Astronomy/Physics question

[Nadrakas]

Re: Astronomy/Physics question

 

584' date='387 times zero is still zero.[/quote']

 

Agemegos,

 

Sorry but I should have explained myself better. The "quote" you are referring to is not a math equation. Instead:

 

- Difference between the Volume & Mass of a 100 km radius "Jump" bubble and the Alien Ship is:

-- Volume = 5,235.98775625 times greater than the volume of the Alien Ship

-- Mass = 584,387.0853125 times greater than the mass of the Alien Ship

 

So...nothing is multiplied by "Zero" so nothing is "Zero" as there is no equation in the above. Just a comparison.

 

 

Granted, whether the volume is 800 km^3 or some bigger volume, the effect would not cause the moon to collapse, explode, fly out of orbit, or crash into the Earth. Anyone living upon the Moon would suffer Moonquakes, which could breach the living structures upon/inside the Moon & there might be some magma escaping upon the surface. Not much effect upon Earth, but the local effect upon the Lunarians would be huge. Plenty of opportunity for PC Hero's to save innocents and do their Superhero Gig.

 

 

Now...Allow me to posit these two scenarios as an aside by taking into account the Energy effect of the FTL drive of the Alien Ship on the moon. I was wondering if the FTL drive might release energy during it's activation or if it might create a Worm Hole...so:

 

Q-1: Does the FTL "Bubble" cause an energy shockwave?

Q-1-2: If so...how much energy release from the FTL "Bubble" would be needed to "crack" the moon (not destroy it...just create huge fissures and the like).

Q-1-3: What other affects do you see happening?

 

 

Q-2: Does the FTL drive create a Worm Hole?

Q-2-1: If so, what affect would that have on the Moon?

Q-2-2: Would it create a distabilizing affect at the core that the simple disappearance of mass would not?

Q-2-3: What other affects do you see happening?

 

 

 

I only ask these in order to give some interesting events for Trebuchet's game. Afterall...a good story needs some challenges!!! And this is an RPG...not a Physics or Math course. Though the Physics & Math are fun (shudder...was that me saying that????)

 

 

Also...another totally unrelated question (Personal campaign thought here): In 3K, Mechanon 3000 (Page 71-74) is listed as being 1,000 km diameter (500 km radius). What effect would he have on the Earth/Moon system? (Besides releasing his robotic forces to destroy all living beings of course...)

 

 

 

Nadrakas...

[SirViss]

Re: Astronomy/Physics question

 

...

v^2 = G.M/r

 

r = v^2/(G.M)

...

 

Just want to point out that is the formula that I arrived at too, on the 1st page, but it should be:

 

r = (GM)/v²

 

shouldn't it?

[keithcurtis]

Re: Astronomy/Physics question

 

Indeed not. To hit the Earth the Moon would need to lose nearly all of its orbital velocity.

Exactly so. It always used to bother me that whenever SA Superman threw an object into the sun, he would aim directly at the sun. You need to aim ~90° away from the sun, back along your orbital path so you cancel your orbital speed. Then it will fall towards the sun.

Of course, SA Superman could throw things at light speed, I suppose, so he could aim "slightly to the left of the sun" as it were and still hit it. Physicists in comic book universes must get headaches a lot.

 

Keith "Science!" Curtis

[austenandrews]

Re: Astronomy/Physics question

 

I think Nadrakas has the right idea. We've more or less established what will happen in minimalist terms. But if you're throwing around enough energy to make an object that big vanish and reappear in a distant star system(presumably that's an astronomical amount of energy) you can arrange for any number of side-effects, and as catastrophic an end result as you care to invent.

 

You may even be able to throw the moon out of orbit or crash it into the Earth, if you want. I bet we could collectively figure out how that could happen.

