The discussion revolves around the rectilinear motion of two attracting masses, focusing on the calculation of the rate of change of distance between them, denoted as dr/dt. The problem involves gravitational forces and the motion of the masses under their mutual attraction.
The discussion is active, with participants offering different perspectives on how to approach the problem. Some suggest using conservation of energy to find dr/dt as a function of r, while others emphasize the need for initial conditions to relate dr/dt to time. There is an ongoing exploration of the implications of the equations presented and the relationships between the velocities of the masses.
Participants note the lack of explicit initial conditions and the ambiguity in the original problem statement, which complicates the ability to derive dr/dt directly. The discussion also touches on the distinction between finding dr/dt as a function of r versus t, highlighting the challenges in each approach.
haruspex said:Are you sure you have stated the question correctly? I would expect it to ask for d2r/dt2. There is no way to know dr/dt without further information.
The acceleration you found, Gm/r2, is only for mass M.
You can solve for dr/dt. The simplest is to find dr/dt as a function of r. You can use conservation of energy for that. Or you can derive it from the force=mass*acceleration equation using (d/dt)(v) = (dr/dt)(d/dr)(v) = v dv/dr. Finding dr/dt as a function of t (or r as a function of t) is harder.H. Leone said:I find it really peculiar how you can solve for d2r/dt2 and not dr/dt.
Yes and no. Your original formulation asked for dr/dt as a function of r. Knowing dr/dt at a given t does not help since we don't what what r is at that time. If you specify dr/dt at some initial r then you can find dr/dt as a function of r. If you want to know r or dr/dt as a function of t then you need an initial r and an initial t.H. Leone said:You said you needed more information. Would it help at all if we said that both masses were stationary at t=0?
haruspex said:You can solve for dr/dt. The simplest is to find dr/dt as a function of r. You can use conservation of energy for that. Or you can derive it from the force=mass*acceleration equation using (d/dt)(v) = (dr/dt)(d/dr)(v) = v dv/dr. Finding dr/dt as a function of t (or r as a function of t) is harder.
Yes and no. Your original formulation asked for dr/dt as a function of r. Knowing dr/dt at a given t does not help since we don't what what r is at that time. If you specify dr/dt at some initial r then you can find dr/dt as a function of r. If you want to know r or dr/dt as a function of t then you need an initial r and an initial t.
no, that's saying KE=PE. Conservation says the total is constant (need not be zero, or even positive).H. Leone said:Oh neat! I hadn't considered conservation of energy. So something like this?
1/2 mv2 = -GMm / r
No, dr/dt is the same as v. Just used v to make the notation easier.H. Leone said:Where v is presumably dr/dt? In your second method you seem to distinguish between dr/dt and v, leading me to think that I'm incorrect here.
dr/dt = sin(r) would be a differential equation with dr/dt as a function of r; dr/dt=kt would be one with dr/dt as a function of t. In the present problem, you can readily find out how dr/dt depends on r, but it is much harder to find how it depends on t.H. Leone said:Also, a question regarding syntax: how is "dr/dt as a function of r," different from "dr/dt as a function of t?"
haruspex said:no, that's saying KE=PE. Conservation says the total is constant (need not be zero, or even positive).
No, dr/dt is the same as v. Just used v to make the notation easier.
dr/dt = sin(r) would be a differential equation with dr/dt as a function of r; dr/dt=kt would be one with dr/dt as a function of t. In the present problem, you can readily find out how dr/dt depends on r, but it is much harder to find how it depends on t.
H. Leone said:Would I be correct in saying that abs (v1) + abs (v2) = dr/dt ?
H. Leone said:1/2 mv12 + 1/2 Mv22 - GMm/ r = c (constant)
I don't see how.theodoros.mihos said:Yes. This equation can produced by time derivation of
Not quite. dr/dt can be negative, and the two velocities could have opposite signs. That equation covers neither of those cases.H. Leone said:The problem now is that we have two velocities to describe the change in r.
Would I be correct in saying that abs (v1) + abs (v2) = dr/dt ?
##x_1##, ##x_2##, ##r## are functions of time.haruspex said:I don't see how.
Sure, but if you differentiate the energy equation m, M and G are not going to disappear. So how will you deduce an equation that contains none of those? The relationship between the two velocities and dr/dt is purely kinematic. It owes nothing to forces.theodoros.mihos said:##x_1##, ##x_2##, ##r## are functions of time.
The relationship between the two velocities and dr/dt has nothing to do with Newton's Laws. It is a simple fact of geometry and kinematics. You could change the gravitational law (and, consistently, energy conservation etc.) and the relationship would still hold. If you have a way of obtaining that relationship by differentiating the energy equation please post it, preferably in a private mail, not on this thread. I can imagine a way, but it goes something like this:theodoros.mihos said:Disapear because ##F_1+F_2=0## by 3o Newton's Law.