Modeling the VLBI delay for Earth satellites

Abstract

Very-long-baseline interferometry (VLBI) observations of satellites orbiting the Earth and emitting an artificial radio signal have the potential of becoming an important technique for improving the frame ties between celestial and terrestrial reference frames. Modeling the delay of the signal reception at one station with respect to the other station of a baseline is a fundamental step for correlation and parameter estimation. The near-field VLBI delay models developed so far include numerical computation, which may become expensive in terms of computation time. This applies especially when partial derivatives are to be computed, which is the normal case for least squares adjustments. Furthermore, all the models are formulated in the barycentric celestial reference system requiring large numbers. Here we present an analytical expression for the VLBI delay for the special case of satellites orbiting the Earth, observed by ground-based radio telescopes. We analytically solve the light time equation for each signal propagation path from the source to receiver one and to receiver two under the simplification of linearizing the trajectory of the satellite. By approximating the motion of the Earth as uniform during the short signal travel times we are able to work in the geocentric celestial reference system. We investigate differences between numerical and analytical solutions by simulating VLBI observations of Earth satellites. These tests reveal that delays computed with the analytical formula are consistent with those computed with the numerical solution below the detection level of VLBI but at less computational cost.

Appendix A: Derivation of the formulas

In this section we give a detailed derivation of the analytical formulas for the signal travel times. Since the final quantity of interest is the difference of the signal travel time, without loss of generality the time of reception of the signal at station one is defined to be zero, and the other times, i.e., the time of emission and time of reception at station two, are then to be understood relative to zero.

In the following all quantities denoted by x (positions) and v (velocities) are elements of ${\mathbb {R}}^3$, while quantities denoted by t (time) are elements of ${\mathbb {R}}$.

1.1 A.1: From sender to station 1

The light time equation for the signal propagation path from the sender to station one is

$$\begin{aligned} t_0 = -\frac{\left| x_0(t_0) - x_1(t_1)\right| }{c} - t_{\mathrm{g}\,01}, \end{aligned}$$

(32)

where $x_1(t_1)$ is the position of receiver one at time $t_1$, both of which are known quantities, $t_{\mathrm{g}\,01}$ is the gravitational delay. This equation can be rewritten as

$$\begin{aligned} t_0 + t_{\mathrm{g}\,01} = -\frac{\left| x_0(t_0) - x_1(t_1)\right| }{c}, \end{aligned}$$

(33)

and we square both sides in order to be able to handle the norm of the right-hand side,

$$\begin{aligned} \left[ t_0 + t_{\mathrm{g}\,01}\right] ^2 = \frac{\left[ x_0(t_0) - x_1(t_1)\right] ^2}{c^2}, \end{aligned}$$

(34)

and we have to keep in mind that this squared equation has one solution which does not satisfy Eq. (32). We linearize the trajectory $x_0(t)$ around $t_1 = 0$,

$$\begin{aligned} x_0(t) = x_0(t_1) + v_0(t_1)[t - t_1] = x_0 + v_0t. \end{aligned}$$

(35)

Inserting (35) into (34) yields

$$\begin{aligned} \left[ t_0 + t_{\mathrm{g}\,01}\right] ^2 = \frac{\left[ x_0 + v_0t_0 - x_1\right] ^2}{c^2}, \end{aligned}$$

(36)

where we omit the argument for $x_0$, $x_1$, and $v_0$, which are from now on to be understood to be evaluated at $t_1 = 0$. With the definition $x_{01} := x_0 - x_1$ Eq. (36) becomes

$$\begin{aligned} \left[ t_0 + t_{\mathrm{g}\,01}\right] ^2 = \frac{\left[ x_{01} + v_0t_0\right] ^2}{c^2}. \end{aligned}$$

(37)

Expanding both sides yields

$$\begin{aligned} t_0^2 + t_{\mathrm{g}\,01}^2 + 2t_0t_{\mathrm{g}\,01} = \frac{x_{01}^2 + v_0^2t_0^2 + 2x_{01}\cdot v_0t_0}{c^2}. \end{aligned}$$

(38)

Bringing everything on the left-hand side,

$$\begin{aligned} t_0^2 + t_{\mathrm{g}\,01}^2 + 2t_0t_{\mathrm{g}\,01} - \frac{x_{01}^2 + v_0^2t_0^2 + 2x_{01}\cdot v_0t_0}{c^2} = 0, \end{aligned}$$

