Two mass points a and b (with large difference of mass) are in motion of two-body problem (a large, b small). One picture is in outer space, and the knot of a,b is gravity (b is in a circular motion). The other picture is on a plane (no friction), and the knot is tension of a string (the mass of the string is zero). In the two pictures, the centrifugal force of b is a vector that is on the extension line of a,b. The reaction is gravity in one picture and tension of the string in the other picture. Where and how is the centripetal force?

Tension of the string between a and b is constant. Gravity is not constant.

]]>The propagation of gravity will be done in an instant. For this, here are two reasons. One reason is that two-body problem, many-body problem are true for celestial bodies. The other reason is that the whole solar system is in an uniform linear motion and planets are in elliptical revolution on their revolution planes.9

]]>A celestial body called Vulcan is revolving on orbit of Mercury. It has the same mass and revolution cycle as Mercury. And diameter is twohold (the both stars are uniform in density). Since the sun’s gravitational field is non-uniform, the sun’s gravity acting on both stars will be slightly larger in Vulcan and smaller in Mercury. The value of perihelion shift also likely will be similar. In short, the size of the celestial body (close to the gravitational source like Mercury) will be the main reason for the perihelion shift.

Imagine a cone with evenly spaced concentric circles on its surface. The non-uniformity of gravity will be exponential non-uniformity.

]]>Your “meaningless” conclusion is, of course, correct in that there is not a single exponential relationship that can used to represent water vapor around the world. Nonetheless, it is still possible at a specific location to calculate a time-averaged exponential relationship for water vapor.

]]>No it’s not

]]>Let me see what I can do

]]>It’s an argument based on thermodynamics. I may do a derivation but it’s not something I teach as the students in my class don’t have enough knowledge of thermodynamics when they learn about blackbody radiation

]]>No they cannot.

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