two massive particles of masses M and m (M>m) are separated by a distance l.They rotate with equal angular velocity under their gravitational attraction.What is the linear speed of the particle of mass m.
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Body moving in circular path..
So, Fg = ma = m v²/r ........(i) [r = radius, v = linear velocity]
And Fg = GMm/L²...............(ii) [ Fg = gravitational force]
We need to equate (i) and (ii)...
∴ m v²/r = GMm/L²
⇒v²/r = G m/L²
⇒v² = G Mr/L²
⇒v = 1/L√GMr
∴ Linear speed = 1/L√GMr
Body moving in circular path..
So, Fg = ma = m v²/r ........(i) [r = radius, v = linear velocity]
And Fg = GMm/L²...............(ii) [ Fg = gravitational force]
We need to equate (i) and (ii)...
∴ m v²/r = GMm/L²
⇒v²/r = G m/L²
⇒v² = G Mr/L²
⇒v = 1/L√GMr
∴ Linear speed = 1/L√GMr
kvnmurty:
perhaps r = L ?
why?
yes
See question
It is by distance L
L is the distance
Answered by
98
see the diagram.
This is the situation when two masses m and M (like a pair of planets/stars) are going around in circular orbits, of radii r1 and r2 respectively. They are always at a fixed distance L (=r1+r2). They are orbiting around their center O of mass at the same angular frequency ω.
Their linear velocities v1 and v2 are always in opposite directions. Their center of mass may be moving with a uniform speed. There is no external force or torque on center of mass. The smaller mass goes in a larger circle. The larger mass M orbits in a circle completely inside the bigger circle. The circles are concentric. They are moving with uniform speeds.
Distance between center of mass and m = r1 = L* M/(m+M)
Distance between center of mass and M = r2 = L *m/(m+M)
The mutual gravitational attraction force is equal to the centripetal force. m and M are always at distance L=r1 + r2 on either side of COM.
m v1²/r1 (r1 /r1) = (GMm/L²) (r1 /r1) = F = m a1 = m d²r1/dt²
or, m v1²/r1 = GM m/L²
v1² = GM r1/L² = GM² / {(m+M)L }
v1 = √[GM² / {L(m+M)} ]
Similarly we can calculate other related quantities also:
v2 = √[Gm² / {L(m+M)} ]
Linear momentum of m = m v1 = mM√[ G/{L(M+m)} ]
of M = M v2 = Mm √[ G / {L(M+m)} ]
Net linear momentum of both = 0, as v1 and v2 are in opposite directions.
Angular velocity of m or M: ω = v1/r1 = √[ G(M+m)/L³ ]
F = GMm/L²
Angular momentum of m = m r1 v1 = m r1² ω
Angular momentum of M = M r2 v2
This is the situation when two masses m and M (like a pair of planets/stars) are going around in circular orbits, of radii r1 and r2 respectively. They are always at a fixed distance L (=r1+r2). They are orbiting around their center O of mass at the same angular frequency ω.
Their linear velocities v1 and v2 are always in opposite directions. Their center of mass may be moving with a uniform speed. There is no external force or torque on center of mass. The smaller mass goes in a larger circle. The larger mass M orbits in a circle completely inside the bigger circle. The circles are concentric. They are moving with uniform speeds.
Distance between center of mass and m = r1 = L* M/(m+M)
Distance between center of mass and M = r2 = L *m/(m+M)
The mutual gravitational attraction force is equal to the centripetal force. m and M are always at distance L=r1 + r2 on either side of COM.
m v1²/r1 (r1 /r1) = (GMm/L²) (r1 /r1) = F = m a1 = m d²r1/dt²
or, m v1²/r1 = GM m/L²
v1² = GM r1/L² = GM² / {(m+M)L }
v1 = √[GM² / {L(m+M)} ]
Similarly we can calculate other related quantities also:
v2 = √[Gm² / {L(m+M)} ]
Linear momentum of m = m v1 = mM√[ G/{L(M+m)} ]
of M = M v2 = Mm √[ G / {L(M+m)} ]
Net linear momentum of both = 0, as v1 and v2 are in opposite directions.
Angular velocity of m or M: ω = v1/r1 = √[ G(M+m)/L³ ]
F = GMm/L²
Angular momentum of m = m r1 v1 = m r1² ω
Angular momentum of M = M r2 v2
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Best answer with best possible explanation! Great answer Sir!
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