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dv/dt=Gm/d2?
By Brad Rodgers (P1930) on Wednesday, May 17, 2000 - 10:16 pm:

In Newtonian physics, it is a well known equation that F=Gm1m2/d2. Substituting, we receive ma=Gm1m2/d2, so a=dv/dt=Gm/d2. My question is, how do we integrate this to find what the velocity is at a given moment or at a given distance?

By Kerwin Hui (P1312) on Thursday, May 18, 2000 - 11:51 am:

By chain rule, we have

a=dv/dt=(dx/dt)(dv/dx)=vdv/dx

Thus

vdv/dx=-Gm/x2

Separating variables and integrate to give v in terms of x. The result is

v2-v02=2Gm(1/x-1/x0)

and thus we can use the fact that v=dx/dt to separate variables and integrate again to give t in terms of x. Unfortunately, this usually does not give a nice formula.

Hope this helps.

Kerwin

By Neil Morrison (P1462) on Thursday, May 18, 2000 - 04:12 pm:

Kerwin is right...
This is a constantly asked CSYS / A-Level Mechanics question. You change the variable from time to displacement as indicated.

Neil M

By Brad Rodgers (P1930) on Thursday, May 18, 2000 - 09:45 pm:

Sorry, but what does the v or x with a zero under it mean. I am fairly unacquainted with physics notatation. Also, when you label the distance traveled as x, the d in my equation represents the distance from the earth (or whatever body we are dealing with). So, how do you derive that the distance from the earth is equivalent to the distance the object has traveled.

Unless of course, it is meant that v=-dx/dt, as the more a body moves, the closer it is to earth. Anyway, you accounted for this in your second line of math but I am not sure that you differentiated with this in mind as I do not know what v0 indicates. Thanks though,

Brad

By Kerwin Hui (P1312) on Friday, May 19, 2000 - 03:41 pm:

Brad,

Sorry for the confusion caused.

v0 denotes the initial speed,

and similarly,

x0 denotes the initial separation of the centres of mass.

The equation of motion is

a=dv/dt=-Gm/x2

where x denotes the separation of centres of mass.

Kerwin

By Brad Rodgers (P1930) on Tuesday, May 23, 2000 - 02:23 am:

I have attempted another integration of this. I would like to know if this makes sense

We start with the equation

v.dv/dx=-Gm/x2

v.dv=(-Gm/x2)dx

Integrating both terms

Gm/x=v2/2

Therefore

v=sqrt(2Gm/x)

This is of course assuming that the object was dropped from "infinite height". I am fairly sure that this is right, but when I try to integrate with v=-dx/dt, I end up with nonsense.

By Michael Doré (P904) on Saturday, May 27, 2000 - 09:47 am:

Hi Brad,

It is worth knowing that the integrated form of your equation, i.e.

½v2 = Gm/x + C

is actually a statement of the law of conservation of energy. In your equation presumably m is the mass of the object which is attracting the object you're considering. Suppose we let the object you are considering (which has position vector x) have mass M. Now multiply both sides of the equation by M and re-arrange.

½Mv2 - GmM/x = CM = constant

The first term is the kinetic energy. The second term is what we call the gravitational potential energy. The statement says that the sum of gravitational and kinetic energies are constant.

The formula -GMm/x for GPE is useful generally. You might like to try to show how to derive the approximation mgdh for small dh.

In fact you can use it to find the escape velocity very easily. Remember the total energy of an object free-falling stays constant. So what minimum total energy does a body need to get to an infinite distance of its attractor? Supposing it starts at a distance r, therefore what KE does it need? So what is the escape velocity? You can use this to calculate the Schwarzchild radius of a black hole of known mass.

Actually you can do this with forces too, but it seems slightly more straightforward this way.

Yours,

Michael