## The Planet Uranus in Opposition

Note: I hope to update this piece when I catch Uranus at the turning point, August 2018. As of right now we are in thunderstorms, so it may take a while.

## Okay Let’s Cut to the Chase

Here is an animated GIF of two astrophotographs I took of the planet Uranus from my house in Virgin, Utah. The first one was taken on October 24, 2017 around 2:00 am, the next on November 19,2017 at 10pm (click on the image for full-size animated versions). See if you can spot Uranus. In the course of that month it has moved a bit, near the center of the image, so you should be able to see a blue-green dot jumping back and forth.

One-month TIme-lapse of Uranus. (Nikon D-5000, F9 3 minute exposure, equatorial mount)

If you still can’t catch it, here is an annotated versions, with labels and stuff (again, click on the images for the full screen version):

In addition to the dated labels, I have put in some graphics showing the constellation Pisces, as well as a chart, showing where computer models say Uranus should be in that part of the sky, for various dates between 2016 and 2020. I had to pull all of these other things in, just to convince myself that I really caught the planet, and not just a random earth satellite or other transient object.

It has taken me quite a bit of work to get to this final product, of which I am quite proud, and happy that it came out so cleanly, riding exactly along those predicted lines. In August of 2018 (now) I hope to capture that endpoint of maximum extent. The rest of this blog piece is the retelling of the story of this image, along with the occasional digressions into the geometry of the whole thing.

## About the Planet Uranus

Uranus (Wikipedia) Voyager II photo 1986

Here is a picture of the planet Uranus, taken by the Voyager II spacecraft in 1986.

By the time I came to work at the NASA Jet Propulsion labs in ’87 the Voyager II probe had already passed by Uranus and was approaching Neptune, so I never got a chance to see these “live” images coming in. It’s not much to look at, and is best described as a large ice ball (unlike Saturn or Jupiter which are mostly gaseous). Even with a really good earthbound 8″ telescope, you’re not going to see much more than a fuzzy dot.

Though it had been seen before (even in ancient times), the object was not identified as a planet until it was observed and reported by William Herschel in 1781, who thought it might be a comet. However, after reporting it to the Astronomer Royal Nevil Maskelyne (who figures prominently in the quest to measure Longitude), Maskelyne concluded that it was probably a planet.

Other than its name (being the only one in the solar system based on the original Greek gods, and not the later Latin names of the Roman gods), Uranus is notable for having its rotational axis nearly horizontal to the orbital plane, so that for half the Uranian year (about 45 earth years) the “north” pole is in perpetual day, and the other half the year is perpetual night.

## Uranus in Opposition

What started all this for me was the announcement last month that the planet Uranus was in what they call “opposition“, meaning that it was on the opposite side of the celestial sphere (as seen from Earth) as the Sun. From the Sun’s perspective, this means that the Earth and Uranus are on the same side of the Sun, and typically on closest approach to each other:

Planetary Opposition (source: Wikipedia)

The news in social media suggested that it would be so close that “it could be seen with the naked eye.” That sounded like hogwash to me, as there have been a lot of viral bogus memes around about being able to see things like the rings of Saturn and such.

Having now tracked down the planet, I can attest that — technically — it would be possible for a young person with excellent sharp eyesight to see the planet Uranus without binoculars … if they knew exactly where to look, and gazed at it out of the corner of their eye, and in a place (like where I live) with extremely dark skies and no cities nearby, but only on a cool clear night. But otherwise, forget about it.

## The Plan

Barn Door Equatorial Mount

Anyway, with the announcement of the opposition of Uranus in October I decided that this was a good opportunity to do some amateur astronomy and try to capture Uranus with some very low-tech equipment, which is a Nikon D-5000 camera mounted on a crude “Barn Door” equatorial mount. Using this mount, I can take long-exposures of up to 15 or 20 minutes, without smearing of the stars due to earth’s rotation.

