- #1
Matter_Matters
- 36
- 2
Hi there guys,
I'm struggling! I've been looking at the International Earth Rotation Services (IERS) "standards" for motion of a satellite in GR. the expression is far from trivial and I'm battling to determine where to even start with this bad boy.
The expression is given by
[tex]
\Delta \ddot{\mathbf{r}} = \frac{GM_E}{c^2r^3} \left\{\left[2(\beta+\gamma)\frac{GM_E}{r} - \gamma \dot{\mathbf{r}} \cdot \dot{\mathbf{r}} \right] \mathbf{r} + 2(1+\gamma)(\mathbf{r}\cdot\dot{\mathbf{r}})\dot{\mathbf{r}} \right\} + (1+\gamma)\frac{GM_E}{c^2r^3} \left[ \frac{3}{r^2}(\mathbf{r}\times\dot{\mathbf{r}})(\mathbf{r}\cdot\mathbf{J})+(\dot{\mathbf{r}}\times \mathbf{J})\right] + \left\{ (1+2\gamma)\left[\dot{\mathbf{R}} \times \left( \frac{-GM_S \mathbf{R}}{c^2R^3} \right) \right] \times\dot{\mathbf{r}} \right\}.
[/tex]
The terms in the expression correspond to the following:
[itex]c = [/itex] speed of light.
[itex] \beta, \gamma [/itex] = PPN (parameterized post-Newtonian) parameters, equal to 1 in General Relativity.
[itex] \mathbf{r} [/itex] is the position of the satellite with respect to the Earth.
[itex] \mathbf{R} [/itex] is the position of the Earth with respect to the Sun.
[itex] J [/itex] is the Earth’s angular momentum per unit mass.
[itex] GM_E [/itex] and [itex] GM_S [/itex] are the gravitational coefficients of the Earth and Sun, respectively.
Now, obviously nobody in their right mind is going to know how to derive this monster off the top of their heads, unless of course you wrote the technical note, BUT, does anyone have experience with Relativistic mechanics of satellite is geocentric reference frame or barycentric for that matter?
I'm struggling! I've been looking at the International Earth Rotation Services (IERS) "standards" for motion of a satellite in GR. the expression is far from trivial and I'm battling to determine where to even start with this bad boy.
The expression is given by
[tex]
\Delta \ddot{\mathbf{r}} = \frac{GM_E}{c^2r^3} \left\{\left[2(\beta+\gamma)\frac{GM_E}{r} - \gamma \dot{\mathbf{r}} \cdot \dot{\mathbf{r}} \right] \mathbf{r} + 2(1+\gamma)(\mathbf{r}\cdot\dot{\mathbf{r}})\dot{\mathbf{r}} \right\} + (1+\gamma)\frac{GM_E}{c^2r^3} \left[ \frac{3}{r^2}(\mathbf{r}\times\dot{\mathbf{r}})(\mathbf{r}\cdot\mathbf{J})+(\dot{\mathbf{r}}\times \mathbf{J})\right] + \left\{ (1+2\gamma)\left[\dot{\mathbf{R}} \times \left( \frac{-GM_S \mathbf{R}}{c^2R^3} \right) \right] \times\dot{\mathbf{r}} \right\}.
[/tex]
The terms in the expression correspond to the following:
[itex]c = [/itex] speed of light.
[itex] \beta, \gamma [/itex] = PPN (parameterized post-Newtonian) parameters, equal to 1 in General Relativity.
[itex] \mathbf{r} [/itex] is the position of the satellite with respect to the Earth.
[itex] \mathbf{R} [/itex] is the position of the Earth with respect to the Sun.
[itex] J [/itex] is the Earth’s angular momentum per unit mass.
[itex] GM_E [/itex] and [itex] GM_S [/itex] are the gravitational coefficients of the Earth and Sun, respectively.
Now, obviously nobody in their right mind is going to know how to derive this monster off the top of their heads, unless of course you wrote the technical note, BUT, does anyone have experience with Relativistic mechanics of satellite is geocentric reference frame or barycentric for that matter?
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