Re: Gravity and the Sun

From: Old Man (nomail_at_nomail.net)
Date: 08/18/04 

Date: Tue, 17 Aug 2004 19:17:13 -0500

"JM Albuquerque" <jm.aREMOV.E@sapo.pt> wrote in message
news:2of9t7Fa9vo1U1@uni-berlin.de...
>
> "Old Man" <nomail@nomail.net> wrote:
>
> > "JM Albuquerque" <jm.aREMOV.E@sapo.pt> wrote in message
> > news:2ocg2lF8rv6bU1@uni-berlin.de...
> > >
> > > I've learned something new here.
> > >
> > > Its OK with the angular momentum translation to linear
> > > momentum with a single mathematic formula.
> > >
> > > Linear motion has the same angular momentum relative to
> > > any point in space.
> >
> > Old Man is impressed with JM's ability and willingness
> > to learn. Old Man withdraws all previous claims to the
> > contrary. Welcome to the club.
> >
> > L = r x p = r (m v) sin(theta)
> >
> > For straight line motion, let r = b (impact parameter) be
> > the distance of closest approach of the line to the origin.
> > Then, for velocity, v, and time, t, the distance from the
> > origin,r, is given by
> >
> > r = sqrt[ b^2 + (vt)^2 ]
> >
> > and sin(theta) = b / r = b / sqrt[ b^2 + (vt)^2 ]
> >
> > Combining equations yields
> >
> > L = m b v = constant (independent of time)
> >
> > [Old Man]
>
>
> OK Old Man, I like to learn and thank you very much.
> But I didn't learn that much, since I already knew that momentum
> is always conserved, as I said previously.
> The news are that angular momentum and linear momentum
> can be related by a single formula:
> L = r x p = r (m v) sin(theta)
>
> But that's not all the story and I want to learn more.
>
> I could be wrong, but this equation ...
> delta_R = (1 / 2) (GM / R^2) (delta_t)^2
>
> ...doesn't follow from these ones (angular momentum):
> L / m = r x v = constant
> L = r x p = r (m v) sin(theta)

JM is correct. It doesn;t follow. Old Man never said it
did follow.

>
> The problem is that angular momentum tells me nothing about
> gravity, forces involved, energy, power, whatever is the reasoning
> that you want to do with the angular momentum conservation.
>
> Angular momentum is conserved but angular momentum doesn't
> mean nothing special about physics. Its just a conservation Law
> which helps to solve some motion problems.

Hogwash.

> Being the term: GM / R^2 = g = gravity acceleration, your above
> equations "delta_R = (1 / 2) (GM / R^2) (delta_t)^2" = 1/2 g t^2
> is the Law of spaces in the uniformly accelerated linear motion:
> x = travelled distance = 1/2 g t^2
> velocity = v = dx/dt = g t
> acceleration = dv/dt = g
>
> That's another aspect of the equivalence principle:
> A system rotating with constant speed is formally equivalent to
> a linear system with uniformly accelerated motion.
>
> Since the gravity of the Sun was switched off, the equivalence
> principle doesn't apply anymore (it makes no sense to talk about
> it) and you came up with an acceleration which has no physical
> meaning anymore (nothing looks to be accelerating), nor you have
> any force to assign that acceleration.

Old Man gave an expression for delta_R, the difference
in radial distance between the circular orbit with gravity
turned-on and the straight-line orbit with gravity turned-
off. GM / R^2 is the acceleration for the initial circular
orbit with gravity turned-on. It is assumed that
delta_R << R and that v*delta_t << R.

Except for the sign of delta_R, the order of events makes
no difference. Gravity can either be turned on or off for
one second. The only difference is in the sign of delta_R.

