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[MetalSlimeHunt]

No, the atmosphere has not been removed. The gist of the question is "under optimal conditions, can the Moon's gravity alone kill things on the surface"?

[Sheb]

Would be interesting to calculate the tides if the moon was, let's say, 400 km from the surface.

[Il Palazzo]

Why this:

Spoiler (click to show/hide)

Explain plox.

Loose screws holding the front wheel. Or, more likely, intentionally removed, so as to make a Jackass-style video.

Edit: Anyway my question. The Earth's rotation is slowing down. Now, this is obviously not very good, so let's stop this. And for fun and profit, let's assume that we're using an equatorial mounted laser to do so. How much power would we need to emit.

Tidal braking currently changes the length of day by ~2 miliseconds per century.
Here, as in all that follows, I'll round off all the numbers to powers of ten, so the above change of the length of day will be: ΔT=10^-3 seconds.

The total change in angular velocity is Δω=ω₁-ω₀
ω₁ = angular velocity after 100 years' braking in radians
ω₀ = initial angular velocity
and since
ω=2∏�/T
then
Δω=2∏�/(T₀+ΔT)-2∏�/(T₀)
T₀ = length of day before braking in seconds
T₀+ΔT = length of day after 100 years' braking
Δω=-2∏ΔT/(T₀^2+T₀ΔT)
which equals roughly -10^-12 radians/sec

The angular acceleration α=Δω/T₁₀₀
where T₁₀₀ is 100 years in seconds, giving roughly α=-10^-21 radians/sec^2

The torque(τ) required to produce such angular acceleration is:
τ=Iα
where I is the moment of inertia of Earth, approximated as 2/5m_ER_E^2, which gives roughly I=10^38 kg m^2
and the torque is then in the vicinity of τ=-10^17 N m

The torque required to counter the tidal torque is then τ=+10^17 N m
at the equator, the lever arm is equal to the radius of the Earth, so
τ=R_EF
(for the laser firing perpendicular to the radius, for maximum efficiency*)
and the force required to negate the tidal torque equals
F=10^10 Newtons

From Newton's second law, force is the rate of change of momentum:
F=Δp/Δt
the momentum of a photon or a flux of photons is given by p=E/c
Power of a laser is the total energy of photons produced by it per unit time P=E/Δt
so, the exiting beam will change the momentum of the laser assembly by Δp=PΔt/c
and the power of the laser required will be:
P=Fc
or something in the order of magnitude of 10^18 Watts (1 exawatt, or ten times the total solar irradiance we get)

Somebody check my work

*assuming no atmosphere. With atmosphere, there's most likely a more efficient angle to aim the laser at, to mitigate the atmospheric attenuation. But then again, the atmosphere in the path of the beam would probably vaporise anyway, so maybe that's no big deal.

[misko27]

Well, if we assume the atmosphere has been removed to simplify calculation, we can safely say that life will suffocate.

This reminds me of my paper proving Geocentrism:
"If the moon can eclipse the sun, how does the sun not eclipse the moon? Clearly, the moon must orbit the earth, and not the sun. Score one for Geocentrism."

Later in the same paper: "There is a very, very large body of evidence originating from man-made satellites that support Heliocentrism. But I already proved the heliocentric model is wrong, so the real question is, why has this date been falsified?"

[Mageziya]

Let me pull a scenario out of the air for this.

In a pure argon atmosphere, at what speed, in nanometers per year, would a tungsten sphere, with a radius of 17 decimeters, need to initially move to pierce through the Rocky Mountains at 300 feet above sea level, starting from the south-east most part and ending at the northwest most part? All of this is assuming that the sphere is indestructible, travels in a straight line, the atmosphere is 2.11 times thicker than normal, and that the gravity is 7/6th of Earth's normal gravity, while Earth is the same size.

Bonus question, how much energy, in Kilocalories, would be required to perform this?

Hey, you asked for insane physics questions.

[Tsuchigumo550]

I wonder if there's a way to calculate how much energy it would take to run a mecha for a mission time of say, an hour, if it would be possible with anything but nuclear, and what kind of hellish weight it would have assuming:

The mecha stands at about 5 meters tall.
The mecha uses some sort of modified bipedal movement system, allowing it to move faster across flat terrain using wheels or skates
The mecha is unarmed/unloaded
The mecha is able to fit within the space of two lanes of highway cleanly
The mecha could be powered by some sort of system able to fit on it's back not accounting for more than 40% or weight
The mecha can utilize materials such as carbon fiber, fiber optics, and technology such as flywheels, but the outer armor must be tank-grade metals

There needs to be more data than that, obviously, like projected weight of parts. I'm sure there's a game out there that says something or someone has a good guess as to what everything weighs.

[wierd]

I would need to know what movement speed you are considering--

But at only 5 meters, without any weapon systems, and without actually hauling/pulling anything-- you could probably power it on diesel electric.

