Gravity issues

[GeeKay]

According to my (often suspect) calculations, it appears that no object is able to orbit Phobos, the inner moon of Mars, without being continually pulled in by Mars' own gravity field. Even the situation with the outer moon, Deimos, is a close call in this respect. I wonder, though, if I've got my maths right. This being more than possible, I would welcome any help or advice. Many thanks.

 

PS. I am aware that maintaining a freefall orbit round either moon would be problematic in any case, given the varying gravitational fields resulting from their irregular dimensions. I'm ignoring that for now.

[imatfaal]

Are you actually trying to solve a full three-body calculation or making simplifying assumptions?

[GeeKay]

Well, let's see: if one accepts that the mass of Mars is 639e21 kg and 1.0659e16 kg is that of Phobos, accepting too that the mean distance between the centres of these two bodies is about 12,7805 km, it turns out (or the calculations appear to indicate) that the gravitational attraction of Mars at this distance is 0.2094 Newtons. Meanwhile, an object orbiting Phobos at an altitude of (say) 31 km from the satellite's centre works out to a measily 0.00074 Newtons. If these figures are correct, this strongly suggests - to me at least - that no stable orbit is possible round Phobos. I did as a control apply the same set of calculations for the Earth and the Moon, and they checked out fine. Of course, the Moon is many times larger than Phobos, but then the distance between the Moon and Earth is in relative terms much greater than that between Mars and Phobos. Indeed, I understand that Phobos is in a dangerously low orbit, that it's not that far off its Roche Limit and that tidal stresses will eventually cause it to break up and form a ring round Mars and/or make a number of impressive dents in the Martian regolith. I hope I've not made too many assumptions - simple or otherwise - in my assessment of the situation thus far. But I do wonder if I've made a fundamental error somewhere along the line. Hence my query.

Correction: a misplaced decimal point. The distance between the centres of Mars and Phobos is actually 12,780.5 km.

[Enthalpy]

In a first approach:

Mars weighs 6.4e23kg, Phobos 1.0e16kg

Their distance is 9377km, so Phobos' gravitation dominates up to 16km from Phobos' centre only.

 

So any orbit around Phobos needs R<16km BUT Phobos is 27*22*18km big, so there's very little room to fit.

 

Then there are the less simple perturbations by Mars and also the Sun...

 

 

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First solution: land on Phobos instead of orbiting it. Take samples, preferably from several sites, do the same at Deimos, bring them back to Earth. Trivial with my Sunheat engine; described there

http://www.scienceforums.net/topic/76627-solar-thermal-rocket/#entry755323

this was before the improved script

http://www.scienceforums.net/topic/76627-solar-thermal-rocket/page-2#entry818683

so the mass back to Earth might be like 4.5t instead of 1.5t - almost too easy. Only sampling Mars is a challenge with the sunheat engine, coming later.

 

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Maybe an other possibility, but I'm not sure at all: use a Mars orbit instead, that is, >19km from Phobos. Phobos' attraction would deform it slightly to pass sometimes below, sometimes above, right, left... The orbit can be slightly tilted and elliptic to help.

 

Tell me if you find one, it would serve at the small moons of Jupiter and Saturn, since the sunheat engine permits to orbit all their equatorial moons. At D~10km it's easier to land.

 

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One good possibility: stay at a Lagrangian point L1 or L2. They're 16km from Phobos, you make nice pictures from there. Very little fuel move the probe between them.

