Question : Two particles of mass m and M are initially at rest at an infinite distance apart. They move towards each other and gain speeds due to gravitational attraction. Find their speeds when the separation between the masses becomes 'd'.
Doubt by Muskan
Solution :
Consider two cases, First when both the particles of masses m and M are very far apart when they have not even started moving towards each other and the second case is that when the separation between them is 'd'.
As per the law of conservation of energy
Total energy of the system in case I = Total energy of the system in case II
KE1 + PE1 = KE2 + PE2
0 (Because They have not started moving) + 0 (Because They are placed at infinity) = (½mV12+½MV22) + (-GMm/d)
Total energy of the system in case I = Total energy of the system in case II
KE1 + PE1 = KE2 + PE2
0 (Because They have not started moving) + 0 (Because They are placed at infinity) = (½mV12+½MV22) + (-GMm/d)
0+0=(½mV12+½MV22) + (-GMm/d)
0=(½mV12+½MV22) + (-GMm/d)
GMm/d = ½mV12+½MV22
GMm/d = ½[mV12+MV22]
2GMm/d = mV12+MV22 — (1)
Since both the bodies are in an isolated system so by using the law of conservation of energy
Initial Momentum in Case I = Final Momentum in Case II
mV1 = MV2
mV1/M = V2
V2 = mV1/M
0=(½mV12+½MV22) + (-GMm/d)
GMm/d = ½mV12+½MV22
GMm/d = ½[mV12+MV22]
2GMm/d = mV12+MV22 — (1)
Since both the bodies are in an isolated system so by using the law of conservation of energy
Initial Momentum in Case I = Final Momentum in Case II
mV1 = MV2
mV1/M = V2
V2 = mV1/M
substituting this value of V2 in equation
(1)
(1)
2GMm/d = mV12+M(mV1/M)2
2GMm/d = mV12+m2V12/M
Find the value of V1 by yourself.
2GMm/d = mV12+m2V12/M
Find the value of V1 by yourself.