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meteor84
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Hi everyone,
I want to discuss the following question:
What would happen to the distance between two objects with masses M and m with M >> m if the central mass M was time-dependent (e.g. the moon's orbit if the mass of the Earth would increase slowly)?
My theory goes like this:
The orbit equation in the above case can be written as
r = \frac{l^2}{m^2 \; \gamma} \frac{1}{1+e}, \ \ \ \ \ (1)where r is the separation distance between the two bodies and \theta is the angle that \mathbf{r} makes with the axis of periapsis (also called the true anomaly). The parameter l is the angular momentum of the orbiting body about the central body, and is equal to mr^2\dot{\theta}.
The parameter \gamma is the constant for which \gamma/r^2 equals the acceleration of the smaller body (for gravitation, \gamma is the standard gravitational parameter, GM). The parameter e is the eccentricity of the orbit.
(See: http://en.wikipedia.org/wiki/Orbit_equation)
For making it easier, as a first approach, I would like to look at the special case of a circular orbit (e = 0) in which equation (1) reduces to:
<br /> r = \frac{l^2}{m^2 \; \gamma}. \ \ \ \ \ (2)<br />
From equation (2) and the definitions of l and \gamma it follows that another valid expression for r is:
<br /> r = \sqrt[3]{\frac{G \; M}{(d\theta / dt)^2} }. \ \ \ \ \ (3)<br />Since the orbital period is T = \frac{2 \pi r}{v} and the tangential velocity v should be unaffected by a change of the central mass it follows that \frac{d \theta}{d t} \propto r^{-1}.
Together with equation (3) I come to the conclusion that the following relation is valid:
<br /> r \propto G \; M <br />
and since G is a constant:<br /> r \propto M. <br />So a change of for example \frac{\Delta M}{M} = 10^{-11} in the mass of the central body should lead to the same relative change in the distance r.
Maybe this is one possible explanation for the 3.8 cm per year increase of the distance between the Earth and the moon, but I'm not sure if my derivation is correct.
Cheers,
S.
I want to discuss the following question:
What would happen to the distance between two objects with masses M and m with M >> m if the central mass M was time-dependent (e.g. the moon's orbit if the mass of the Earth would increase slowly)?
My theory goes like this:
The orbit equation in the above case can be written as
r = \frac{l^2}{m^2 \; \gamma} \frac{1}{1+e}, \ \ \ \ \ (1)where r is the separation distance between the two bodies and \theta is the angle that \mathbf{r} makes with the axis of periapsis (also called the true anomaly). The parameter l is the angular momentum of the orbiting body about the central body, and is equal to mr^2\dot{\theta}.
The parameter \gamma is the constant for which \gamma/r^2 equals the acceleration of the smaller body (for gravitation, \gamma is the standard gravitational parameter, GM). The parameter e is the eccentricity of the orbit.
(See: http://en.wikipedia.org/wiki/Orbit_equation)
For making it easier, as a first approach, I would like to look at the special case of a circular orbit (e = 0) in which equation (1) reduces to:
<br /> r = \frac{l^2}{m^2 \; \gamma}. \ \ \ \ \ (2)<br />
From equation (2) and the definitions of l and \gamma it follows that another valid expression for r is:
<br /> r = \sqrt[3]{\frac{G \; M}{(d\theta / dt)^2} }. \ \ \ \ \ (3)<br />Since the orbital period is T = \frac{2 \pi r}{v} and the tangential velocity v should be unaffected by a change of the central mass it follows that \frac{d \theta}{d t} \propto r^{-1}.
Together with equation (3) I come to the conclusion that the following relation is valid:
<br /> r \propto G \; M <br />
and since G is a constant:<br /> r \propto M. <br />So a change of for example \frac{\Delta M}{M} = 10^{-11} in the mass of the central body should lead to the same relative change in the distance r.
Maybe this is one possible explanation for the 3.8 cm per year increase of the distance between the Earth and the moon, but I'm not sure if my derivation is correct.
Cheers,
S.
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