# What would it take to make a scientific second equal exactly one traditional second?

A scientist is resetting the clock on his microwave one day as he considers the hopelessness of keeping the time exactly right. Not for the reasons we worry about like power outages and daylight savings times, but because in the back of his head, he knows that Earth's movements through space are not properly standardized for good time keeping system.

This makes him unreasonably mad, so he decides that the best way to correct for this aberration is to simply alter the spin and orbital period of the earth so that a day is exactly 86400 seconds and a year is exactly 365 days based on caesium frequencies so that he never has to worry about converting units again.

Although the historical definition of the unit (seconds) was based on this division of the Earth's rotation cycle, the formal definition in the International System of Units (SI) is a much steadier timekeeper: it is defined by taking the fixed numerical value of the caesium frequency ∆νCs, the unperturbed ground-state hyperfine transition frequency of the caesium 133 atom, to be 9192631770 when expressed in the unit Hz, which is equal to s−1.1[2] Because the Earth's rotation varies and is also slowing ever so slightly, a leap second is periodically added to clock time[nb 1] to keep clocks in sync with Earth's rotation. ~ https://en.wikipedia.org/wiki/Second

So, our mad scientist devises a two step plan to unify metric and traditional time once and for all! The first stage it to use a series of powerful explosions to speed up/ slow down the Earth's movements to make days and seconds the right lengths, the second it to install propulsion systems on the Earth to keep it moving at these speeds indefinitely.

The Question: How much force (and in what directions) does the scientist need to exert on the Earth to achieve his goals?

Bonus points if new seconds actually = scientific seconds when rounded to the level of a double floating-point number, but I would be surprised if anyone could actually find measurements on the Earth accurate enough to do this, so no pressure.

How long is indefinitely?

If additional thrust needs to be added over time, the scientist trusts future generations (assuming he hasn't killed everyone) to continue his work. His initial thruster just needs to be strong enough to make sure that he doesn't start seeing desync start creeping back in before his own end-of-life. If thrust needed decreases over time, assume his thruster can throttle down to compensate.

the explosions and "propulsion" would probably devastate the biosphere at worst it might even generate enough energy to melt the crust and boil the oceans to a significant extent.

Devastation does not necessarily need to be addressed for purposes of this question unless it involves there not being an Earth left to have a day/night cycle.

• I see you've asked for precision rather than accuracy. I'm pretty certain I can manage that ;-) Nov 21, 2019 at 18:21
• It won't be pretty, at best the explosions and "propulsion" would probably devastate the biosphere at worst it might even generate enough energy to melt the crust and boil the oceans to a significant extent. I won't bother with the calculation as Starfish Prime has it in hand, we shall see. Nov 21, 2019 at 18:32
• @Slarty oh gosh, don't hold out on my account; I'm doing a bunch of other things at the same time, so it'll take mne a while to assemble my answer. Nov 21, 2019 at 18:33
• World devastation is an acceptable consequence when it comes to solving important issues such as this. Nov 21, 2019 at 18:38
• The real problem isn't fixing the Earth's rotation now, it's how to keep the rotation constant after you've fixed it. Otherwise you're going to have to keep making adjustments... Nov 21, 2019 at 18:44

Lets first think about how much energy this needs. You've asked for spurious precision, but I'll save that til the end because no-one wants to see all the tedious decimal places in the workings (and if they do, they can repeat the process themselves).

You want an orbit with a period of precisely 365 days, each 24 hours long. Via Kepler's third law, we can see that this will need to reduce the semimajor axis of Earth's orbit by about 71950km.

At its perihelion of its current orbit, earth has a velocity of about 30.2868km/s. In its new orbit, with the same perihelion (and hence a reduced aphelion) it will have a velocity of more like 30.2797km/s. Given Earth's mass, that will require its kinetic energy be changed by a bit over 9.02 x 1031 joules. I'm not really certain where you'd get this much energy from... it is nearly two orders of magnitude more energy than the kinetic energy of Mars if you crashed it into earth at 4km/s (about the speed of the hypothesized Theia impact), and more energy than you'd get from all the solar radiation falling onto Earth in about 16 million (old-style) years. If anyone has any suggestions on where to source 500 billion tonnes of antimatter, that'd be great.

It is also about 2/5ths of Earth's gravitational binding energy, meaning that it if it were not released carefully over an extended period of time you'd reduce the planet to a ring of gravel orbitting the sun. Releasing the energy slowly and carefully enough will probably take entirely too long for anyone's attention span.

