Janus

Actually [math]\frac{0.99c+0.5c}{1+ \frac{0.99c(0.5c)}{c^2} }= .9967c[/math]

Not for the occupants of the ship, they feel a constant 20m/sec² acceleration the whole time.

Taking Relativity into account: [math]t_{ship time} = \frac{c}{a} \cosh^{-1}\left ( \frac{ad}{c^2}+1 \right ) = 2.24 yr[/math] Earth time would be 26.39 yr

Atmospheres can get thicker or thinner or change composition, but they don't "rupture".

Well, that can depend on who you ask. Some have the position that SR only holds in intertial reference frames, and since it deals with acceleration, it is GR Other would say that it is a result The Realtivity of Simultaneity and thus an SR effect. At the moment of the first passing the distance between the two twins is small so the acceleration effect will be small. The velocity difference will be large, so each twin would see the other's clock running slow. As time goes by, the distance will increase, and the velocity difference will decrease, each twin sees the other's clock as running fast. When the two twins reach the turnaround point (that instant when their velocities stasrt to change direction) The total time each twin will have seen the other's clock gain and lose due to running fast and slow will cancel out, and the accumulated time on both of their clocks will be the same. The two twins start to come back together and the reverse of the outbound leg happens. At the instant they pass each other, their clocks will once again agree. (but only for that instant, for each twin will see the other's as runing slow at this point.)

Acceleration has an effect on what each twin sees. After factoring out time dilation due to velocity: If your accelerate away from a clock, you see it running slower. If you accelerate towards a clock, you see it running faster. How fast or how slow depends on the magnitude of the acceleration, and the distance to the clock as measured along the line of acceleration. (Note, the actual distance between the observer and the clock does not have to be increasing or decreasing. If you have two clocks accelerating in the same direction at the same rate, the leading clock will run faster than the trailing clock, according to both clocks.) If you factor in this effect, it turns out, that each twin will see the other's clock as running fast during a portion of the trip, and this will cancel out that portion when he saw it running slow, so that, at the end, each twin will have aged the same.

Don't count on FTL to save us from population growth. Even if we had it right now, in order to maintain a fixed population on the Earth, we would have to load people on ships at a rate of 91 million per day.

The Great Bear, aka the Big Dipper, more properly known as Ursa Major, is a Northernly constellation of 7 stars. The front two stars of the dipper are called the pointers, because if you extend the line joining them it will point to the last star in the handle of the Little Dipper, aka the Small Bear, or Ursa Minor (another 7 star constellation). This star, named Polaris, is the North star.

Okay, now that I understand where you are coming from, let's attack this from a different direction. Start with a a given mass (M) that can be converted to energy, either in whole or in part. If we covert it all to energy in the form of photons this energy will be: [math]E=Mc^2[/math] (The total energy equivalence of M) and the momentum: [math]P_p= \frac{E}{c} = \frac{Mc^2}{c} = Mc[/math] If we convert all but a part (m) to energy and that energy is in the form of the kinetic energy of m, the total energy is: [math]E= \gamma mc^2 [/math] Since the total energy involved is conserved, it must stay the same both before and after conversion, thus. [math]E=Mc^2[/math] and [math]E= \gamma mc^2 [/math] are the same and so, [math]Mc^2= \gamma mc^2 [/math] [math]M= \gamma m [/math] Solving for gamma: [math] \gamma = \frac{M}{m} [/math] The momentum will be [math]P_m = \gamma mv[/math] substituting from above: [math]P_m =\frac{M}{m} mv[/math] [math]P_m =Mv[/math] Comparing Pm to Pp: [math]\frac{P_p}{P_m} = \frac{Mc}{Mv} = \frac{c}{v}[/math] Which is the same relationship you arrived at from a different direction. Here, however we can see why the gamma factor doesn't come into play in the final analysis: If we go back to the equation: [math]M= \gamma m [/math] and re-arrange: [math]m= \frac{M}{\gamma} [/math] We see that as you convert more and more of M to energy to increase the velocity of m(the rest mass of the remaining matter), m decreases such that that amount of mass that gamma has to work on in [math]P_m = \gamma mv[/math] decreases at the same rate as gamma increases. As a result, the momentum increases in a steady fashion until it reaches a maximum when all the matter is converted to photons.

You mean that you have x amount of energy that you can either release as photons, or partially convert to matter while using the rest to provide the velocity? This second choice seems a little silly, as it would be easier to store the mass that you are ejecting as matter rather than trying to store it as energy and then converting. Also, in this case, the momentum you will get will depend on how much of the energy is converted to matter. From 0 if you converted all of it to matter (leaving none for kinetic energy and giving it 0 velocity) to approaching that of the photons as the amount of matter approaches 0 and its velocity approaches c.