 

Also...another totally unrelated question (Personal campaign thought here): In 3K' date=' Mechanon 3000 (Page 71-74) is listed as being 1,000 km diameter (500 km radius). What effect would he have on the Earth/Moon system? (Besides releasing his robotic forces to destroy all living beings of course...)[/quote']

What's the mass of the thing? That's a big robot, obviously, but again if it's mostly empty space, it might affect the planet more like a big meteor swarm than a planetoid.

[Phil]

Re: Astronomy/Physics question

 

I'm sorry, but that is rubbish. The centripetal force required to keep an object moving along a give curve is proportional to its mass: the formula is f© = m.v^2/r. The force of gravity on the object is also proportional to its mass: the formula is f(g)= GMm/r^2. Since gravity is the centripetal force:

 

f© = f(g)

 

m.v^2/r = G.M.m/r^2

 

The two 'm's cancel out, as do two of the 'r's.

 

v^2 = G.M/r

 

r = v^2/(G.M)

 

The [instantaneous) radius of curvature of an object's orbit is independent of the object's mass.

 

My turn to step in. OK, I understand this, but the query I have is whether the sudden loss of mass does not of itself constitute the acting of a force upon the moon? I can accept that in two parallel universes a hollowed moon and a non-hollowed moon would behave identically, but I'm it certainly seems counter intuitive (a term therefore suggesting that it's probably good physics!) that the transition of solid to hollowed would have no effect.

 

For example, any change in velocity owing from a reduction on mass is not going to be instantaneous (or will it?), whereas a reduction in mass does have an instantaneous gravitational effect. Now, I'm not a physicist, so I'm merely asking the questions, but I just wanted to make sure we werent comparing two steady-states when we need to consider the transition.

 

Also, just a minor point, but the calculations as to moon mass at the centre are surely likely to be slightly out, because of the greater pressure and thus greater mass density at the centre of an object? I'm sure it's calculable, and possible on an object the size of the moon it isnt significant, but wanted to flag it up!

[keithcurtis]

Re: Astronomy/Physics question

 

Being able to cause a mass of that amount to just... disappear is a devastating weapon in its own right. If you could do that to the surface of the earth, you could certainly destroy civilization, at least cause a mass-extinction event and possibly end all multicellular life on earth.

The Death Star always seemed like a ridiculous amount of overkill to me. Why destroy the whole planet when you only need to destroy the outer 1-2% to effectively kill everyone and destroy every human artifact? The difference in energy consumption has got to be by several orders of magnitude.

 

Keith "leaning on the derail switch" Curtis

[austenandrews]

Re: Astronomy/Physics question

 

For example' date=' any change in velocity owing from a reduction on mass is not going to be instantaneous (or will it?), whereas a reduction in mass does have an instantaneous gravitational effect. Now, I'm not a physicist, so I'm merely asking the questions, but I just wanted to make sure we werent comparing two steady-states when we need to consider the transition.[/quote']

That's where the fun begins. Even if we discount side-effects from the method of vanishing, there are still some weird little details to consider. Whether they amount to any changes in the noticable decimal places, I don't know. But consider:

 

- The vanishing ship is not spherical, but a rectangular slab. The collapse of core material into the resulting void will be more energetic along the axis of greatest length. If any "bubbles" are going to rise from the center, they will probably do so along this axis. This will likely mark the areas of heaviest moonquake.

 

- If this axis is oriented along at an angle to the moon's rotational axis, a tiny wobble might result. If it's oriented at a certain angle to the moon's orbital path, expelled debris might collect at L5.

 

- The seismic disturbances will probably be greater on the side of the moon facing the Earth. The crust is thinner there (for what I presume are the obvious reasons). Also the core collapse will be slightly more energetic from the side falling toward the Earth than from the side falling away from the Earth. If more debris is expelled Earthward than elsewhere, it may slightly counter an Earthward nudge or exaggerate an anti-Earthward nudge (though I'm positive any debris plumes would have a negligible effect on the moon's basic numbers).

 

Also, just a minor point, but the calculations as to moon mass at the centre are surely likely to be slightly out, because of the greater pressure and thus greater mass density at the centre of an object? I'm sure it's calculable, and possible on an object the size of the moon it isnt significant, but wanted to flag it up!