## The Warp Equation

I plan on using the trekkie terminology (and standard relativity) to state and prove the following interesting fact:

 The Warp Equation If you have a payload with mass $m_{payload}$, and a means of converting matter into kinetic energy with 100% efficiency, then the mass $m_{fuel}$ of fuel needed for you to travel at an effective speed of Warp $\omega$ where $\omega > 0$ is given by$$m_{fuel} = {\omega}^2 m_{payload}$$

So for example, in order to travel at Warp 2, a person of mass 80 kilograms would require 320 kilograms of (say) a proton-antiproton fuel in order to travel at that effective speed. That is roughly equivalent to 6,400 Megatons of TNT. Coincidentally, that is almost exactly the combined explosive power of all nuclear weapons now on our planet. That is a hell of a lot of energy, but the point to be made is that is within the bounds of our current technology.

The fact that you have to square the warp factor to get the amount of energy to go that speed makes perfect sense. Even in classical Newtonian physics, the energy related to going at velocity $v$ is given by

$$E = \frac{1}{2}mv^2$$

so doubling the velocity $v$ on the right hand side multiplies the energy by four. The fact that the energy happens to be equivalent to four times your payload’s mass comes from Einstein.

The way in which we’ll prove this is to first calculate how much matter is needed to attain an observed velocity v, and then figure out what the relationship is between the observed velocity, and what effective velocity the passenger actually experiences. Note: I have no doubt that there is probably an easier way to derive this formula. But this is the one I came up with and it isn’t all that complicated.

## Conversion of Matter to Kinetic Energy

$$E=mc^2$$

What we are going to do is to use this equation, together with the law of conservation of energy, to compute how much matter it takes to accelerate a payload $m$ to (observed) velocity $v_{o}$. Now as the observed velocity $v_o$ approaches the speed of light, the relativistic mass of the payload becomes:

$$m_{relative} = \frac{m_{payload}}{\sqrt{1-(\frac{v_o}{c})^2}}$$

Now Einstein’s equation for energy represents both the energy of the mass at rest, together with the (kinetic) energy of the mass in motion. And so, if this mass was put into motion by the conversion (at rest) of a certain mass $m_{fuel}$, where

$$m_{fuel} = \alpha m_{payload}, where \alpha > 0$$

Then since energy is conserved we can relate the conversion of the mass $m_{fuel}$ into motion $v_o$ by:

$$(m_{payload}+m_{fuel})c^2 = E_{rest} = E_{moving} =\frac{m_{payload}}{\sqrt{1-(\frac{v_o}{c})^2}} c^2$$

so dividing both sides by $m_{payload}c^2$

$$1 + \alpha = \frac{1}{\sqrt{1-(\frac{v_o}{c})^2}}$$

squaring both sides and solving for $v_o$ we get the following rule:

 Matter to Velocity Conversion For a payload of mass $m$ and a ratio $\alpha > 0$, if fuel $m_{fuel}=\alpha m$ is converted to kinetic energy, the observed velocity $v_o$ of the body will be$$v_o = (\sqrt{\frac{\alpha}{1+\alpha}})c$$

This jibes with what Einstein said about observed velocities, as the right hand side will never be greater than the speed of light $c$. As the ratio $\alpha \rightarrow \infty$, the velocity goes to $c$, so we can get as close to $c$ as we like — but no further.

## Velocity – Observed and Effective

So now we come to the idea of “effective” velocity. The weirdness of relativity comes from the fact that as the observed  velocity $v$ of ship approaches the speed of light, the passenger’s own time-scale is compressed by what’s called the Lorentz-FitzGerald contraction, according to the formula

$$t_{effective} = t_{observed}\sqrt{1-(\frac{v_o}{c})^2}$$

(From this point on we will just write $t_e$ and $t_o$ for $t_{effective}$ and $t_{observed}$ respectively) Then given a fixed distance $\Delta x_o$ as measured by the observers on earth, the effective velocity as experienced by the passengers when traversing that segment of space over their time $\Delta t_e$  is:

$$v_e = \frac{\Delta x_o}{\Delta t_e} = \frac{\Delta x_o}{\Delta t_o\sqrt{1-(\frac{v_o}{c})^2}}$$

which in turn simplifies to this formula for converting observed to effective velocity:

 Observed to Effective Velocity $$v_e = \frac{v_{o}}{\sqrt{1-(\frac{v_o}{c})^2}}$$

Tech Note 2: “Effective” velocity is non-standard terminology. In the literature, this would be the velocity as measured by the passenger’s Proper Time.

