- 08 Mar '10 23:51http://www.universetoday.com/2008/05/13/phobos-might-only-have-10-million-years-to-live/

This piece suggests Phobos will crash into Mars in 10 million years but will reach the Roche limit of about 7000 Km from the center of mars, 3620 Km above the surface, in about 7.6 million years and be torn to bits by tidal forces.

So here it is: Say it's a couple thousand years from now, Mars is fully inhabited, huge cities circle the globe, billions of people live there. Now they come to realize all that about Phobos is true. They also have many gambling casino's there and VERY high rise high cost hotels. So they are saying to themselves, how much force do we need to apply to Phobos to keep it from falling any more?

It now orbits about 9380 Km from the center of Mars now, so it only needs to lose about 2380 Km of altitude to get to the Roche limit. It masses 1.08 E16 Kg and if the number of 7.6 million years to reach Roche limit is correct, it is losing about 313 meters per year altitude. Is it possible to show how much force (choose your own units, Newtons, Kg, Tons, whatever) or thrust continuously applied like maybe a rocket on a circumferential railroad, constantly pointing to Mars or some like technology to provide Mars facing thrust, how much thrust is needed to keep it in a steady orbit? - 09 Mar '10 08:31 / 1 edit

Pointing the rockets downwards, to the surface of Mars, is totally in vain.*Originally posted by sonhouse***Is it possible to show how much force (choose your own units, Newtons, Kg, Tons, whatever) or thrust continuously applied like maybe a rocket on a circumferential railroad, constantly pointing to Mars or some like technology to provide Mars facing thrust, how much thrust is needed to keep it in a steady orbit?**

You have to point the rockets backwards, gaining "orbital energy", according to Newtons action-reaction law. - 09 Mar '10 15:17

But first you have to slow down its rotation to zero, or to a bounded rotation. In that way you can attach your propulsion device fixed to its ground.*Originally posted by sonhouse***Ok, backwards then. I was thinking the wrong vector, eh.**

Now it's time to fire her up, and let her fire, feeding her with fuel, until you get high enough orbital energy. - 09 Mar '10 17:53 / 1 edit

That is the question, how much thrust? I guess the thrust could be minimized by the fact you have a few million years in which to act so if you had to do it in one year, you would clearly need 7.6 million times the thrust if you could do it in 7.6 million years, whatever that would turn out to be.*Originally posted by FabianFnas***But first you have to slow down its rotation to zero, or to a bounded rotation. In that way you can attach your propulsion device fixed to its ground.**

Now it's time to fire her up, and let her fire, feeding her with fuel, until you get high enough orbital energy.

One thought: lets assume you do just that, get a rocket going, having stopped the rotation relative to Mars, wouldn't the same forces keep rolling the orbit back towards Mars even if it was in a higher orbit, in other words, just delaying the bad day? - 10 Mar '10 12:00 / 3 edits

I don't know about the thrust but it is possible to clculate the power (in watt/horsepower/whatever).*Originally posted by sonhouse***http://www.universetoday.com/2008/05/13/phobos-might-only-have-10-million-years-to-live/**

This piece suggests Phobos will crash into Mars in 10 million years but will reach the Roche limit of about 7000 Km from the center of mars, 3620 Km above the surface, in about 7.6 million years and be torn to bits by tidal forces.

So here it is: Say it's a coupl hnology to provide Mars facing thrust, how much thrust is needed to keep it in a steady orbit?

If I'm not mistaken it's not 3.13m per year but 0.313m per year.

The total energy (Potential+kinetic) of Phobos is -GMm/(2r), where G is the gravitational constant, M and m are the masses of Mars anf Phobos, r is the distance between their centers. Calculate the loss of energy (in joules) when r decreases by 0.313 meters and divide by the number of seconds in a year. This is the rate (in watts) of loss of energy. So the attached engines must compensate for this.

