Solve Relativity Q: Speed of a Satellite & GPS Implications

[madorangepand]

A space shuttle is in orbit traveling with velocity v with respect to the Earth, along the x direction. The crew launch a satellite forward with velocity u in their own rest frame. The speed of the satellite relative to the Earth is w. A beacon on the satellite emits photons of wavelength $$\lambda$$ in all directions in the satellite rest frame.

i) Define the z axis as the axis joining the photon to the Earth, i.e perpendicular to its velocity in the rest frame of the satellite. Give an expression in terms of w and $$\lambda0$$ for
1. the wavelength $$\lambda1$$, of a photon emitted along the x axis

2. the wavelength $$\lambda2$$, of a photon emitted along the z axis

as seen by an observer stationary with respect to the Earth.ii)A realistic value for w is v=30m/s. comment on the values of $$\lambda1$$ and $$\lambda2$$ in this case, and on the implications of your answer for GPS satellites which emit a signal at a particular frequency.

iii)Write down the equation for time dilation, and comment on the effect on a clock on the satellite which has to stay in time with one on the Earth.

iv)Wrtie down the four-vector momentum, p, of a photon which is emitted along the x direction, in terms of $$\lambda0$$

Give a matrix equation for the four-vector momentum, p', of this photon in the rest frame of the Earth, in terms of u, v and p.

Use this equation to show that the velocity, w, of the satellite in the rest frame of the Earth is given by

$$w=\frac{u+v}{1+\frac{uv}{c^2}}$$

ii) and iii) I can do with ease (they are the easy bits after all)

I can't even seem to start parts i) or iv)

 

[BigJDubs]

though. Any help would be great.i) 1. The wavelength of a photon emitted along the x axis, as seen by an observer stationary with respect to the Earth, is \lambda1 = \lambda_0/\gamma, where \gamma = (1 - v^2/c^2)^(-1/2) is the Lorentz factor.2. The wavelength of a photon emitted along the z axis, as seen by an observer stationary with respect to the Earth, is \lambda2 = \lambda_0/\gamma w/c, where \gamma = (1 - v^2/c^2)^(-1/2) is the Lorentz factor and w is the velocity of the satellite relative to the Earth.ii) A realistic value for w is v = 30 m/s. In this case, \lambda1 = \lambda0/(1 - 0.001)^(-1/2) and \lambda2 = \lambda0/(1 - 0.001)^(-1/2) * 0.03. This implies that the wavelength of a photon emitted along the x axis is longer than that of a photon emitted along the z axis. This has implications for GPS satellites which emit a signal at a particular frequency, as the signal will appear to be shifted in frequency when viewed from Earth due to the Doppler effect.iv) The four-vector momentum, p, of a photon which is emitted along the x direction, in terms of \lambda0, is p = (h/\lambda_0, 0, 0, c).The four-vector momentum, p', of this photon in the rest frame of the Earth, in terms of u, v and p, is given byp' = \gamma(c(u + v), p_x, p_y, c(u + v))where \gamma = (1 - v^2/c^2)^(-1/2).Solving for w, we havew = \frac{u + v}{1 + \frac{uv}{c^2}}.