(39)

and collecting all coefficients of the quadratic and linear terms,

$$\begin{aligned} \left[ 1- \frac{v_0^2}{c^2}\right] t_0^2 + \left[ 2t_{\mathrm{g}\,01} - \frac{2x_{01}\cdot v_0}{c^2}\right] t_0 + t_{\mathrm{g}\,01}^2 -\frac{x_{01}^2}{c^2} = 0.\nonumber \\ \end{aligned}$$

(40)

Introducing the definition $\gamma _0 := [1 - v_0^2/c^2]^{-1/2}$, this becomes

$$\begin{aligned} \gamma _0^{-2}t_0^2 - 2\left[ \frac{x_{01}\cdot v_0}{c^2} - t_{\mathrm{g}\,01}\right] t_0 -\frac{x_{01}^2}{c^2} + t_{\mathrm{g}\,01}^2 = 0, \end{aligned}$$

(41)

or equivalently

$$\begin{aligned} t_0^2 - 2\gamma _0^{2}\left[ \frac{x_{01}\cdot v_0}{c^2} - t_{\mathrm{g}\,01}\right] t_0 - \gamma _0^{2}\left[ \frac{x_{01}^2}{c^2} - t_{\mathrm{g}\,01}^2\right] = 0, \end{aligned}$$

(42)

which has the solutions

$$\begin{aligned} t_0= & {} \gamma _0^{2}\left[ \frac{x_{01}\cdot v_0}{c^2} - t_{\mathrm{g}\,01}\right] \nonumber \\&\pm \sqrt{\gamma _0^{4}\left[ \frac{x_{01}\cdot v_0}{c^2} - t_{\mathrm{g}\,01}\right] ^2 + \gamma _0^{2}\left[ \frac{x_{01}^2}{c^2} - t_{\mathrm{g}\,01}^2\right] }. \end{aligned}$$

(43)

Remembering that $t_0$ is the time of emission of the signal before reception at station one, we choose the minus sign here,

$$\begin{aligned} t_0= & {} \gamma _0^{2}\left[ \frac{x_{01}\cdot v_0}{c^2} - t_{\mathrm{g}\,01}\right] \end{aligned}$$

(44)

$$\begin{aligned}&- \sqrt{\gamma _0^{4}\left[ \frac{x_{01}\cdot v_0}{c^2} - t_{\mathrm{g}\,01}\right] ^2 + \gamma _0^{2}\left[ \frac{x_{01}^2}{c^2} - t_{\mathrm{g}\,01}^2\right] }. \end{aligned}$$

(45)

1.2 A.2: From sender to station 2

Being the signal emitted at the previously determined time $t_0$ the light time equation for the signal propagation path from there to station two reads

$$\begin{aligned} t_2 = t_0 + \frac{\left| x_0(t_0) - x_2(t_2)\right| }{c} + t_{\mathrm{g}\,02}, \end{aligned}$$

(46)

where $x_0(t_0) = x_0$ is known now and $x_2$ has to be linearized,

$$\begin{aligned} x_2(t) = x_2(t_1) + v_2(t_1)[t - t_1] = x_2 + v_2t. \end{aligned}$$

(47)

Since we are only interested in the light travel time, we rewrite Eq. (46) as

$$\begin{aligned} {\varDelta }t_2 := t_2 - t_0 = \frac{\left| x_0(t_0) - x_2(t_2)\right| }{c} + t_{\mathrm{g}\,02}, \end{aligned}$$

(48)

where ${\varDelta }t_2$ is defined as the light travel time. Inserting (35) and (47) into Eq. (48),

$$\begin{aligned} {\varDelta }t_2= & {} \frac{\left| x_0(t_1) + v_0(t_1)t_0 - [x_2(t_1) + v_2(t_1)[{\varDelta }t_2 + t_0]]\right| }{c} \nonumber \\&+ t_{\mathrm{g}\,02}\end{aligned}$$

(49)

$$\begin{aligned}= & {} \frac{\left| {\tilde{x}}_{02} + v_0(t_1)t_0 - [v_2(t_1)[{\varDelta }t_2 + t_0]]\right| }{c} + t_{\mathrm{g}\,02} \end{aligned}$$

(50)

$$\begin{aligned}= & {} \frac{\left| {\tilde{x}}_{02} + v_{02}t_0 - v_2{\varDelta }t_2\right| }{c} + t_{\mathrm{g}\,02} \end{aligned}$$

