A particle is moving with a uniform speed in a circular orbit of radius r in a central force inversely proportional to the nth power of r. If the period of rotation of the particle is t, then,
Answers
t is proportional to r^[(n+1)/2]
The centripetal force for a circular object of radius 'r' is
F = mv²/r
Also it is supplied in the question that force is inversely proportional to the nth power of r. So,
F ∝ 1/(rⁿ)
Considering the proportionality constant to be k, then,
F = k/(rⁿ)
Or, mv²/r = k/(rⁿ)
⇒ v² = kr/mrⁿ
As mass m is a constant, we consider (k/m) to be a new constant K.
So, v² = K/mrⁿ.r⁻¹ [r = 1/r⁻¹ ]
⇒ v² = K/mrⁿ⁻¹
⇒ v = √k/[mr^(n-1/2)]
⇒ v = k'/[mr^({n-1}/2)] [√k is a constant, representing it by k']
We know, time period for rotation t is given as:
t = 2Пr/v
Substituting the value of 'v' we got earlier, we get:
t = 2Пr[mr^({n-1}/2)]/k'
t = 2Пmr^({n-1}/2)+1)/k' = 2Пmr^({n+1}/2)]/k' [(n-1)/2 + 1 = (n+1)/2]
As 2Пm/k' is a constant,
So, t ∝ r^[(n+1)/2]