Too Close for Comfort: Newsletter Challenge (03/01/11)

Yes. If g' is the acceleration of gravity at the planet's surface, a velocity just greater than √(rg') will do the job.

That's the answer giving a circular orbit, but any velocity between √(rg') and √(2rg') gives a possible elliptiptical orbit, of increasing eccentricity as velocity increases. Of course it would only be closest to the planet's surface at 2 points of the orbit.

Above √(2rg') (the escape velocity) it doesn't orbit but leaves the planet altogether.

Cheers...........Codey

velocity between √(rg') and √(2rg')

Above √(2rg') (the escape velocity) it doesn't orbit

?

Not clear what you're saying. Agree or disagree? Please elaborate.

Cheers..........Codey

any velocity between √(rg') and √(2rg') gives a possible elliptiptical orbit

Above √(2rg') (the escape velocity) it doesn't orbit

read error, root sign looks like reziproke ( √(2rg') 1/(2rg) )

OK I think I understand. But you're a man of few words!

Cheers........Codey

Ummm... Isn't this Physics 101? Didn't Newton give an example of a (theoretical) canon at the top of a mountain doing exactly this?

And based on the latest definition of a 'planet' the planet would have to be generally spherical (not some oblong shape that might complicate the answer).

Yep. Here it is from Wikipedia:

The Answer will be posted right here on CR4 on April 5th February 27.

(Gavel drops.)

ξ

I suspect that the questioner is thinking more of a shell fired from a barrel, not a maneuverable rocket. If so, the answer is no, it will hit the gun barrel at the end of the first orbit.

Nah, you're supposed to duck before the bullet makes its way all around....

(v/2πr for those who wish to calculate.)

Not if the planet is rotating. By the time the shell completes one orbit the canon would have rotated eastward (or westward, depending on the planet's rotation direction) and the shell will miss the canon. When a typical LEO satellite launched from the KSC facility in Florida completes one orbit the launch point has moved 22.5 degrees eastward.

There are specific cases where the shell might hit the canon, but there are many possibilities where the shell would not.

Correct and it is more complex than that.

KSC is 28.39° N latitude and not on the rotational equator. The result of the orbit, if mapped on a globe and the rotation of the Earth taken into account, produces a ground track that looks more like a sinusoidal wave across the map.

To have the projectile return to the same spot at launch would require that the planet is not rotating or that the launch take place precisely on the equator and launch due east or west of the rotational pole.

"To have the projectile return to the same spot at launch would require that the planet is not rotating or that the launch take place precisely on the equator and launch due east or west of the rotational pole."

Exactly what I had in mind, but it is immaterial because in my version of this hypothetical problem, the planet is not rotating. You can do what you wish with yours

I really would have like to, but when I went back to check on my version of the hypothetical planet I discovered to my utter dismay that it had been vaporized by a Vogon Constructor Fleet just minutes before.

Of course you have your towel!

Yup. Haven't thrown that in, yet.

Your scaring the "Straights"! Doubt if a lot of readers hitch hike and carry a towel.

Better to bring a towel than to listen to some Vogon poetry.

Uh.... If the planet is not rotating, are there rotational Poles?

Exactly. That's where the word "or" comes into it.

Yes, provided it is launched from the highest point on the great circle that its orbit will follow.

I would agree, but that is only true in a perfectly isolated system and the bodies are of perfect spheroid dimensions and homogeneous density. That only exists in theory.

Any irregularity or external influence will, over time, alter the perfect orbit and will end as a crash. This would also include rotation of the bodies, solar winds, etc. Even the launcher represents a mass that is not homogeneous to the rest of the planet.

Remember the laws of thermodynamics?

The way I see it, the original post uses the word planet, which has a common meaning of a rocky or gaseous body. None are of perfect construction.

If the problem was stated as spheroid masses in an isolated universe or system, then it would be a different matter, but the problem was not framed that way.

As usual for Challenge questions there's the Information Deficit in operation. A 'stable' orbit is not specified, neither is the number of orbits beore the inevitable.

I agree that in the 'real universe' a very low altitude stable orbit is not on.

Nor a high altitude orbit, although, a high altitude orbit will be less impacted by small variations to its environment.