(51)

$$\begin{aligned}= & {} \frac{\left| x_{02}(t_0) - v_2 {\varDelta }t_2\right| }{c} + t_{\mathrm{g}\,02}, \end{aligned}$$

(52)

with ${\tilde{x}}_{02} = x_0(t_1) - x_2(t_1)$, $v_{02} = v_0(t_1) - v_2(t_1)$, and $x_{02}(t_0) = {\tilde{x}}_{02} + v_{02}t_0 =: x_{02}$. Rewriting yields

$$\begin{aligned}&{\varDelta }t_2 - t_{\mathrm{g}\,02} = \frac{\left| x_{02} - v_2{\varDelta }t_2\right| }{c} \end{aligned}$$

(53)

$$\begin{aligned}&\quad \Rightarrow \left[ {\varDelta }t_2 - t_{\mathrm{g}\,02}\right] ^2 = \frac{\left[ x_{02} - v_2{\varDelta }t_2\right] ^2}{c^2} \end{aligned}$$

(54)

$$\begin{aligned}&\quad \Leftrightarrow {\varDelta }t_2^2 + t_{\mathrm{g}\,02}^2 - 2{\varDelta }t_2t_{\mathrm{g}\,02}\nonumber \\&= \frac{x_{02}^2 + v_2^2{\varDelta }t_2^2 - 2x_{02}\cdot v_{2}{\varDelta }t_2}{c^2} \end{aligned}$$

(55)

$$\begin{aligned}&\quad \Leftrightarrow {\varDelta }t_2^2 + t_{\mathrm{g}\,02}^2 - 2{\varDelta }t_2t_{\mathrm{g}\,02}\nonumber \\&\qquad - \frac{x_{02}^2 + v_2^2{\varDelta }t_2^2 - 2x_{02}\cdot v_{2}{\varDelta }t_2}{c^2} = 0 \end{aligned}$$

(56)

$$\begin{aligned}&\quad \Leftrightarrow \left[ 1 - \frac{v_2^2}{c^2}\right] {\varDelta }t_2^2 - 2\left[ t_{\mathrm{g}\,02} - \frac{x_{02}\cdot v_{2}}{c^2}\right] t_2 - \frac{x_{02}^2 }{c^2} \nonumber \\&\qquad + t_{\mathrm{g}\,02}^2 = 0 \end{aligned}$$

(57)

$$\begin{aligned}&\quad \Leftrightarrow \gamma _2^{-2}{\varDelta }t_2^2 - 2\left[ t_{\mathrm{g}\,02} - \frac{x_{02}\cdot v_{2}}{c^2}\right] {\varDelta }t_2 - \frac{x_{02}^2 }{c^2} + t_{\mathrm{g}\,02}^2 = 0 \nonumber \\ \end{aligned}$$

(58)

$$\begin{aligned}&\quad \Leftrightarrow {\varDelta }t_2^2 - \gamma _2^{2}2\left[ t_{\mathrm{g}\,02} - \frac{x_{02}\cdot v_{2}}{c^2}\right] {\varDelta }t_2 \nonumber \\&\qquad - \gamma _2^{2}\left[ \frac{x_{02}^2 }{c^2} - t_{\mathrm{g}\,02}^2\right] = 0 \end{aligned}$$

(59)

$$\begin{aligned}&\quad \Rightarrow {\varDelta }t_2 = \gamma _2^{2}\left[ t_{\mathrm{g}\,02} - \frac{x_{02}\cdot v_{2}}{c^2}\right] \nonumber \\&\qquad \pm \sqrt{\gamma _2^{4}\left[ t_{\mathrm{g}\,02} - \frac{x_{02}\cdot v_{2}}{c^2}\right] ^2 + \gamma _2^{2}\left[ \frac{x_{02}^2}{c^2} - t_{\mathrm{g}\,02}^2\right] }. \end{aligned}$$

(60)

And this time we have to choose the positive sign,

$$\begin{aligned} {\varDelta }t_2= & {} \gamma _2^{2}\left[ t_{\mathrm{g}\,02} - \frac{x_{02}\cdot v_{2}}{c^2}\right] \nonumber \\&+ \sqrt{\gamma _2^{4}\left[ t_{\mathrm{g}\,02} - \frac{x_{02}\cdot v_{2}}{c^2}\right] ^2 + \gamma _2^{2}\left[ \frac{x_{02}^2}{c^2} - t_{\mathrm{g}\,02}^2\right] }. \end{aligned}$$

(61)