Ok, here goes the calculation:

G=6.67*10^-11

M=6.42*10^23

m=1.07*10^16

r=9380000

(All is units of kg, m and sec)

so -GMm/(2r)=2.44*10^22 joules

A decrease of 0.313m in r changes r by 1 part in 30000000 and this is also the change in energy, so the energy loss in a year is 8.1*10^14 joules, that is 26 megajoule per second, so the effective power of the engines must be 26 megawatt or about 35000 horsepower. - 10 Mar '10 21:30

35,000 hp. Ok. I think 1 hp is the energy needed to lift 555 pounds one foot in one second or 32 Hp to lift 555 pounds at one G.*Originally posted by David113***I don't know about the thrust but it is possible to clculate the power (in watt/horsepower/whatever).**

If I'm not mistaken it's not 3.13m per year but 0.313m per year.

The total energy (Potential+kinetic) of Phobos is -GMm/(2r), where G is the gravitational constant, M and m are the masses of Mars anf Phobos, r is the distance between their centers. Calcul ...[text shortened]... second, so the effective power of the engines must be 26 megawatt or about 35000 horsepower.

I guess that 35K hp would have to be continuously applied forever to keep it in that same exact orbit, or double it to 70,000 hp to make it go UP 313 Cm/year?

What bugs me is why is it not a stable orbit? Aren't the satellites in Geo-synchronous orbits totally stable and meteors notwithstanding and solar storms, etc., wouldn't those sats be still there a billion years from now? Or is there some instability that would bring even those down at over 36,000 Km up?

Earths moon Luna is actually receding from earth at some small rate, an inch a year or something. What is making the difference for Phobos? - 10 Mar '10 23:00

http://en.wikipedia.org/wiki/Tidal_acceleration*Originally posted by sonhouse***35,000 hp. Ok. I think 1 hp is the energy needed to lift 555 pounds one foot in one second or 32 Hp to lift 555 pounds at one G.**

I guess that 35K hp would have to be continuously applied forever to keep it in that same exact orbit, or double it to 70,000 hp to make it go UP 313 Cm/year?

What bugs me is why is it not a stable orbit? Aren't the satellites i ...[text shortened]... rth at some small rate, an inch a year or something. What is making the difference for Phobos? - 11 Mar '10 05:40

It would be stable if there were no outer forces in action. And there are lots of them acting on the satellite. Solar wind, radiation, other bodies, like Jupiter, etc.*Originally posted by sonhouse***What bugs me is why is it not a stable orbit?**

Only if the universe consisted only of these two bodies, the satellite and the planet (witho no atmosphere), no sun, no galaxy, no nothing else, the orbit would be matematically stable.

In fact the entire solar system is an caotic orbital system. Nothing is stable. It's stable enough, that's what counts. - 11 Mar '10 13:14 / 1 edit

Yes, I've read the article. Well written but old stuff.*Originally posted by David113***Not true. Read the wikipedia article.**

Don't you believe in the two-body orbital system theory?

If not, read http://en.wikipedia.org/wiki/Gravitational_two-body_problem and tell us where you find any errors? - 11 Mar '10 20:16 / 1 edit

Very interesting article. One thing that fascinated me, the estimate of the energy in the lunar/Earth tidal effect: almost 4,000,000 megawatts! Too bad we can't tap into that somehow!*Originally posted by David113***http://en.wikipedia.org/wiki/Tidal_acceleration**

So with Phobos, is it an exercise in futility to have rockets with that 35,000 hp figure keeping it in the same orbit or using more to get it higher?

So how high would the orbit have to be to make the spiral in motion negligible?

On the Terawatt Earth/moon system power, I see a way to capture some of it:

Lets assume in the future, cable has been made strong enough for the space elevator to work, we have a bunch of them Ala Arthur C. Clarke.

We could use that cable in a system that would have a cable stretched from Earth to moon, then the 33 mm/year of the recession of the moon could be tapped by say, compressing air or something, converting that slow motion to something usably fast.