Hi AH, you wrote: "Nor a high altitude orbit, although, a high altitude orbit will be less impacted by small variations to its environment."

I suppose one can argue that high altitude circular orbits around planets may last virtually forever, although not at exactly the same shape. Many of the effects on them cancel out over time.

For surface-skimming orbits, I agree that they won't last very long, although pretty low orbits can last for a long time around our moon if they are chosen correctly, at the so-called "frozen orbit" inclinations (Nasa).

-J

Here in lies the problem. There are no definitions as to what is meant by "orbit". Is that one revolution, two, ten, or infinite?

I think infinite is a definite impossibility, but to what degree of stability does the challenge demand to be called a success? My guess is that someone did not consider that requirement when crafting the question.

"Here in lies the problem. There are no definitions as to what is meant by "orbit". Is that one revolution, two, ten, or infinite?"

I reckon "orbit" must be taken to mean a practical number of orbits, without being boosted in any way. Due to drag, most very low Earth satellites need a boost every few months, otherwise they reenter the atmosphere. But they are still classified as 'orbits'.

This ultra-low orbit with no atmosphere may have some other problems, which need to be overcome, somehow...

-J

Define close.

And define the barren planet!

No planet is going to be:

1. Perfectly spheroid in shape.

2. Perfectly homogeneous in density.

Both factors will contribute to orbital perturbation as the gravitational field varies from point to point along the orbital path.

This will produce an imperfect orbit and will not produce a true circular orbit, which will get more eccentric over time. The satellite will eventually contact the planet surface.

This doesn't even take into account any other distant bodies that may have an effect on the orbital status of the satellite.

Does this planet have a magnetic field?

What's the composition of the satellite?

I would say no, based on the assumption that the vacuum of space continues right down to the surface of the planet, any satellite launched would continue in a straight line from the launch point...............away from the planet.

Only if you exceed the escape velocity for the planet

Ve = √(2GM/r)

It's not the atmosphere that causes things to fall back to Earth.

I'm still sticking with no. If it doesn't exceed escape velocity, any gravitational pull is going to crash it.......pretty quick. I would think.

Atmosphere doesn't hold things in orbit - gravity does.

This sounds a lot like this discussion.

Correct! This question was asked and answered weeks ago. Search CR4 must not be working again .

Based on the little info given, the answer is Not a chance. Even the moon has a gravitational force that will bring down an object after some time. If this planet has no gravitational force acting upon it, Not having an atmosphere = nothing. The satellite will just take off in a straight line and, like the Energizer Bunny, just keep going and going and going until it hits another object. Bounce off and keep on keepin on.

??? "If this planet has no gravitational force acting upon it, Not having an atmosphere = nothing"

You do realize that every object has gravity, right?

The planet has gravity, therefore it pulls the object to it, the speed of the object is only slowed by atmosphere (not having any is an important part of the question) so there is no slowing. if launched at exactly the escape velocity you would get a perfect orbit (until some other force acted on it - such as an imperfect surface or magnetic field)

The velocity necessary to keep it in orbit doesn't need to be anything like escape velocity. Do this experiment in your head:

• imagine a satellite in a stable orbit;
• now give it a boost in the direction it is going (relative to the object it is orbiting);
• it will move into a slightly higher orbit (may not still be the same shape , and, may cross the original orbit, but on "average" will be higher).
• You will need to give it a lot of "boosts" before it escapes.

The velocity necessary to keep it in orbit doesn't need to be anything like escape velocity.

That's right, to be more precise, the velocity needed to keep it in orbit just above the surface is escape velocity/√2.

Cheers........Codey

Given the dearth of details given in the original question I suspect that the hook here is the Roche Limit.

I'll go with no.

http://en.wikipedia.org/wiki/Roche_limit

Hmmmm

I wonder if there are any planets in the universe which spin at a speed which makes the equator effectively geostationary? For smaller bodies I guess this is inevitable, but, I'm thinking of something with a gravitational pull at the poles comparable with the earth.

Interesting! The inhabitants of such a planet would experience no gravity!

To go a little bit further, even the surface material of that planet would have no reason to stay in place, right? Such a planet would soon disintegrate, as layer by layer of its crust would leave to the outer space, while the planet would spin faster and faster (as the angular momentum would have to stay constant.)