> The equivalence principle is gone because a rotating system once
> before equivalent to a linear system uniformly accelerated motion,
> in the reality ends up with constant linear motion.
> Conclusion:
> The acceleration is gone because gravity is gone too.
> So gravity is an acceleration and that acceleration is gone.
>
> The Sun gravity is switched off during one second and then switched
> on again, so that you said that the new orbit radius is:

No. The "new orbit radius" is

R_new = R + delta_R = R + (1 / 2) (GM / R^2) (delta_t)^2

>
> delta_R = (1 / 2) g (delta_t)^2 = 0.003 m
> g = G M / r^2 = 0.006 m/s^2
>
> Earth:
> M = 5.98 x 10^24 kg
> r = 6.37 x 10^6 m
> g = (6.668x10^-11 x 5.98x10^24) / (6.37x10^6)^2 = 9.827 m/s^2
>
> Sun:
> M = 1.9891 x 10^30 kg
> r = 6.96 x 10^8 m
> g = (6.668x10^-11 x 1.9891 x 10^30) / (6.96 x 10^8)^2 = 273.8 m/s^2
>
> The above values agree with known references and fact ***.
>
>
> You have said before:
> "Old Man" <nomail@nomail.net> wrote:
> news:_oOdnYwRgOMFr4TcRVn-pA@prairiewave.com...
> >
> > angular momentum, L, being conserved,
> >
> > L / m = r x v = constant,
> >
> > the Earth would travel along a tangent to its orbit for one
> > second. As expected, this would result in a small, permanent,
> > increase in the Earth-Sun distance,
> >
> > delta_R = (1 / 2) (GM / R^2) (delta_t)^2
> >
> > wherein delta_t = 1 second. The radial acceleration of the
> > Earth in its orbit about the Sun is GM / R^2 = 0.006 m / s^2
> > or about 0.0006 Earth g. Thus, delta_R = 0.003 meters.
>
> The value delta_R = 0.003 meters is the correct one, but I'm
> still wondering where did you find g = GM / r^2 = 0.006 m/s^2 ???

Old Man gets his solar system data from:

http://nssdc.gsfc.nasa.gov/planetary/planetfact.html

> Let me see this one:
> Earth Sun distance = R = 150,000,000,000 m = 1.5 x 10^11 m
> Earth mass = M = 5.98 x 10^24 kg
> Centripetal acceleration = g =
> = (6.668x10^-11 x 5.98x10^24) / (1.5x10^11)^2 = 1.77x10^-8 m/s^2
> Nope.

Nope is right. To get the orbital acceleration of Earth
in its orbit about the Sun from a = - GM / R^2, M must
be the Sun's Mass, not the mass of the Earth.

> The solution is:
> radial acceleration in constant circular motion = g = v^2 / r
> Earth velocity = v = 30,000 m/s (linear tangential velocity)
> So that:
> g = (30,000 m/s)^2 / 1.5 x10^-11m = 0.006 m/s^2
>
> Which means that the following equality must hold:
> GM / r^2 = v^2 / r = 0.006 m/s^2
> And:
> GM / r = v^2
> So:
> M = r x v^2 / G = 1.5x10^-11 x (30,000)^2 / 6.668x10^-11 =
> M = 2 x 10^8 kg (very small value)
> Nope, can't be.

JM is in error. r = 1.5 x 10^(+11), not r = 1.5 x 10^(-11).
 Old Man gets M_sun = 2 x 10^30 kg which is about right

> Let me assume your value:
> GM / r^2 = 0.006 m/s^2
> Let me assume the mass of the Sun = 1.9891 x 10^30 kg
> r = sqrt (GM /0.006) = 2.2 x 10^22 m (Wrong, too big).

wrong again:

>From http://nssdc.gsfc.nasa.gov/planetary/fact***/sunfact.html

Sun: GM = 1.3 x 10^20 m^3 / s^2
thus, r = sqrt (GM / 0.006) = 1.5 ^ 11 m

> Now the mass of the Earth = 5.98 x 10^24 kg
> r = sqrt (GM /0.006) = 3.99 x 10^14 m (Wrong, still too big).

You must use the mass of the Sun.

> The big question is:
> How do you get GM / r^2 = 0.006 m/s^2 ???

>From http://nssdc.gsfc.nasa.gov/planetary/fact***/sunfact.html

Sun: GM = 1.3 x 10^20 m^3 / s^2

>From http://nssdc.gsfc.nasa.gov/planetary/fact***/earthfact.html

Earth's orbit: R = 1.5 x 10^11 m

Thus a = - GM / R^2 = - 0.006 m / s^2

> No value of M and r seams to satisfy your above statement.
> Can you explain please?

JM must explain it. He has made numerous arithmetic errors.

[Old Man]

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