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

[imatfaal]

Well, let's see: if one accepts that the mass of Mars is 639e21 kg and 1.0659e16 kg is that of Phobos, accepting too that the mean distance between the centres of these two bodies is about 12,7805 km, it turns out (or the calculations appear to indicate) that the gravitational attraction of Mars at this distance is 0.2094 Newtons. Meanwhile, an object orbiting Phobos at an altitude of (say) 31 km from the satellite's centre works out to a measily 0.00074 Newtons. If these figures are correct, this strongly suggests - to me at least - that no stable orbit is possible round Phobos. I did as a control apply the same set of calculations for the Earth and the Moon, and they checked out fine. Of course, the Moon is many times larger than Phobos, but then the distance between the Moon and Earth is in relative terms much greater than that between Mars and Phobos. Indeed, I understand that Phobos is in a dangerously low orbit, that it's not that far off its Roche Limit and that tidal stresses will eventually cause it to break up and form a ring round Mars and/or make a number of impressive dents in the Martian regolith. I hope I've not made too many assumptions - simple or otherwise - in my assessment of the situation thus far. But I do wonder if I've made a fundamental error somewhere along the line. Hence my query.

Correction: a misplaced decimal point. The distance between the centres of Mars and Phobos is actually 12,780.5 km.

 

Phobos orbits mars at about 9500km.

 

To give a gravitational attraction in Newtons you have to specify both objects - the formula is F_grav = G M_1 m_2/r^2 Even for a unit kilo mass I couldnt get your figure (although close).

 

I make the roche limit (assuming that phobos is spherical which it isnt) just over 10000 km - which Phobos is within. So it is clear that the actual structural integrity of Phobos is holding it together rather than being able to rely on gravity alone (ie outside the roche limit self gravity will hold something together)

 

I would take a step into the dark and agree with you - an orbiter that cannot maintain a test mass on its surface (ie is within the rigd body roche limit) is not going to be able to maintain an object in simple orbit. But, and it is a big but...this is a three body problem and simple solutions are often obvious and wrong. There may exist a complex orbit around both bodies (mars and phobos) - or one that looks like it is around one but is in fact unstable eventually etc.

 

I would read up on the three body problem and Poincare, Newton, Lagrange, d'Alembert (which is a damned impressive group) and all the other great names of Physics who have thought about it

[swansont]

In a first approach:

 

First solution: land on Phobos instead of orbiting it.

 

!

Moderator Note

That's not the topic under discussion. You have a thread where you can talk about this, so please confine the discussion to that thread and also not respond to this modnote in the thread.

[imatfaal]

In a first approach:

Mars weighs 6.4e23kg, Phobos 1.0e16kg

Their distance is 9377km, so Phobos' gravitation dominates up to 16km from Phobos' centre only.

 

So any orbit around Phobos needs R<16km BUT Phobos is 27*22*18km big, so there's very little room to fit.

 

Then there are the less simple perturbations by Mars and also the Sun...

 

.../snipped well off topic stuff

 

Could you explain that calc Enthalpy? By equating the force under newtons law of universal gravitation (which seems simplest way - hope not too simple) I get much less than that. Which makes sense as phobos is within the simple rigid sphere roche limit.

[GeeKay]

it would appear that any spacecraft being inserted into a fairly stable and fuel-efficient orbit round Phobos (and possibly Deimos too) requires a non-Keplerian orbit, such as are described in the link below.

 

http://trs-new.jpl.nasa.gov/dspace/bitstream/2014/42478/1/12-3014.pdf

[Enthalpy]

 

Could you explain that calc Enthalpy? By equating the force under newtons law of universal gravitation (which seems simplest way - hope not too simple) I get much less than that. Which makes sense as phobos is within the simple rigid sphere roche limit.

 

Compensing Mars' attraction forces is not necessary, because Phobos "falls" (orbits) freely in Mars' attraction. So this is not a case where Mars and Phobos are pinned to the floor and both attract a test mass in between, for instance a pebble at Phobos' surface to compare with the Roche radius. It is a case where Mars attracts the whole Phobos, and we wonder by how much the acceleration resulting from Mars is stronger at Phobos' surface than at its center of inertia, and whether the acceleration produced at Phobos' surface by its own mass can overwhelm it.

 

So as a first correction, Mars acts only by its tidal effect. I write M, m the masses of Mars and Phobos, R the distance to Phobos, r the distance from Phobos to the questionable position.