(energy change would be ~9.015096928089181 x 1031 joules)

So much for the year. What about the day?

The angular kinetic energy of the Earth (using the Lambeck 1980 figure for the Earth's moment of inertia about its polar axis from here) is about 2.136 x 1029J. Speeding up the Earth to give it a nice round 24 hour day requires an angular KE of more like 2.124 x 1029J, giving a required oomph of about 1.165 x 1027J, a much more manageable figure as I'm sure you'll agree. Please use caution releasing this much energy in the atmosphere all at once, because whilst it isn't quite enough to vapourise our oceans, it is more than enough to boil them, and the clouds of hot steam will spoil the view.

I did have a look at imparting this energy using a train of carefully aimed asteroids, with trajectories in the Earth's equatorial plane, hitting the equator at an optimal 15 degrees angle. Unfortunately the plan started to resemble a re-run of the Hadean era, and the inefficiencies of using explosions or rocks to change the Earth's rate of rotation resulted in a lot of waste heat and seemed lamentably inefficient. There may or may not be any oceans or atmosphere afterwards, but the clouds of dust and debris and subsequent re-entry probably preclude any appreciation of a day-night cycle for some time (possibly thousands of years).

(energy change would be ~1.1648246454801083x 1027 joules)

So much for the day. Can we just deal with the changing day length, if nothing else?

Turns out that no-one can seem to say anything useful about exactly how much deviation you'd need to correct for... the state of ΔT is woefully confusing. The day length only changes by milliseconds per century, but the leap seconds keep on coming.

Lets just look at a system that can manage to change earth's day length by a second (because I'm despairing of getting anything to work).

This requires adding ~4.96 x 1024 joules of angular kinetic energy. By a happy coincidence, this is a little under the total amount of solar energy that strikes the Earth every year (more like 5.5 x 1024 J). Using a rocket to do this needs 1.57 x 1017W of useful thrust. Given efficiency issues, it will not alas be practical to resurface the Earth with solar panels and use the Earth's own oceans as reaction mass, but it is soooo close.

I have a final alternative plan for you. A lot of our problems are caused by the moon. It takes ~7.62×1028J to throw that rock into deep space, where it will never offend your eyes or your day length again. Just say the word, and we'll draw up a plan for you...

(rocket thrust power would be ~1.571089676036397 x 1017 watts)

• Bravo @Starfish Prime, gorgeous answer. Nov 22, 2019 at 1:22
• I love the well thought out logic of this answer, but there seems to be a math error in step one that makes things look more bleek than they are. 30286.8 - 30279.7 = 7.1m/s. Earth's mass is ~ 5.972 × 10^24kg so then energy change should "only" be 1.20524*10^26J . I have not checked all your math, but you may want to double check it all just to make sure. Nov 22, 2019 at 15:20
• @Nosajimiki So, I think I see where the issue arises.The energy I quoted was the difference between the kinetic energy at the perihelion in the old orbit, and perihelion at the new orbit, and I'm reasonably certain that this change in energy is the correct figure to think a about. The energy cost of a given $\Delta_v$ depends on the object's current velocity... have a look at this physics.SE question covering the same subject. Nov 22, 2019 at 17:55
• Ah, I see where that is coming from now. Thanks. Nov 22, 2019 at 22:07

Explosives on the Earth's surface, no matter their power level (short of ejecting significant chunks of the crust) will never change the Earth's rotation rate or orbit. Nor will a reaction drive of any kind -- with the exception that if its exhaust, after exiting the atmosphere, is still above Earth's escape speed, some tiny fraction of its thrust will act to change the Earth's velocity.

Modern times changes in Earth rotation rate have been attributed to changes in the amount of water captured behind dams (hence further from the Earth's centroid than its natural height average), melting of glaciers or ice caps, and (very rarely) land movement due to tectonic events (major eruptions and earthquakes). Your mad scientist need "merely" alter the proportion of water trapped in the polar ice caps relative to the oceans in order to take and maintain very fine control (on the order of microseconds alteration in the day length) of the Earth's rotation.

Now, to change the orbit will require going off Earth. The most likely way to accomplish this (to cut around a quarter day off the period -- ideally without changing the eccentricity or ecliptic plane) would be to attach large mass drivers to a biggish asteroid (Vesta, perhaps), use them to drive it around the Solar System, and then use the asteroid as a gravity tug to subtly change the Earth's orbit.