I'm too bored (and too busy) to check the maths of it, so I cannot tell if this process will ever stop, though

"Interesting! The inhabitants of such a planet [spinning effectively at surface geostationary rate] would experience no gravity!"

Only at the equator, though...

"Such a planet would soon disintegrate, as layer by layer of its crust would leave to the outer space, while the planet would spin faster and faster (as the angular momentum would have to stay constant.)"

Nope, the constant angular momentum includes the bits that fly off. What remains should not spin faster, I think. It is when stars collapse under gravity and then explode (i.e. super novae) that the core spins faster, not when the outer layers are thrown off due to spinning too fast in the first place.

Me too haven't done the math, so perhaps you are right, after all.

-J

I umm'd and arr'd about the spin rate. Started with the same idea as you (the bits coming off each take their chunk of momentum with them), then thought about Iω, then decided the change in moment of inertia was irrelevant.

Imagine a skater spinning with arms outstretched, a brick in each hand. Pull the bricks into the body - it speeds up. But if the arms are kept outstretched and the bricks are released, I think it's clear (if only intuitively) that the skater's spin rate wouldn't change.

Still haven't done the sums, tho'!

The case of the odd planet is slightly easier to analyse intuitively. With the skater they are having to hold on to the bricks, which will fly off at a tangent when the skater lets go. On the planet a loose brick on the equator would just stay where it was: incidentally fulfilling the requirements of the challenge; and thereby demonstrating that I wasn't flying off at a tangent when I posted the reply.

Hi JohnDG, you wrote: "I umm'd and arr'd about the spin rate. "

We should do more of that, because I think this where the solution lies.

A planet with a reasonably uniform equatorial surface (no high mountains), spinning at just below the 'geostationary surface rate', will have a tiny positive 'g' at its equatorial surface. Hence, no pieces will 'fly off' due to centrifugal forces. A little above that surface, there will be zero 'g'.

If you place a small satellite there, stationary relative to the surface, it will be in a geostationary orbit. If there is no atmosphere, so no winds or changing pressure (buoyancy), the satellite can stay there for a long time. Inhomogeneous density and tidal forces caused by the 'sun' and any moons should not have any significant effect in this particular situation - surface and satellite are affected equally.

I haven't done the calcs, but this seems entirely possible somewhere in the cosmos...

-J

Aside - back-of-envelope suggests that for this condition to obtain on Earth, a day would be about 85 minutes long (I think!). Certainly not unimaginable.

"Aside - back-of-envelope suggests that for this condition to obtain on Earth, a day would be about 85 minutes long (I think!). Certainly not unimaginable."

I got the same value, but the problem is that Earth's oblateness will increase quite a bit, meaning the radius at the equator will increase significantly, complicating the issue.

Not quite sure how to solve for that - a bit late in my valley (after 11 pm) for starting such math. Will try tomorrow.

-J

I wrote: "... the problem is that Earth's oblateness will increase quite a bit, meaning the radius at the equator will increase significantly, complicating the issue. Not quite sure how to solve for that ..."

I've done some calcs now and if correct, one needs an Earth-mass & density planet, spinning about 12 times faster than Earth, with its 'day' ~ 118 minutes long. This will increase the hypothetical planet's equatorial 'bulge' to about 7983 km radius, with a surface speed of about 7.08 km/s. Here the centrifugal force just about cancels the gravitational force (if my sums are right).

So, spin Earth up to about 12 times its present rate, put your satellite on top of an equatorial tower of a height that just makes you weightless at the top (some tens of meters high). Take away the tower and you have a geostationary satellite - or not?

-J

I wrote: "So, spin Earth up to about 12 times its present rate, put your satellite on top of an equatorial tower of a height that just makes you weightless at the top (some tens of meters high). Take away the tower and you have a geostationary satellite - or not?"

The 12 times spin rate will only work if the final 'earth' has the same density as Earth today, which will not be the case. The density of a fast spinning sphere goes down, because the volume of an oblate sphere is V = 4/3 pi a²b, which is larger than the non-spinning sphere, where a is the equatorial radius and b the polar radius.

