r^2
v․dv = -GM/r^2 dr
∫v․dv = ∫-GM/r^2 dr
1/2v^2 = GM/r + c
The above equation also indicate the conservation of energy
1/2 mv^2 – GMm/r = constant
v inversely proportional to √r
Therefore the answer is (b)
希望幫到你
2008-11-18 14:43:11 補充：
The orbit of case (2) need "not" be circular
2008-11-18 22:26:33 補充：
The major difference = the effects of gravitational force
(1) motion not affected by gravitational field and become uniform circular motion. The gravitation force on that ring was balanced due to its symmetry about the planet
The centrifugal force F = m a = m v^2/r
2008-11-18 22:28:06 補充：
But how can the acceleration be rewritten in terms of the r and the period, T, i.e. the time required to go around once?
The period can be related to the speed as follows:
v = circumference / T
= 2πr/T
Substituting into the equation a= m v^2/r gives
a = (4π^2r)/(T^2)
2008-11-18 22:31:58 補充：
(2) non-uniform circular motion affected by the gravitational force.
you may think the above equation (1/2 mv^2 – GMm/r = constant) be simplify to
Total Energy = Kinetic Energy + Gravitational Potential Energy = Const
Gravitational force between 2 bodies
F = GMm / r^2
2008-11-18 22:32:27 補充：
and the potential energy of the mass
P = mgh = mgr
Since (mg) also equal to GMm / r^2
Therefore, the Gravitational Potential Energy = r ․GMm / r^2 = GMm/r
Therefore,
Total Energy = Kinetic Energy + Gravitational Potential Energy = Constant
Total Energy = 1/2 mv^2 – GMm/r = constant
2008-11-18 22:33:07 補充：
Therefore, v ^2 is proportional to 1/r
v is proportional to 1/ √r