It would still be very heavy.

The problem with "Gigazord" mechas is that the metal itself has a sag weight. There is a point of in-feasibility for function after certain scales. You would need a non-existent material from which to construct the Gigazord.

For a long time in avionics, a material with the properties to titanium was GREATLY sought after but was unavailable. As such, it was colloquially called "Unobtanium", because it could not be obtained. Now it is commonly used, and its availability has opened the doors to such things as "Super jumbo jets", like the 787.

A 5 meter mecha is within the realm of "Possible", as long as it isnt doing anything particularly crazy (which would have crazy shear and flex forces applied to the materials).

Granted, such a diesel-electric mecha would not be very fast moving, and would employ electric motors to drive hydrualic systems to actuate the limbs--- very slowly.

[Tsuchigumo550]

The idea is that they would have a secondary movement system allowing them to move faster than a tank across flat concrete, and be able to climb uneven terrain and cross shallow water more easily than a tank would.

Meaning, they should have at least tank speed and a half in urban enviroments and be able to move at least half tank speed elsewhere.

This is assuming "heavy tank" speed, so a lighter tank that's faster than other tanks isn't baseline.

Let's assume we're using mounted weapons- no arms. Less actuators. Sag weight could be a problem, so let's also assume that these are drones rather than human piloted and the "core" contains only the control software and the ilk, and therefore, the entire build is basically legs.

[Karlito]

Would be interesting to calculate the tides if the moon was, let's say, 400 km from the surface.

See, at that point there wouldn't be any earthly tides (except those caused by the sun), because the earth's gravity would rip the moon apart and we'd get a nice moon rock ring around the planet.

[alway]

Would be interesting to calculate the tides if the moon was, let's say, 400 km from the surface.

See, at that point there wouldn't be any earthly tides (except those caused by the sun), because the earth's gravity would rip the moon apart and we'd get a nice moon rock ring around the planet.

The minimum distance from earth would thus be approximately 1.06 * earth's radius; center point would be 1.33. Assuming stable orbit of center of mass and no rotation, you would get an acceleration towards earth of 3.18m/s^2; subtract out the moon's surface gravity, and you get about 1.5 m/s^2 acceleration away from the moon's surface and towards Earth (approximately the moon's surface gravity, actually), at the closest point. This decreases the closer to orthogonal you get. So large quantities of material would be pulled inwards to Earth's atmosphere.

On the far side, you are 1.6 times Earth's radius out. As you are travelling faster than the required orbital velocity, assuming the center of mass is in a stable (lower) orbit of 1.33 times Earth's radius, you would thus experience an outwards pull of 1.7m/s^2. Again, this decreases the closer to orthogonal you are. Subtract out the moon's gravity, and you would be very slowly drifting away from the moon's surface at  0.1m/s^2.

All in all, you would get a very large quantity of the closer face of the moon pulled into the earth's atmosphere. A small amount of material would be flung outwards away from the earth on the far side, but it likely wouldn't have much effect. Especially because....
By asymetrically losing mass that was closer to the earth, the Moon's center of gravity would move outwards. The stable orbit at a constant altitude would become more elliptical. After a little while, most of the moon's mass would be lost, and a much smaller moon would eventually exist in a stable orbit, after losing sufficient mass that its surface gravity was equivalent to the difference from the gravity gradient.

At this point, the surface of the earth would probably be superheated along a wide, and expanding path, seeing as much of the moon will have just decelerated from 17km/s in the atmosphere.

[Sheb]

the atmosphere in the path of the beam would probably vaporise anyway

Our laser is so powerful it instantly turn the atmosphere into a gas!

[Andrew425]

How many calories a day would someone like Captain America be forced to eat to be able to fulfil a combat role?

On a calorie per pound weight, who can carry more food to sustain themselves, a normal fit person or the Captain?

[Tarran]

Question: Let's say the dwarf planet Pluto suffered some catastrophic event and was teleported next to Earth (with no changes at all to the make of Pluto). Let's also say that Pluto somehow got the right speed and travel direction to orbit the Earth.

How close to Earth can Pluto relatively stably orbit (I.e, nothing bad will happen in--and Pluto won't be flung away within--and Pluto won't disintegrate within--a million years or more) the Earth, without adversely affecting life?



Second question: Let's say Pluto was teleported right onto the surface of Earth so as a relatively flat part of its surface touches a relatively flat surface of the Earth and it is (initially) moving in the right direction and right speed to stay roughly where it was teleported to (relative to the Earth) for at least three seconds. What would happen?

[wierd]

Since earth already has a moon that is of considerable mass compared to the earth, this is going to become a 3 body solution problem.

[MaximumZero]

What would happen to Pluto if it were put in Earth's range from the sun, say, in a Lagrangian point? Would it suddenly have shitloads of water and a thick atmosphere due to the sun's heat that it was deprived of as a child?