The difference is not very large, since 11 times spin rate (period ~ 130 minutes) will still do the trick of (almost) balancing the surface gravitational force with the centrifugal force at the bulging equator (radius ~ 8480 km). Like all artificial satellites, this tiny geostationary satellite just above our heads would also need occasional thrusters for station keeping. The moon's tidal effect will likely be the biggest cause for instability of the orbit.

-J

PS: I've used the 'flattening' equation from http://en.wikipedia.org/wiki/Equatorial_bulge, together with standard Newton forces. It is a headache to solve, because both radii (a and b), period T and density ρ are variables. I cheated by employing the 'Goal Seek' tool in MS Excel.

Not quite the same thing, but an interesting side-note I think, in that they are 'geo'-stationary to each other: [From Wikipedia]

Charon and Pluto revolve about each other every 6.387 days. The two objects are gravitationally locked, so each keeps the same face towards the other. The average distance between Charon and Pluto is 19,570 kilometres (12,160 mi). The discovery of Charon allowed astronomers to accurately calculate the mass of the Plutonian system, and mutual occultations revealed their sizes. However, neither indicated the two bodies' individual masses, which could only be estimated, until the discovery of Pluto's outer moons in late 2005. Details in the orbits of the outer moons reveal that Charon has approximately 11.65% of the mass of Pluto.^[4] This shows it to have a density of 1.65 ± 0.06 g/cm³, suggesting a composition of 55 ± 5% "rock" to 45% ice, whereas Pluto is somewhat denser and about 70% "rock".

Hi Usbport, very interesting lateral thinking!

Apart from the fact that the 'launch' of such a satellite will be a bit difficult (although I suppose 'launch' can mean many things^(a)), this will surely do the same trick as a planet that spins fast enough to keep a very low satellite geostationary.^(b)

How low (or "close to the planet's surface", as the challenge states) you can come before tidal gravity breaks up one or both the bodies is also a bit of a question. But again, the question leaves it wide open. Pluto and Charon surely co-exits...

-J

Footnotes:

(a) Charon may have formed by another body colliding with Pluto, with the debris field coalescing into a moon, much like our own moon is thought to have formed. I suppose that's a form of 'launch'.

(b) For those who have joined in late and could not not read the whole lengthy thread, the reason for 'geostationary' is to make the gravitational field that the satellite 'sees' as uniform as possible. This also avoids obstacles most effectively. Neither is the case for satellites moving across a non-homogeneous planet's surface and they may prematurely crash into the planet (or need an excessive amount of station keeping fuel).

...but I wonder if the planet would disintegrate anyway, due to the centrifugal forces. For example, let g be the acceleration of gravity at the surface. Then g = ω²r. If now we take a point inside the planet, at r/2 from the centre, then the gravitational acceleration there would be 8 times lower (assuming constant specific weight) but the centrifugal force will be just twice as less. Similarly for other points inside the planet, it seems that gravity always falls short to hold the planet together.

I did some math in a rush; I hope I'm not ridiculously wrong!

If this argument were to hold, would there not be points inside the Earth where the centrifugal force is greater than the gravitational attraction, so making the Earth fly apart?

Maybe we rely on the Stickiness factor to keep us together (see Appendix IV, Table (c) in the reference I gave in #57 ).

If this argument were to hold, would there not be points inside the Earth where the centrifugal force is greater than the gravitational attraction, so making the Earth fly apart?

Don't forget that the layers above those points you mention will themselves apply a pressure to the points below. In the case of the super-fast spinning planet though, all points would be loose.

Of course, the assumption that in such a case all points will rotate at same angular velocity ω might be wrong. Sooner or later the various parts of the planet will move in different ways. Maybe we will also see the sphere getting distorted to a number of rings...

Ok, what I was doing was a bad job of trying to point out was an error in your sums.

You said "... at r/2 from the centre, [then] the gravitational acceleration there would be 8 times lower ... ".

For the whole planet, the acceleration due to gravity at the (equatorial) surface is:

g_p = G.M_p / R_p²

where M_p is the mass of the planet, and R_p is the equatorial radius.