 

From GM/R² the difference of attraction by Mars is r*2GM/R³ (write signs where you want if it helps you, here's a stock - - - -), to be compared with the full Gm/r², which gives 2M/R³ = m/r³, or (r/R)³ = m/2M.

With M=6.4e23kg, m=1.0e16kg R=9377km this brings us a first Roche limit of r=18.6km, slightly more than Phobos' biggest radius.

 

A second correction (often included in the term "tidal effect") is that the centrifugal force also differs between Phobos' centre of inertia and the questionable position; it varies as r/R, and equates at the centre of inertia the attraction GM/R², introduces again a r*1*GM/R³, so that the limit is now

3M/R³ = m/r³, or (r/R)³ = m/3M. Now r=16.3km, still a bit more than Phobos' biggest radius.

 

This is for a moon in synchronous rotation (probable if we wonder about Roche's limit) and holds as well for the Lagrangian points L1 and L2

 

A computation at Wiki if someone finds it any understandable - they make it with density instead of mass

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

the magic word "tidal" is in the first line of the "explanation" paragraph. This is for a stiff body; a deformable one is more complicated.

 

This computation is exactly the same as for the Lagrangian points L1 and L2, these ones are less complicated at Wiki

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

Edited August 24, 2014 by Enthalpy

[Janus]

Could you explain that calc Enthalpy? By equating the force under newtons law of universal gravitation (which seems simplest way - hope not too simple) I get much less than that. Which makes sense as phobos is within the simple rigid sphere roche limit.

He's basically using the equation for the hill sphere:

 

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

 

However, it should be noted that while the outer limit for a orbit around Phobos is 16 km, for long term stability, you need to be at ~1/2 this distance.

[Enthalpy]

Yes, that' it, Hill sphere. The Roche limit is essentially the same if treating a non-deformable body, but astronomers define it versus the densities rather than masses, and seek the limiting distance to the more massive body. It boils down to the same condition.

 

I certainly agree that any orbit so near to the limit wouldn't last.

 

Though, pebbles just resting on Phobos would stay there, for being nearer to Phobos' center than the Lagrangian points L1 and L2 are - which is the case because only tidal forces from Mars act on them.

[GeeKay]

So I gather then that a fairly largish object (an astronaut, say) could safely stand on that part of Phobos facing Mars and not be borne away by the Martian gravity field - or its tidal forces, I mean?

[fiveworlds]

Explain Mars weighs 6.4e23kg

[Janus]

So I gather then that a fairly largish object (an astronaut, say) could safely stand on that part of Phobos facing Mars and not be borne away by the Martian gravity field - or its tidal forces, I mean?

Yes, using the rigid body formula for the Roche limit, the distance from Mars where tidal force from Mars and the gravity of Phobos would cancel out is ~5450 km, and Phobos orbits at over 9000 km.

[imatfaal]

Thanks for the Explanations - Enthalpy and Janus

[fiveworlds]

It's okay nasa put it up https://www.google.com/url?q=http://spacemath.gsfc.nasa.gov/Insight/Insight14.pdf&sa=U&ei=kAL9U_CGOsrb7Ab6woDoCQ&ved=0CAwQFjAA&usg=AFQjCNGIj0HDxXQmXqUTxoT3G2xxnzZwhA

But there is no sun's gravity there.

[Enthalpy]

So I gather then that a fairly largish object (an astronaut, say) could safely stand on that part of Phobos facing Mars and not be borne away by the Martian gravity field - or its tidal forces, I mean?

 

Yes. An astronaut, a pebble... Phobos is still far enough from Mars - but not by much - to retain its integrity through its own gravitation.

 

An orbit is more difficult, because it deforms over time du to Mars' influence, so I bet an orbit good enough would be very difficult. But that's only a guess, and there must be other possibilities serving the same purpose - say, an orbit formally around Mars that stays near Phobos and permits to observe Phobos from varied directions.