It's kind of the opposite. The density of a spaceship is bound to be far less than the solid iron scientists have presumed is there (unless the ship is made of neutronium or something). The mass lost will be significantly less than an equal volume of iron. This would also mean, btw, that the iron core is denser than current calculations predict.

 

I wonder if suddenly churning up the iron core would heighten the moon's magnetic field?

[Nadrakas]

Re: Astronomy/Physics question

 

What's the mass of the thing? That's a big robot' date=' obviously, but again if it's mostly empty space, it might affect the planet more like a big meteor swarm than a planetoid.[/quote']

 

Here you go.

 

Mechanon Statistics:

Total Volume in KM = 523,598,775.6 km^3

Total Volume in Hexes = 2.617993878e11 hexes^3 (261,799,387,800 hexes^3) (Total Area in Hexes = 196,349,540,800 hexes^2)

Total Mass = 25 pton (25,000,000,000,000,000, metric tons!!! or 25,000,000,000,000,000,000 kg)

Vehicle Size Category = 58 (See Ultimate Vehicle, page 9)

STR Required to Lift = 280

 

 

Yeah...he's really BIG. And he comes with his own Legions & Space Armada. All for the low, low, low price of...nothing. He'll exterminate ALL organic life for Free!!!

 

You know...one could say that Mechanon has a really BIG head

 

 

Nadrakas...

 

P.S.: Again...thanks to my wonderful wife (aka: Isis) for assisting with all the darn math.

[Nyrath]

Re: Astronomy/Physics question

 

The Death Star always seemed like a ridiculous amount of overkill to me. Why destroy the whole planet when you only need to destroy the outer 1-2% to effectively kill everyone and destroy every human artifact? The difference in energy consumption has got to be by several orders of magnitude.

Yeah, I'm with you. The Earth is about 8000 miles in diameter, I'm sure you could effectively exterminate the human race by just turning the surface of the Earth into molten lava down to the depth of about a mile, leaving the rest of the Earth untouched.

 

My slide rule say that will take roughly 0.00075 as much energy as melting the entire Earth, and even less than blowing the Earth to smithereens.

[Nadrakas]

Re: Astronomy/Physics question

 

Ahhh...but the Death Star had another purpose -- Terror!!! Destroying the surface of the planet would be a powerful symbol of the Empires might, yet turning a planet into dust shows the Invincibility of the Empire. Thus...the Empire strikes Terror into the hearts of those would resist.

 

Besides, the Empire already had something that could destroy the surface of a Planet -- a Sector Fleet (shrug) -- but that's so "normal seeming." Now a Moon sized star base that can wipe out an entire planet in one shot...a little frightening.

 

Besides, which is more dramatic? Blasting the top layer off of a planet or turning that same planet into dust. Of course...spinning the planet would be truly horrifying, not to mention nauseating.

 

 

Nadrakas..."Let's play spin the planet Princess Bunhead!!!"

[Supreme]

Re: Astronomy/Physics question

 

...The Jewish Calender will be screwed...

As would the Buddhist, but that's OK because everything ends.

[Agemegos]

Re: Astronomy/Physics question

 

Just want to point out that is the formula that I arrived at too, on the 1st page, but it should be:

 

r = (GM)/v²

 

shouldn't it?

 

Quite correct.

[Agemegos]

Re: Astronomy/Physics question

 

My turn to step in. OK' date=' I understand this, but the query I have is whether the sudden loss of mass does not of itself constitute the acting of a force upon the moon?[/quote']

 

No, it does not. The drive might exert such a force: it is fictional, it might by hypothesis do anything. But simply taking some mass away is simply taking some mass away.

 

For example, any change in velocity owing from a reduction on mass is not going to be instantaneous

 

There isn't any such change in velocity.

 

whereas a reduction in mass does have an instantaneous gravitational effect.