With half the radius, the mass (assuming uniform density) and radius are given by:

M_x = M_p/8

R_x = R_p/2

so the surface gravitational acceleration would be:

g_x = G.M_x / R_x² = G.(^M_p/₈) / (^R_p/₂)² = G.M_p/2.R_p²

=> g_x = g_p/2

OK right, I better return to my chores, otherwise I will make fool of myself again! Indeed, since my university years I remember the gravity being reduced linearly inside a planet as you approach the centre...

Anyway, it seems that if the spin is too fast at the surface, it is too fast for any point inside the planet: Both gravity and centrifugal are inversely proportional to the distance from the centre. So in the one thing I was right (?) is that the planet will disintegrate.

tkot wrote: "Anyway, it seems that if the spin is too fast at the surface, it is too fast for any point inside the planet: Both gravity and centrifugal are inversely proportional to the distance from the centre."

I think this is only true for uniform density, which is not the case for real planets - they are less dense at the surface than at the center. So a progressively increasing spin will probably throw off the surface layers first...

This same effect probably also further complicates the calculations that I made above on the zero-g equatorial surface.

-J

Or maybe it just looks like

Hi, what is the reason for some 2.2 your weight at the poles ?

The corpus is flat in your consideration, has an elliptical profile. With less mass under our feet gravity should be significantly less than 1 (normal).

And I'm going to place my guess to main question just here for convenience (and off-topic points perhaps, I'm not so familar with the rules here):

Yes. Because all contributors seam to agree to the anticipation of that fast spinning planet, where the satellite is only to be held up a little bit high to destination height and released there.

If the planet would not be spinning that fast, the satellite would tear the planet in its spin/direction and thereby looses its own velocity and descends. Its Einsteins distorting of time/space by moving masses. Gravity spirals. But don't know about the maths here.

Hi Jo (?), welcome.

Without going into any maths or technical discussion - standing at the equator you'd be like a bob-weight on a long string (equatorial radius of the planet), but near the poles you'd be like the same bob-weight (same mass), going round at the same rate, but on a short string. Spot the difference in string tension?

There are also some differences in the distance from the centre of mass to consider.

(Few rules here - check out the CR4 rules of conduct and anything else goes).

_{Jeez I've just read 'em - there's hunerts of 'em. Just be careful.}

Jo Yardman wrote: "what is the reason for some 2.2 your weight at the poles ?"

An Earth-like planet spinning at 11 times Earth's rate would bulge out to about 8,481 km radius at the equator and 'flatten' at the poles to about 4,254 km radius. The gravitational acceleration is g = -GM/r², with GM for Earth ~ 3.9866 x 10¹⁴ m³/s². This gives g ~ -22 m/s² at the poles, which is about 2.2 average Earth gravities (of -9.81 m/s²).

-J

And there I was thinking gravity related to mass - no matter what the shape or velocity.

Hi 34point5, actually, shape has more to do with it than what I gave it credit for above. The value of g = -GM/r² is only strictly valid for homogeneous, spherical symmetrical masses. Earth is close enough to that for all practical purposes, but my '11-times-spin' planet is probably not.

My intuition is that such a planet's polar gravity will be slightly less than the 2.2g of the equation, but we'll have to research that further.

-J

Many moons [!] ago, I saw a derivation showing that for a homegeneous spherical solid, the integral of gravitation of all particles equals that of a concentrated mass at the center. I don't recall what would happen with a sphere as graded shells of varying density, or a hollow sphere. It seems that an oblate spheroid would spread out the vectors so as to decrease the total gravitation from a pole to the center. I'm too rusty to tackle it now, but I would guess you are right.

This has gotten far afield from the original problem. I took it as a "pure" exercise assuming a non-lumpy spherical planet, and ignoring the perturbations of other bodies. Tidal effects would also degrade a simple circular orbit over a long time, but to me the "spirit" of the problem avoided such complexities; hence my answer in post 1.

Usually these things are "answered" within a week or month, about half of the time incorrectly; but in this case the OP says April 5, so we haven't heard yet....

Well let's hope the 'math answer' doesn't vary too much from what is apparent in that Saturn eclipse shot ...

I look forward to your re-think as my intuition says; 1 or less, due to increased distance of the equatorial mass - like tending to zero g in the center of a doughnut.

