You asked me to "sober up enought to learn basic Newtonian gravity" and asked me to "start showing at least some grasp of basic concepts."

High Seas wrote:Read the damn book, Brandon, exact reference was posted above. Sober up enough to learn basic Newtonian gravity (infinite GRAVITY propagation speed MUST apply IF angular momentum is to be conserved, and the conservation of angular momentum wasn't doubted by Newton!) first, though.

I really don't see where you add value to this discussion if you can't even grasp complete BASICS. Sorry I'm not going to waste any more time until you start showing at least SOME grasp of basic concepts, never mind relativistic gravity field equations.

I acceded to your wishes, and asked you to demonstrate the basic familiarity you were asking me to show. I posed the following high school physics problem in Newtonian mechanics and gravity to you.

Brandon9000 wrote:Although Newtonian mechanics has been superseded by relativistic mechanics, and, therefore, is not relevant here, I am willing to respond to this post of yours. Neither of us can add value unless he understands the basics. With two physics degrees, I think I do. Do you? I'll ask you an elementary question about gravity and circular motion in Newtonian mechanics, which anyone who completed a high school physics class with good grades could work out in 10 or 15 minutes. If you can answer, we can continue the discussion. If not, I'll give the solution in detail, and you can stop talking about physics.

Your post was number 3519687. I'll use that number. A satellite has a circular orbit around the Earth at a height of 3519 miles. What is its speed and kinetic energy? Give your units, and show your work.

and

Brandon9000 wrote:For the kinetic energy question, you can assume that the satellite weighs 10 pounds, or else keep your equations in symbolic form.

Certainly a great expert on Newtonian mechanics and satellite orbits like yourself should be able to operate at the high school level. According to your own criteria:

High Seas wrote:I really don't see where you add value to this discussion if you can't even grasp complete BASICS.

We aren't discussing philosophy or sociology. There actually is an absolute correct and incorrect here. You shouldn't claim expertise on a subject you know little to nothing about. Certainly, you cannot say much about a subject so complicated that it's over the heads of most physicists, if you lack even the physics knowledge of a bright high school student.

Here is the solution of the problem I posed above to you. Please forgive any arithmetic errors it might contain.

I'll start by defining my terms:

Me = mass of the Earth

Ms = mass of the satellite

G = the gravitational constant

r = the distance between the satellite and the center of the Earth

v = the velocity of the satellite

F = the force exerted by gravity to hold the satellite in orbit

v and F refer to the magnitude of these vector quantities, since this is a scalar calculation. As stated above, the Force between two masses, in this case, the Earth and the satellite, is:

G Me Ms / r ^ 2

This is equal to the centripetal force on the orbiting satellite:

G Me Ms / r ^ 2 = Ms v ^ 2 / r

Therefore,

v = SQRT ( G Me / r )

Before I go much further, I'm going to convert everything to metric units, so I don't have to convert the gravitational constant into imperial units.

I indicated a height of 3519 miles

3519 mi = 3519 mi x 1.6093 km/mi = 5663 km = 5.663 x 10 ^ 6 m

However, presumably this is above the surface of the Earth, not its center, so the radius of the Earth must be added in. The radius of the Earth is 6.38 x 10^6 m.

r = 5.663 x 10 ^ 6 m + 6.38 x 10^6 m = 1.2043 x 10 ^ 7 m

The weight of the satellite was given as 10 lbs. Obviously, I meant it's weight on the Earth's surface. The mass in a 1 lb weight is equivalent to 0.45359 kg. Note that pounds are units of force, but a kilogram is a unit of mass.

Ms = 4.5359 kg

The mass of the Earth, Me, is 5.97 x 10^24 kg. The gravitational constant, G, is 6.67 x 10 ^ -11 N-m^2/kg^2 (N is the unit Newtons).

Therefore,

v = SQRT ( G Me / r ) = SQRT ( 6.67 x 10 ^ -11 N-m^2/kg^2 x 5.97 x 10^24 kg / 1.2043 x 10 ^ 7 m ) = SQRT( 3.3065 x 10^ 7 m^2 / s^2 ) = 5.75 x 10 ^ 3 m/s

The kinetic energy is 1/2 m v ^2 = 1/2 ( 4.5359 kg) ( 5.75 x 10 ^ 3 m/s )^2 = 7.4984 x 10 ^ 7 kg - m^2/s^2 = 7.4984 x 10 ^ 7 joules