# Satellite orbiting around Earth - Spacetime Metric

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1. May 10, 2015

### unscientific

1. The problem statement, all variables and given/known data

The metric near earth is $ds^2 = -c^2 \left(1-\frac{2GM}{rc^2} \right)dt^2 + \left(1+\frac{2GM}{rc^2} \right)\left( dx^2+dy^2+dz^2\right)$.

(a) Find all non-zero christoffel symbols for this metric.
(b) Find satellite's period.
(c) Why does $R^i_{0j0} \simeq \partial_j \Gamma^i_{00} - \partial_0 \Gamma^i_{0j} = \frac{1}{2}\partial_i \partial_j g_{00}$?
(d)Why does the separation grow in time?

2. Relevant equations

3. The attempt at a solution

Part(a)
Let lagrangian be $-c^2 d\tau^2 = -c^2 \left(1-\frac{2GM}{rc^2} \right) \dot {t}^2 + \left(1+\frac{2GM}{rc^2}\right)(\dot x^2 + \dot y^2 + \dot z^2)$. The corresponding geodesic equations are
$$\ddot t + \frac{\left( \frac{GM}{r^2c^2} \right)}{1-\frac{2GM}{rc^2}} \dot r \dot t = 0$$
$$\ddot r - \frac{\left( \frac{GM}{r^2c^2} \right)}{1+\frac{2GM}{rc^2}} (\dot r)^2 + \frac{\left( \frac{GM}{r^2} \right)}{1+\frac{2GM}{rc^2}} (\dot t)^2 = 0$$
The christoffel symbols are given by $\Gamma^t_{rt} = \Gamma^t_{tr} = \frac{\left( \frac{GM}{r^2c^2} \right)}{1-\frac{2GM}{rc^2}}, \Gamma^r_{rr} = -\frac{\left( \frac{GM}{r^2c^2} \right)}{1+\frac{2GM}{rc^2}}, \Gamma^r_{tt} = \frac{\left( \frac{GM}{r^2} \right)}{1+\frac{2GM}{rc^2}}$.

Part (b)
Given $x=R \cos(\omega \tau)$ and $y = R \sin (\omega \tau)$ and $z=0$ we have $dx^2 + dy^2 + dz^2 = (R\omega)^2 d\tau^2$. The metric now becomes
$$-c^2 d\tau^2 = -c^2 \left(1-\frac{2GM}{Rc^2} \right)dt^2 + \left(1+\frac{2GM}{Rc^2} \right)\left( (R\omega)^2 d\tau^2 \right)$$
This relates the time duration on earth $dt$ with proper time on satellite $d\tau$.
$$dt = \sqrt{\frac{1+ \left( \frac{R\omega}{c}\right)^2 \left( 1 + \frac{2GM}{Rc^2} \right) }{1 - \frac{2GM}{Rc^2}}} d\tau$$

Part(c)
Under the approximation $\frac{GM}{rc^2} \ll 1$ the christoffel symbols become $\Gamma^t_{rt} \approx \frac{GM}{r^2c^2}, \Gamma^r_{rr} \approx -\frac{GM}{r^2c^2}, \Gamma^r_{tt} \approx \frac{GM}{r^2}$.
Not sure why $R^i_{0j0} \simeq \partial_j \Gamma^i_{00} - \partial_0 \Gamma^i_{0j} = \frac{1}{2}\partial_i \partial_j g_{00}$ holds. Is there some trick here?

Regardless, we have
$$\frac{1}{2}\partial_i \partial_j g_{00} = -\frac{GM}{c^2} \partial_i \partial_j \left[ (x_kx_k)^{-\frac{1}{2}}\right]$$
$$= \frac{GM}{c^2} \partial_i \left[ (x_kx_k)^{-\frac{3}{2}} x_j \right]$$
$$= \frac{GM}{c^2} \left[ \delta_{ij} (x_kx_k)^{-\frac{3}{2}}-3(x_kx_k)^{-\frac{5}{2}} x^i x^j \right]$$
$$\frac{1}{2}\partial_i \partial_j g_{00} = \frac{GM}{c^2 r^3} \left[ \delta_{ij} -3\frac{x^i x^j}{r^2} \right]$$

Part(d)
I suppose one is nearer to the earth and experiences a stronger field than other. Thus the one closer to earth experiences a greater acceleration, so the spatial distances between them increases.

Last edited: May 10, 2015
2. May 14, 2015

### unscientific

bumpp

3. May 16, 2015

### unscientific

bumpp on part (b) - satellite's period. The time experienced on earth is $dt$ while proper period is $d\tau$, so is the period of the satellite observed on earth simply
$$T = \sqrt{\frac{1+ \left( \frac{R\omega}{c}\right)^2 \left( 1 + \frac{2GM}{Rc^2} \right) }{1 - \frac{2GM}{Rc^2}}} \left(\frac{2\pi}{\omega}\right)$$

?

4. May 18, 2015

### unscientific

bump on period in (b)

5. May 22, 2015

solved.