## All Together Now

So if we start with fuel $\alpha m$ which we use to accelerate our mass $m$ to the observed velocity $v_o$, we can use the two formulas we just derived to express the effective velocity $v_e$ as a function of $\alpha$. We can rewrite the “Matter to Velocity” formula as

$$(\frac{v_o}{c})^2 = \frac{\alpha}{1+\alpha}$$

So our effective velocity formula simplifies the bottom of the fraction to

$$v_e = v_{o}\sqrt{1+\alpha}$$

and then substituting the formula again for  $v_o$ we see that our fuel mass $\alpha m$ gives us an effective velocity of:

$$v_e = c\sqrt{\alpha}$$

Thus if we have defined velocity “Warp $\omega$” to be $\omega c$, then we can write

$$\omega = v_e / c = \sqrt{\alpha}$$

So that to attain an effective velocity of Warp $\omega$ we must use a fuel-payload ratio of $\alpha = \omega^2$, ie

$$m_{fuel} = \omega^2 m_{payload}$$

which is exactly the “Warp Equation” we were to prove. QED

## Let’s Do the Time-Warp Again

It should be pointed out that of course to the observers on earth, even though you are going at an effective speed of Warp $\omega$ you will never appear to be going faster than $c$ and so it will take you a long time to get where you are going. You, however, will not experience that, and so you will effectively be travelling through time much faster than your friends at home. How much faster? According to our formula above relating $t_e$ to $t_o$, and expressing that in terms of the warp factor $\omega$, we can show that the time-warp you experience will be:

$$t_e = \frac{t_o}{\sqrt{1+\omega^2}}$$

And so, in our example, the 80kg person travelling at Warp 2 will feel like they’ve reached their destination in $1/ \sqrt{5}$ of the earth time, ie getting them there in about 0.44 of the time observed on earth, and exactly twice the time it would take light to appear to get there.

So, not only can you go as fast as you like, you can also travel as far in the future as you like. For example, to travel 100 years into the future, just get in a spaceship armed with 10000 times your own mass in matter-antimatter fuel, and then travel at Warp 100 for one year. When you reach your destination, one year will have passed for you, and 100 years (plus a little bit) will have passed on earth.

Of course, then you’ll have to get back to earth, so good luck with that.

## The Buckaroo Banzai Principle

 The Buckaroo Banzai Principle No matter where you go — there you are. — Buckaroo Banzai

The point of this exercise is that if you really understand what Einstein said, the idea should be that there is no absolute frame of reference. What this means is that even if you are travelling at 99.999 % the speed of light relative to the earth, as far as you know everything still looks and feels like Newton’s physics, where F = ma and you can always accelerate faster and faster. And not only that, but if you are heading for a specific location, the faster you go, the faster you will get there.

## No Free Lunches

Now having said that, there are some consequences that the universe may unleash should you decide to try to go Warp 100. This is because even though the physics of your spaceship will be the same even at this insane speed, you are also surrounded by the gases in your local galaxy, as well as all of the light from stars that are visible to you. And even though from the earth much of this light is nice, low-energy visible spectrum, and even though that light will still be reaching you at the speed of light, it’s relative energy is radically different when you are plowing through that light at Warp 100. In fact, what you will be observing is a massive Doppler-shifting into the deep blue/ultraviolet of all light coming at you in the direction you are headed (and conversely, red-shifted looking back towards earth). Some of this light may be equivalent to the powerful cosmic rays that hit the earth, and which were generated by massive explosions or quasars just after the Big Bang. The energy in these photons may be enough to kill you all by themselves, especially at Warp 100. You may need a very large and thick radiation shield, along with all the extra energy to carry that shield along with you and your ship.

And so as we already should have known, there are no free lunches. At least it is nice to know that a faster-than-light lunch is available, should one choose to pay the price.

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