Mind you, you would be 'closer' to the 'math resolved' center of gravity ...

I wonder if things get lighter or heavier when lowered down a deep mine?

I don't think rotation has anything to do with it - or a revolving flywheel would be 'heavier at the axle', then again it might be and I just haven't noticed.

Hi 34point5, the effective gravity at the poles are surely always stronger than at the equator. However, when the proper ellipsoidal shape is taken into account, I think we get a value for the poles slightly less than for a spherical shape (as I indicated in #99).

This seems to be confirmed by the WGS 84 Geoid, but I must first find time to study it more deeply...

-J

In your hypothetical faster spinning Earth half radius model I do not see the planet flying apart by centrifugal force. I see the square of the velocity varying with the radius, and the gravitation so it requires geometrically more velocity as one goes deeper into the Earth to maintain zero g. As you have noted, gravitation varies directly with the radius of a body. And if the surface velocity is such that an object is weightless on the surface, the velocity must be lower anywhere within the same rigid body.

However, in your model the outer shell surrounding your half radius sphere also has gravitational force, as if you have a two body relationship. Somewhere in between the center and the surface may be a point of lowest "weight?" Possibly a future "Challenge?"

Finally, any body with the gravitation of the Earth must have attracted an atmosphere, perhaps ruling it out as the "planet" in the problem?

I agree with Jorrie that one requirement of the thought exercise is that one must specify that the object must have sufficiently altitude to clear all surface eccentricity.

"In your hypothetical faster spinning Earth half radius model I do not see the planet flying apart by centrifugal force ..." - neither do I. Please re-read, checking the context.

My error. I see I was associating your reply.

Somewhere in between the center and the surface may be a point of lowest "weight

this point is equal the center of the planet/body where all the gravitaitonal forces will be equal from all sides.

The Moon, relative to the Earth?

or a black hole?

I guess that this could be possible in a pure "Newtonian" world. But in real world, every object which is in orbit around a planet, transmits gravitational waves. This leads to a gradual reduction of its speed and the object starts to get closer to the planet. (However, this loss of energy is reeealy small and this procedure will take reeealy long time.) [There are other factors, of course, that could disturb its orbit, like e.g. cosmic radiation.] As a result, its orbital motion can not last forever.

So how come the Moon's getting further away? .

So how come the Moon's getting further away?

Due to tidal effects, the spin of earth slows down, therefore in order to keep the angular momentum of the system constant, moon has to accelerate, so it moves to a higher orbit.

(Well I know that - 'cos I heard a program about it on the wireless, and Googled it to check. I was questioning the assertion in the post to which I was replying.)

Since it is a planet, it revolves around a sun (or star). The effects of such sun's gravitational pull on the planet and the fired shell must be taken into account.

Further, if the mass of the satellite, in relation to the planet, is too high, then the revolution of the latter will cause a certain amount of wobble in the system.

Of course it could. Since "close" was not given a specific number i.e. feet, yards, km or miles, it obviously can be done.

Exactly the critical point IMHO. This challenge has again posted a nebulous set of criteria so that any result could be the intended answer. Too close and the satellite collides with a mountain or tidal forces cause the satellite to morph into a ring. Too far away and and the gravity well of the star is more dominant than the planet.

To be fair to the author of this puzzle, crafting a puzzle that contains all of the critical information to solve the puzzle is a difficult task. This is particularly difficult to craft for an audience of engineers for we solve puzzles as a living.

Yes but! Any surface or density irregularities will cause perturbations in the orbit. The orbit will not be self sustaining over time.

This seems like a no-brainer. What does the atmosphere, absence or presence, have to do with it? There are rockets right now functioning in places with no atmosphere, some of them designed to, eventually, orbit small and bare planets, like Mercury, "close to the ... surface." The fact that they are not "launched" from the planet is immaterial. So the answer seems to be, simply, "Yes," but it raises another question, namely, "why is this a question?" or, perhaps, "is there any reason it could not?" to which the answer is "No." Jwatersphd

You wrote, "What does the atmosphere, absence or presence, have to do with it?"

Everything. Atmospheric drag will force a stable orbit to decay. Even our own low-Earth satellites require small periodic corrections to remain in orbit due to minute drag of air at those altitudes.