Space oddity radio

I will explain what is going on in the question. Radio signal of frequency $f_0$ is emitted from the Earth. One photon in such a signal thus has the intrinsic energy (excluding the potential energy) of $hf_0$ . Since the spaceship is "very far away", I assume that the photon travels a distance long enough to break free from the gravitation of the Earth. Hence the poor photon spent some of its intrinsic energy overcoming the Earth's gravitational pull, i.e. gravitational potential energy, and hence the frequency drops to $f'$ . However, the spaceship is traveling towards the source of the signal, so according to the doppler effect, the apparent frequency to the people in the spaceship should be higher than the actual frequency of the signal reaching them. Therefore, $f'$ increases to $f_0$ again by doppler effect when the spaceship is travelling at a speed of $v$ . Now what we are going to do is to find $v$ .

Initial energy of the photon on Earth=intrinsic energy + gravitational potential energy. i.e.

Initial energy $=hf_0 + (-\frac{GMm}{R})$ , where $M$ is the mass of the Earth, and $m=\frac{hf_0}{c^2}$ is the "mass" of the photon according to $E=mc^2$ , and $R$ is the radius of the Earth.

When the signal reaches the spaceship, the gravitational potential energy of the photon becomes 0 because we assume that the photon is now at a "infinite" distance away from Earth and free of gravitational interaction with any other objects. The photon now has the new frequency, $f'$ and the new energy $hf'$ . By conservation of energy, $hf'=hf_0 + (-\frac{GMm}{R})$ . Sub $m=\frac{hf_0}{c^2}$ to get:

$hf'=hf_0 - \frac{GMhf_0}{c^2 R}$

$f'=f_0 - \frac{GMf_0}{c^2 R}=(1-\frac{GM}{Rc^2})f_0$

$\frac{f'}{f_0}=1-\frac{GM}{Rc^2}$

According to the formula of the doppler effect, $f_0=\frac{c+v}{c}f'$

$\frac{f'}{f_0}=\frac{c}{c+v}=1-\frac{v}{c+v}$ .

Since c>>v, $c+v \approx c$ , and sub $\frac{f'}{f_0}=1-\frac{GM}{Rc^2}$ back to get:

$1-\frac{v}{c}=1-\frac{GM}{Rc^2}$ and thus $v=\frac{GM}{Rc}$ .

When we sub all the values in, we get $v\approx 0.209$ , which is the answer.

Appendix: Actually when dealing with light, we have to use relativistic doppler effect, because speed of light is the same in any inertial frame, but since v is so small, the relativistic doppler effect can be simplify to the normal doppler effect. Here is the proof:

According to the formula of the relativistic doppler effect, the apparent frequency, $f_0=\sqrt{\frac{c+v}{c-v}}f'$

and $\frac{f'}{f_0}=\sqrt{\frac{c-v}{c+v}}=\sqrt{1-\frac{2v}{c+v}}$

, where c is the speed of light and v is speed of the spaceship towards the Earth. Since c>>v, $\frac{2v}{c+v}$ is very small, so by the binomial expansion to the first two term, $f_0=\sqrt{1-\frac{2v}{c+v}}f' \approx (1-(\frac{1}{2})(\frac{2v}{c+v}))f'=(1- \frac{v}{c+v})f'=\frac{c}{c+v}f'$ , in accordance to the normal doppler effect equation. Hence the result will be the same.

Using the information given, we find that the effective mass of the photon, m {eff} is given by M {eff}=E/c^2=h*f/c^2

where E is the energy of the emitted photon at Earth's surface

Now we consider the energy lost by this photon as it travels out of the gravitational field. The new energy E' is given by

E'=E- G M_{earth} M {eff}/R h*f'=h*f-h*f*G*M {earth}/(c^2 R) f'=f (1-G M_{earth}/(c^2 R)) Where R=6370km

We now know the frequency of the carrier waves at a distance far away from earth from a reference frame stationary relative to the Earth.

Since the photon is at a lower frequency that it was originally, we now transform the frequency via a blueshift to obtain the original freqeuncy

f=f' \sqrt{(c+v)/(c-v)} f=f (1-G M_{earth}/(c^2 R))*\sqrt{(c+v)/(c-v)}

we shall now use b to represent v/c

\sqrt{(1-b)/(1+b)}=1-G M_{earth}/(c^2 R)

Substituting the values on the right hand side, we find that it is small, and we thus reduce the left hand side to 1-b for simplicity.

In conclusion, since the radius of the earth is large compared to it's Schwarzschild radius, the velocity of the ship can be approximated very well by the formula v=G M_{earth}/(c R)

This gives us the (very small) value of 0.20914m/s as the ship's velocity.

In truth, the change is frequency due to the gravitational redshift is negligible enough such that even if the space ship were stationary relative to earth, it would still receive the signal rather clearly

As the photons travel from Earth to the spaceship, two redshifts appear, one due to the gravitational pull from the Earth, and other due to the ship moving towards the Earth.

The formulae for each of the redshifts are ( $z$ is the relative change in wavelength) $z_s=\sqrt{\frac{c+v}{c-v}} -1$ for the ship's redshift

$z_e=\frac{GM}{Rc^2}$ for the gravitational redshift, but this only holds when the ship's distance is much larger than the Earth's radius.

These 2 must be the same in order to balance out the 2 effects, and from that we easily get $v=0.21\frac{m}{s}$ .

The kinetic energy of the photon when emitted by the radio station is $KE_e=h f_e$ . The total energy is this energy plus the gravitational potential energy, so

$E_e=KE_e-G \frac{KE_e}{c^2} \frac {M_{earth}}{R_{earth}}$ .

Since energy is conserved, the photon energy observed by myself in the spacecraft is just $E_o=KE_o=E_e$ - there is no gravitational part because the gravitational potential energy is effectively zero very far away from earth. We can therefore solve for the observed frequency using our expression for $KE_o,KE_e$ and $E_e$ ,

$f_o=f_e(1-G \frac {M_{earth}}{c^2R_{earth}})$ .

However, I'm traveling towards the earth, and hence this frequency is also Doppler shifted! The true observed frequency is

$f_o=f_e(1-G \frac {M_{earth}}{c^2R_{earth}})(1+\frac{v}{c})$ .

Since I tuned my radio to the same frequency, i.e. $f_o=f_e$ , these factors must cancel out, which allows us to solve for v=0.209 m/s.