Whether the mad scientist can gain the required precision in altering either the Earth's rotation or its orbital period is up to him/her -- but with good enough computers and software, and a willingness to spend multiple decades on the project, he can quit having to deal with messy numbers of seconds in a day or year -- at the expense of making every astronomer alive an enemy. If he's careful, he could probably accomplish the whole project without a single (attributable) casualty. If not, he might kill a few million with the mass driver exhaust, and a hard-to-count number due to climate changes produce by or required to manage the ice cap project.

The Earth is rotating too slowly for our scientist's liking, and it's also getting slower all the time due to gravitational tidal drag and other factors. This is currently happening at a rate of about $$\mathrm{7.3×10^{−13} day/day}$$, which is also the fraction by which the Earth's angular momentum needs to be topped up. Let's specify the propulsion system to be able to compensate for drift up to $$\mathrm{10^{−12} day/day}$$ for proper redundancy and future-proofing.

The Earth's total rotational angular momentum is

$$L = I \omega = \frac{2}{5}MR^2 \times \frac{2\pi}{86400} \approx \mathrm{7.2\times 10^{33}\ kg \ m^2 \ s^{-1}}$$

And we need to be able to change this by one part in a trillion. In order to get on to torque we need to decide how long the scientist is willing to wait to apply this correction. Let's say he's moderately impatient and wants it to apply over 1000 seconds, or $$\mathrm{10^{−15} day/day/s}$$ (yes those units are getting a bit crazy now). We must therefore be able to apply a torque to the Earth of approximately $$\mathrm{10^{19} N \ m}$$. As noted the best way to do this is actually to move large masses of water closer or further from the Earth's rotation axis, but you've specified explosions, so let's go with that.

We do our explosion on the correct place on the equator and somehow manage to focus it so that all the debris is ejected directly backwards. It's really important that all the debris reaches escape velocity or it doesn't actually contribute a net change to the angular momentum, just sloshes it around in time, so let's say all our debris ends up moving at $$\sim \mathrm{10^4\ m \ s^{-1}}$$, so from our launch site at $$R \approx \mathrm{6.4 \times 10^{6}\ m}$$ each kilo of debris contributes $$\sim \mathrm{10^{10}\ N \ m}$$, meaning we only need to launch $$\sim \mathrm{10^{9} kg}$$ of material at each correction. Simple, that's just a lump . Handling the collateral damage from that is left as an engineering exercise.

• Don't forget the year adjustment -- he wants to get rid of the .2404 (IIRC) extra day in the year, too, so as to eliminate pesky leap years (never mind having to remember which century years are, and which aren't leap years). Nov 21, 2019 at 20:10

Simply put, sane or not, if he were any scientist worth his salt, he'd understand that he cannot make a day any closer to the 86400 seconds that it currently is defined as.

How precise can we be?

The length of the year is ~365.2422ish days. This is the oft-cited duration of a tropical year, the "mean time between between vernal equinoxes", but is in fact arguably wrong (see https://www.hermetic.ch/cal_stud/cassidy/err_trop.htm for the drama inherent in this issue). The time to orbit the sun is ~365.2564 days, and now you're already deeply into the weeds and your scientist is going mad asking himself "so what IS a year? What do I want to align TO?"

And you can't get more precise than those four significant figures (well, five if you're optimistic, some use 365.24219, to specify it to the nearest second, but then it depends where you measure from). Any more digits simply aren't meaningful, because it varies by a few fractions of a second each year, due to chaotic atmospheric effects, geological effects (mantle convection, glacial rebound, etc), the rotation of the earth slowing (due to tidal friction, etc), and more.

This is why we occasionally get leap-seconds.

How much energy do we need?

The rotational kinetic energy of the planet is around 2.138 * 10^29 Joules.

To convert 365.2422ish days to to 365.0000 precisely, would require about a 0.066% change to that energy. We need to find 1.4186948 * 10^26 J, and apply it to slow down the rotation of the planet.

Can we do it with a gravity tug?

A gravity tug seems to have the same problem as moving the earth using mass-drivers to eject mass: it requires more energy than we have either on the earth or on the tug.

Can we do that with a solar sail?

Sunlight hits every square meter of our planet at ~1000 Joules/second, so mirrors placed around the equator so that there was always a 1m square mirror reflecting the sun on the side seeing the sunrise, you'd get a retardation force of that much. Over a year, that'd be 3.1536 * 10^10 J per year, which means you'd have it to the right speed in 10^14 years. If you made the mirrors 10km square, or 100,000,000 square m, it would still take a million years. OK, solar sails aren't the answer.