How to Find Instantaneous Velocity Using Force

AI Thread Summary
To find the instantaneous velocity of object m, the gravitational force equation F = (mMG) / R^2 is referenced, but initial conditions are crucial. Without knowing the initial speed or other boundary conditions, it's impossible to determine the velocity using work-energy principles or kinematic equations. The discussion emphasizes that while instantaneous acceleration can be calculated, the lack of historical data prevents the determination of velocity. Integration of acceleration requires constants that cannot be established without initial conditions. Thus, the conclusion is that velocity cannot be found in this scenario.
Arman777
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Homework Statement


Let's suppose we have two objects on ##x## axis.They have mass ##m## and ##M##, They are exerting a gravitational force on each other and its magnitude can be written as ##F=\frac {mMG} {R^2}##.
So my question is can we find Instantaneous velocity of ##m##, ##{V_m}## ?
We only know ##R##, ##m## and ##M##.

Homework Equations


##W=ΔKE##
##F=\frac {mMG} {R^2}##.

The Attempt at a Solution


I tried to use ##W=ΔKE## but it didnt make sense cause we don't know inital conditions, like inital speed of the object etc. and cause of that I also couldn't use kinematic equations.
 
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Exactly. Without the initial conditions we can't work out the velocity. We can work out the instantaneous acceleration of m, but without knowing its past history we can't say what its velocity is. (Remember V = ∫adt, and any integration has an integration constant which must be determined with reference to the boundary conditions, which you don't know.)
 
mjc123 said:
Exactly. Without the initial conditions we can't work out the velocity. We can work out the instantaneous acceleration of m, but without knowing its past history we can't say what its velocity is. (Remember V = ∫adt, and any integration has an integration constant which must be determined with reference to the boundary conditions, which you don't know.)

Make sense..Thanks
 
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