 

Not necessarily. According to the General Theory of Relativity (which is the current state of the art) gravitational changes propagate at c. But since we are positing an instantaneous drive we are rejecting relativity: we aren't bound to assume that gravity propagates at c in teh game world, but we aren't bound to assume that it propagates instantaneously, either.

[austenandrews]

Re: Astronomy/Physics question

 

Now a Moon sized star base that can wipe out an entire planet in one shot...a little frightening.

So in the post-Empire age, did every planet build a couple superlasers with which to defend itself? I bet you could build it even bigger on a real, live planet.

[tgrandjean]

Re: Astronomy/Physics question

 

Off topic but relating the the Death Star:

What's the purpose of the Superlaser of the Death Star?

To overcome planetary class shields.

Simple.

[tgrandjean]

Re: Astronomy/Physics question

 

So in the post-Empire age' date=' did every planet build a couple superlasers with which to defend itself? I bet you could build it even bigger on a real, live planet.[/quote']

 

You would have a limited arc of fire and no mobility. Possible and possibly even useful, but somewhat limited use. Also, each of the Death Stars took the resources of many systems and several years to build.

.... Just thinking on what it would take to defend against the Death Star, figure that you would need to match the output of the superlaser to protect an area matching the surface area of the Death Star. How many times this output would you need to cover the surface area of the Earth?

[austenandrews]

Re: Astronomy/Physics question

 

You would have a limited arc of fire and no mobility. Possible and possibly even useful' date=' but somewhat limited use.[/quote']

And you may have to blow away your own moons to clear the field of fire. Otherwise a Death Star could approach from cover. Sneaky bastards, those Death Stars.

 

Also, each of the Death Stars took the resources of many systems and several years to build.

Sure, but that's like the difference between building a missile silo and a nuclear submarine.

 

.... Just thinking on what it would take to defend against the Death Star, figure that you would need to match the output of the superlaser to protect an area matching the surface area of the Death Star. How many times this output would you need to cover the surface area of the Earth?

About eleven thousand times, going strictly by surface area. But that's if you have to hold off numerous Death Stars at once. Against one, you can just protect the reach of its single superlaser.

 

God, I'm a geek.

[Curufea]

Re: Astronomy/Physics question

 

And you may have to blow away your own moons to clear the field of fire. Otherwise a Death Star could approach from cover. Sneaky bastards' date=' those Death Stars.[/quote']

 

One laser at each pole, on turrets.

[Basil]

Re: Astronomy/Physics question

 

I would ,like to point out that the mass of an object has no real bearing on where it would orbit. I had too get my astronomy book out, but the radius of an orbit is calculated thus:

 

r = (GM)/v²

 

r -> Radius of the orbit

G -> Gravitational constant

M -> Mass of the planet orbited

v -> orbital velocity

This formula is wrong on two counts. First (which I admit is somewhat of a quibble); one does not speak of the radius unless the orbit is a perfect circle. For an ellipse, one uses the "semimajor axis" (half the distance between the farthest-apart points on the orbit), which is symbolized "s" *. The semimajor axis is also the average distance, averaged across the whole length of the orbit.

 

Second, and much more to the point, the formula is:

s = G(M+m)/v²

where M is the larger, and m is the smaller mass.

 

The formula SirViss gave is a useful (and often used) approximation when M is much larger than m. However, it is an approximation.

 

As you can see from the complete formula, any change in M or m will change the semimajor axis; specifically, a reduction in m will reduce s.

 

Note, though, that this is a change to the average distance. The distance at the moment of removal will not change (unless there's more than a Teleportation going on). Thus, the orbit will "dip" more towards the Earth than if the removal had not occured. If the removal is at perigee, apogee will be reduced, and the orbit will be less eccentric. If the removal is at apogee, perigee will be reduced, and the orbit made more eccentric. Between those two points it gets more confusing.

 

 

*BTW, the semimajor axis is sometimes symbolized "a". As well, "r" can mean the distance from the orbited body at a particular point in the orbit.