Question

2) Prove that the total energy E^/i2 - CNH you derived above is actually half of the time-averaged potential energy of the system, i.e. E 흘 < U >. This is a property of gravitationally bound system in equilibrium, known as the virial theorem, which we will uswe later when discussing the collapse of a cloud Jutjdt. Hint: First find what < U > is over one orbital period, following the formila < f(t) >= You might find this useful as well:

Answer 1

Total energy E = 1/2 mu v^2 - G Mu/v Kinetic energy = 1/2 mv^2. = 1/2 m_1 v^2 + 1/2 m_2 v^2 log (k) = log (mv^2/2) log k = log m + 2 log v -log 2 log k = log (m_1 + m_2) + 2 log v - log 2 dk/k = dm/m_1 + m_2 + 2dv/v dt = k [dm/m_1 + m_2 + 2dv/v] Assume the value s v = 5 m_1 = 64 m_2 = 136 k = 1/2 (m_1 + m_2) v^2 \r\n k = 1/2 (m_1 + m_2) v^2 = 1/2 (64 + 136) (5)^2 = 1/2 (200) (25) 100 (25) k = 2500 J dk = [2500 [3 + 4/64 + 136 + 2 times 1.5/5] dk = [2500 [7/200 + 3/5] dk = 1587.5 I K.E = 2500 plusminus 1587.5.