It's okay nasa put it up https://www.google.com/url?q=http://spacemath.gsfc.nasa.gov/Insight/Insight14.pdf&sa=U&ei=kAL9U_CGOsrb7Ab6woDoCQ&ved=0CAwQFjAA&usg=AFQjCNGIj0HDxXQmXqUTxoT3G2xxnzZwhA

But there is no sun's gravity there.

 

Nasa, Wiki and many more reveal Mars' mass. It doesn't exactly need an explanation.

 

Please feel free to provide us your estimate of the Sun's tidal effect near Mars. Hints: at Earth, the tidal effects of the Sun and our Moon are nearly as strong, Mars is farther to the Sun than Earth is, Mars is heavier than our Moon, and Phobos is much nearer to Mars than Earth to our Moon. And tidal effects vary as distance^-3.

[Strange]

But there is no sun's gravity there.

 

Huh? Where do you think the Sun's gravity stops!?

[fiveworlds]

Hmm I tried this using Halley's comet.

M=[4*(massive pi)2/6.67*10^-11](26692748122530)^3/2365200000=3.1744701576e+32 KiloGrams mass of sun.

I am a bit off 1.9891 × 10³⁰ kilograms

Edited August 27, 2014 by fiveworlds

[Strange]

Hmm I tried this using Halley's comet.

M=[4*(massive pi)2/6.67*10^-11](26692748122530)^3/2365200000=3.1744701576e+32 KiloGrams mass of sun.

I am a bit off 1.9891 × 10³⁰ kilograms

 

I get something pretty close (1.9818 x10³⁰ kg)

 

Your value for the radius (26692748122530) seems about 10 times too big. And you haven't squared the period (2365200000).

[fiveworlds]

17.843 AU in meters

1 AU = 149597871 kilometers

17.843*149597871000=26692748122530 metres

Though you're right I forgot to square silly me

M=[4*(3.141)2/6.67*10^-11](26692748122530)^3/5594171040000000000

Edited August 27, 2014 by fiveworlds

[Strange]

17.843*149597871000=26692748122530 metres

 

You definitely have an extra zero there:

http://www.wolframalpha.com/input/?i=17.843*149597871000

[Janus]

Besides the arithmetic mistakes, there is the issue of significant digits. From the time you gave for the period for Halley's comet you are using the value of 75 years ( and multiplying by 60 x 60 x 24 x 365 to get the time in seconds). Now while there is nothing inherently wrong with this, it should be noted that Halley's comet actually has a period closer to 75.3 years. What this means is that one of your values is only accurate to 2 significant digits. Any the accuracy of any equation is limited by the value with the least significant digits.

 

Put another way, it makes no sense to use a value for an AU that is accurate to 9 digits when you are using a value for the period that is limited to 2 significant digits and value for G that is only accurate to 3 digits.

 

Nor should you expect an answer that is accurate out to 5 digits.

[imatfaal]

Besides the arithmetic mistakes, there is the issue of significant digits. From the time you gave for the period for Halley's comet you are using the value of 75 years ( and multiplying by 60 x 60 x 24 x 365 to get the time in seconds). Now while there is nothing inherently wrong with this, it should be noted that Halley's comet actually has a period closer to 75.3 years. What this means is that one of your values is only accurate to 2 significant digits. Any the accuracy of any equation is limited by the value with the least significant digits.

 

Put another way, it makes no sense to use a value for an AU that is accurate to 9 digits when you are using a value for the period that is limited to 2 significant digits and value for G that is only accurate to 3 digits.

 

Nor should you expect an answer that is accurate out to 5 digits.

 

Completely agree. I have studied no science at an advanced/university level - but recently I have audited a few courses and one of the major surprises was the emphasis and time spent on accuracy/precision. In the end science needs experimentation and empirical observation - but there is no point in looking at experimental data without knowing what the level of uncertainty is. Science must resolve eventually to a prediction based on a set of parameters - and that must entail an understanding and use of uncertainty and imprecision