While moving clocks do in fact record time moving at different rates, the time dilation works both ways. This means that a stationary person will view a moving clock ticking slower, but at the same time, a person moving alongside the moving clock will see the stationary clock ticking slower. However, clocks experiencing great accelerations will be permanently changed, "losing" time relative to a clock not being accelerated. Thus, the age difference occurs during the portion of the journey when the traveler accelerates at a great rate in order to return to Earth.

To derive the Schwarzschild radius of a black hole, set gravitational potential energy equal to kinetic energy at escape velocity:

\(\displaystyle \frac{GMm}{r}=\frac{1}{2}mc^2\)

Solving for mass of the black hole:

\(\displaystyle M=\frac{rc^2}{2G}=\frac{(6.963\times 10^8 \textup{m})(2.998\times 10^8 \textup{m\,s}^{-1})^2}{2(6.674 \times 10^{-11}\textup{m}^3\,\textup{kg}^{-1}\textup{s}^{-2})} =5\times10^{35}\,\textup{kg}\)

The equation for time dilation is given by:

\(\displaystyle T'=\frac{T}{\sqrt{1-\frac{v^2}{c^2}}}\)

In this problem v=0.8c T=12.  Using this equation, we get:

\(\displaystyle T'=\frac{12}{\sqrt{1-(\frac{4}{5})^2}}=\frac{12}{\sqrt{1-\frac{16}{25}}}=\frac{12}{\sqrt{\frac{9}{25}}} =\frac{12}{\frac{3}{5}}=(12)(\frac{5}{3})=20\)

Adding 20 years to the age initial age of 30:

\(\displaystyle 30+20=50\)

The Earth-twin is now 50.

Length contraction only occurs in the direction of motion.  This means that the x component of the length, which is cos(30), does not change; length contraction only occurs the the y component, which is sin(30).

First, we find the Lorentz factor:

\(\displaystyle \gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{1}{\sqrt{1-\frac{9}{25}}}=\frac{1}{\sqrt{\frac{16}{25}}}=\frac{5}{4}\)

Next, we apply the time dilation equation to the length in the y direction:

\(\displaystyle L'_y=L_{y}/ \gamma=\frac{4}{5}\sin(30)=\frac{2}{5}\)

Finally, we find the total length by combining the length-contracted y component and the unchanged x component:

\(\displaystyle L=\sqrt{L_x^2+L'_y^2}=\)\(\displaystyle \sqrt{\frac{3}{4}+\frac{4}{25}}=\frac{\sqrt{91}}{10}\)

Length contraction is given by 

\(\displaystyle L_{rest}=\gamma L_{move}\)

Where in this case,

\(\displaystyle L_{rest}=5\textup{m}, \,\,\,L_{move}=3\textup{m}\)

The Lorentz factor is given by:

\(\displaystyle \gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\)

Combining these two equations, we get:

\(\displaystyle L_{rest}=\frac{L_{move}}{\sqrt{1-\frac{v^2}{c^2}}}\)

Solving for v:

\(\displaystyle v=c\sqrt{1-\left ( \frac{L_{move}}{L_{rest}}\right )^2}= c\sqrt{1-\left ( \frac{3}{5}\right )^2}= c\sqrt{1-\left ( \frac{9}{25}\right )}\)

\(\displaystyle = c\sqrt{\left( \frac{16}{25}\right )}=\frac{4}{5}c=0.8c\)

The total energy E of a relativistic particle is related to its rest mass energy E_o by:

\(\displaystyle E=\gamma E_{o}\)

Where gamma is related to the momentum by:

\(\displaystyle p=\gamma mc\)

Combining the equations and solving for p, we get:

\(\displaystyle p=\frac{E}{E_{o}}mc=37 mc\)

Which, in the units specified, is 37.

For a relativistic particle, momentum is given by:

\(\displaystyle p=\gamma mc\)

from which we can solve for gamma:

\(\displaystyle \gamma=\frac{p}{mc}=\frac{50 mc}{mc}=50\)

Total energy of a relativistic particle is given by:

\(\displaystyle E=\gamma E_{o}=50E_{0}\)

The question tells us:

\(\displaystyle E=4E_{o}\)

Where the rest mass energy of a particle is:

\(\displaystyle E_{o}=mc^{2}\)

Using the equation for the total energy of a particle, we can substitute:

\(\displaystyle E=\sqrt{p^{2}c^{2}+m^{2}c^{4}}\)

\(\displaystyle 4mc^2=\sqrt{p^2c^2+m^2c^4}\)

Solving this for \(\displaystyle p\), we find:

\(\displaystyle p=\sqrt{15m^2c^2}=\sqrt{15}\,\,mc\)

Which, in units of mc, gives us the correct answer.

The total energy of a relativistic particle is given by:

\(\displaystyle E=\sqrt{p^2c^2+m^2c^4}\)

Substituting the momentum, we get:

\(\displaystyle E=\sqrt{9m^2c^4+m^2c^4}=\sqrt{10}mc^2\)

Because the rest mass energy of a particle is given by:

\(\displaystyle E_{o}=mc^2\)

The total energy is:

\(\displaystyle E=\sqrt{10}E_{o}\)

The relativistic Doppler shift equation is given by:

\(\displaystyle \lambda_{moving}=\frac{\sqrt{1+\beta}}{\sqrt{1-\beta}}\lambda_{stationary}\)

Where \(\displaystyle \beta\) is defined as:

\(\displaystyle \beta=\frac{v_{source}}{c}\)

Because the stationary wavelength is shorter than the moving wavelength, the object must be receding from the Earth, eliminating two answers.

The speed of light is approximately \(\displaystyle 3*10^8 \frac{m}{s}\), so \(\displaystyle 3.2*10^8 \frac{m}{s}\) is not a possible answer.

Making the approximation that

\(\displaystyle \frac{\lambda_{m}}{\lambda_{s}}=\frac{643.7}{212.5}\approx3\)

Combining this with the first equation:

\(\displaystyle \frac{\lambda_m}{\lambda_s}=\frac{\sqrt{1+\beta}}{\sqrt{1-\beta}}=3\)

\(\displaystyle 3\sqrt{1-\beta}=\sqrt{1+\beta}\)

\(\displaystyle 9(1-\beta)=1+\beta\)

\(\displaystyle 9-9\beta=1+\beta\)

\(\displaystyle 10\beta=8,\,\,\,\beta=0.8\)

From beta, we can find the velocity:

\(\displaystyle \beta=\frac{v_{source}}{c}\rightarrow v_{source}=\beta c=(0.8)(3\times10^8 \frac{m}{s})=2.4\times10^8 \frac{m}{s}\)