[math] \gamma mc^2[/math] is the total energy equivalence of the matter, which includes the energy equivalence of the rest mass of the matter (The energy released if you convert this matter into energy) But you are not converting this matter into energy, you are just shootiing it out the back as reaction mass. What you want here is the kinetic energy of the matter alone, which is found by [math]c^2( \gamma m -m)[/math] [math] p = \frac{c^2( \gamma m -m)}{c}[/math] [math] p = c( \gamma m -m)[/math] Now we have [math] c( \gamma m -m)[/math] compared to [math]\gamma mv[/math] let's assume that v = .866 c, which gives a gamma of 2. Then for photons: [math]p= c( 2m -m) = cm[/math] and and for matter [math]p= 2 m (.866c) = 1.732cm [/math] matter has more momentum at the same energy. at .1c we would get for photons [math]p = .005cm[/math] and matter [math]p= .1005cm[/math] at .99c we would get for photons [math]p = 6cm[/math] and matter [math]p= 6.93cm[/math] Actually, it has more to do with the relationship between energy and momentum. You get the same type of effect in non-relativistic physics. consider [math]p=mv[/math] and [math]E= \frac{mv^2}{2}[/math] Now, lets take two masses m1 and m2, with equal kinetic energy. so that [math]\frac{m_1 v_1^2}{2} = \frac{m_2 v_2^2}{2}[/math] [math]m_1 v_1^2 = m_2 v_2^2[/math] [math]\frac{m_1}{m_2}=\frac{v_2^2}{v_1^2}[/math] [math]\frac{m_1}{m_2}=\left (\frac{v_2}{v_1}\right )^2[/math] Thus the ratio of the masses is equal to the inverse of the square of the ratio of the velocities. If m1 is twice the mass of m2 then v1 is .707 that of v2 if v1 is 1 m/s and m1 = 1kg then v2 is 1.414 m/s and m2 =.5kg comparing momentums [math]1kg(1\frac{m}{s}) = 1\frac{kgm}{s}[/math] and [math].5kg(1.414\frac{m}{s}) = .707\frac{kgm}{s}[/math] The larger mass (m1) has a greater momentum than the smaller mass (m2) when both are at the same energy. No relativistic effects involved.

It depends on how much matter you are talking about. For instance, if you had 9^16 joules to work with, you could have one electron moving at just slightly less than c, 1kg moving at .866c or 100 kg at .14c.

When talking about the efficiency of a rocket, we are generally dealing with how much delta V we can get from our fuel/reaction mass. For this we use the rocket equation [math]\Delta V = V_e ln(MR)[/math] where Ve is the exhaust velocity and MR is the mass ratio, (the mass of the fully fueled rocket divided by the mass of the unfueled rocket) Note that for any given mass ratio, the dela v depends on the exhaust velocity. The above equation works well for every day velocities, but for relativistic velocities you need to use the Relativistic rocket equation: [math]\Delta V = c \tanh{\left (\frac{V_e}{c} \ln{(MR)}\right)}[/math] Again we note that the delta v depends on the exhaust velocity. We get the best efficiency if we have an exhaust velocity of c. With such as "photonic" drive one could expect the following maximum delta v: for a MR of 2, we can get .6c for 3, .8c 4, .882c 5, .923c 6, .946c 7, .96c ... 100, .9998c

Yes it is. Or more correctly, if the mass of the planet is very small as compared to the mass of the Sun then the contribution of the planet's mass can be safely ignored. When the planet's mass becomes large enough you need to use the following formula to get an accurate answer. [math] P = 2\pi\sqrt{\frac{a^3}{G \left(M_1 + M_2\right)}} [/math] How large is "large enough" depends on what degree of accuracy you need. For instance, with the Earth's orbit, ignoring Earth's mass causes a difference of only about 70 seconds or about 0.00015%. If you needed to calculate Earth's orbital period finer than this then you need to include its mass. (However, in this case you also need to know the Earth's average dist from the Sun, the mass of the Sun, etc to equal accuracies first)

What you are describing is known as a Doppler Shift. While light and sound both exhibit such shifts, it has nothing to do with the Relativistic effects described by Einstein. The effects of Relativity are what are left over after the Doppler shift has been accounted for. So, no, your method is not the easiest way to describe Relativity as it actually leads to an erroneous conclusion as to the nature of the theory.

Actually it would be about 1.1547 years later on Earth.

Er, that would have been a little difficult, since Kepler died in in 1630 and Newton wasn't even born until 1642.

Kepker's Law actually were purely empirical. They were based on observation of the planets in their orbits. He observed that the planets followed ellipitical orbits and he observed that their periods and distances from the sun were related in a particular way. He then formulated these observations as laws. It wasn't until Newton that these laws were mathemathically derived from his laws of motion and universal gravity.

Actually, you would expect it to work out that way. As has been pointed out, the force acting on the falling body decreases porportionally with the decrease in distance from the center. This implies harmonic motion on the part of the object. Harmonic motion can also be described as a projection of circular motion on the diameter of the circle. If you start with an object moving in a circle with a centripetal force of [math]f= \frac{mv^2}{R}[/math] You will find that the component of that force acting along the line of the diameter varies linearly. (as per harmonic motion). Since the centripetal force acting on a surface orbiting object and the intial restoring force of our falling object are the same, if you start the two objects at the same point at the same time, you will find that a line drawn through either object, and perpendicular to the path of the our falling object will always pass through both objects. Thus when the orbiting object reaches a point 90° from its starting point, the falling object has reached the center and both objects at that moment are heading in the same direction at the same speed.

Just one more point. It turns out that the speed at the center (for uniform density) equals [math]\sqrt{\frac{GM}{R}}[/math] Where M is the mass of the Earth and R its radius. This is also equal to the orbital velocity for an object at the surface of the Earth (ignoring air resistance)

No, CPL's figure for the speed is correct. It's what you get when you take the potential energy difference between the center of the Earth and the surface, and then figure out the speed needed to equal this in kinetic energy for a given mass.

Here's the problem: Recent evidence points to Mars still being geologically active. Any past civlization on Mars would have likely died out before all the water left Mars. erosion and geological activity mean a rock cycle, which means that Mars' crust could have gone through a complete recycle since said possible civilization thrived, leaving no traces.