Satellite Mass: Newsletter Challenge (02/01/11)

If the satellite had any significant mass, it would be orbiting the barycenter of the Earth-satellite system. Since you say it is orbiting geocentrically, the mass of the satellite must be extremely small. (I've assumed that your value for the orbital speed is correct for the 2,000 km altitude you've placed it at. I'll run the numbers for that and check back to see if I need to make any additional comments.)

So, the mass is as small as you want it to be.

What you are saying is true. I ran the numbers and the mass of the satellite is not insignificant in this case.

The use of the word satellite was probably used as a deception since most people think of some communication satellite rather than an orbiting companion, such as our own Moon.

The crux of the problem is that any satellite of any significant mass will force the Earth to wobble as the other object orbits around it. The barycenter is not given in the problem.

Use Kepler's Third Law of Orbital Motion:

p^2 = [(4 * π^2)/G(m1+m2)] * a^3

m1 = mass of earth in kg = 5.9742 × 10^24

m2 = ??? Unknown mass

G = 6.67 x 10^-11

P = period, which we can calculate given R = 6378137 + 2,000,000 = 8,378,137 meters.

Velocity is 6,500 m/s.

Circumference of a circular orbit is 2πR = 5.264 x 10^7 meters...

P = 8099 seconds for one orbit.

Now just rearrange the equation and solve for m2.

6.559 x 10^7 = [39.478/(6.676 x 10^-11(5.9742 × 10^24 + m2))] * 5.881 x 10^20

I'll do the equation unravelling later...

Whatever the mass is, it won't stay up there for long at that speed.

Equation for orbital velocity (when the mass of the satellite is insignificant to the parent body):

V = √[(G * M)/R]

Running the numbers V = √[(6.67 x 10^-11 * 5.97 × 10^24) / 8.38 x 10^6 meters = 6.89 km/s

So, you are right, the velocity is too slow for a tiny mass. So, I think this is solved as a two-body problem where the mass of m1 ≃ m2.

Actually, the question doesn't say it's a circular orbit. A vanishingly small mass in an elliptical orbit could have that speed near apogee, while also orbiting geocentrically.

I would throw out the idea that it is eccentric because the problem, if stated properly, should contain all the information needed to be solved.

If it was an elliptical obit more information would have been required and the complexity of the problem jumps exponentially.

I did consider that - but (as usual) there's a vanishingly small amount of data given the the question, so I chose to ignore it as a possibility.

BTW - have you actually done the sums? What major & minor radii did you come up with? I haven't gone into it, but I wonder if the minor radius (semi-minor-axis) is greater than the radius of the Earth.

I haven't had much time to look at this in detail, but I did some quick (rough) calculations.

A tiny mass in an elliptical orbit that varies in altitude from roughly 1100 km to 2000 km above the Earth will have a speed at apogee of around 6.5 km/sec and a speed at perigee of 7.3 km/sec. These are rough calculations because the equations I used assume an orbit that is near-circular, whereas this orbit would have a measurable eccentricity. This, at least, satisfies the condition of an orbital speed of 6.5 km/sec at an altitude of 2000 km.

{e = 0.057 if I remember the formula correctly; e = [(8370 - 7470)/(8370 + 7470)], using R = 6370 km as the radius of the Earth. }

Another solution, as I alluded to initially, would be to assume the question intended to say the satellite orbits the barycenter of the Earth-satellite system. Other responders have taken this approach.

Hi Usbport, you wrote: "I haven't had much time to look at this in detail, but I did some quick (rough) calculations. A tiny mass in an elliptical orbit that varies in altitude from roughly 1100 km to 2000 km above the Earth will have a speed at apogee of around 6.5 km/sec and a speed at perigee of 7.3 km/sec."

A more accurate calculation gives a perigee about 684 km altitude and a speed of ~ 7.7 km/s, which is a nice, stable orbit. However, this does not quite solve the question, where a definite, presumably constant, altitude of 2000 km is given. This is the only interpretation which allows the mass of the satellite to be calculated from the info given.

It also seems very clear that it implies geocentric ("geocentrically orbiting 2,000 km above the surface of the Earth"), for which the only 'valid' solution looks like a negative mass (~ -5 x 10²³ kg). If the orbital speed was intended as barycentric, the situation gets worse, with even less geocentric orbital energy available.

So, unless there is something that everyone missed so far, I can only see a typo as the source of the frustration. :-)

-J

Jorrie,

Thanks for taking the time to refine my rough calculations. I still haven't had time to go find my old book on celestial mechanics and plug the equations into Mathcad. Guess I won't bother to, now.

As my early post noted, the question said geocentric, so I figured that only an insignificant 'test mass' would satisfy that requirement -- which also implied to me that asking for the mass was a red herring, and the real problem was to find a solution that permitted the given orbital speed at the given altitude. A circular orbit didn't work; that left only an elliptical solution, which I attempted to provide an example of with my quick and dirty method.

You are probably correct though, that there was some misinformation (or missing information) in the challenge as it was written.

Regards,

- U.

Hi USB, you wrote: "...that there was some misinformation (or missing information) in the challenge as it was written."

This and my prior reply to guest made me think that maybe they just left out one bit of info: the actual eccentricity(e) of the orbit (or equivalently, the perigee distance). If e was given as a little different from what I calculated for the 'negligible mass case', we would have been in business with calculating a mass.

Maybe we should suggest it, although it is a somewhat involved 2-body problem then...

Gentlepeople,

What if it is also geostationary?

Regards

Bill

Sorry,

a silly question.....

Then it cannot be in circular orbit.

Then how can it be geocentric?

If the orbital speed has the right value, it could be a relatively massive body in perfectly circular orbit around the barycenter of Earth and the mass. This would also give a perfectly circular orbit in geocentric coordinates, just with different orbital speed and radius.

The 6.5 km/s orbital speed seems like an error, because a massive body at a geocentric radius of 8375 km (equatorial) should require a geocentric orbital speed larger than one with negligible mass, which is already ~6.8 km/s. I haven't made the 2-body calculations, but that's my gut feel (like many others above). But, maybe a surprise lurks in the calculations. :)

-J

Oh no! surly not an error!

Hey, possibly it's anti-matter!

"Not to spoil it for those who would like to give it a go, I'll just give my result roughly: ~ 9% of Earth's mass."

Oops! There is a sign problem, the barycenter lies on the wrong side of Earth's center in my solution. So, I think the gut-feelings are still right - no valid solution.

-J

Now you tell me! I've been staring at this negative mass result for hours!

(It's what you get for M_s using:

V_o = √((M_e².G)/(M_e+M_s).r)

where V_o = orbital speed (6.5kms^-1), M_e = Earth mass, M_s = satellite mass and r = radius of Earth + 2,000km).

Is the 6.5 Km/s relative to the centre of the earth or the barycentre.

The question implies that the orbital speed is also geocentric and not barycentric, but it is not specifically stated. This complicates the calculations considerably. :(

I saw that, but figured I was doing something wrong.

Could this be made of that mysterious Dark Matter? ;-)

Hi AH,

"Could this be made of that mysterious Dark Matter? ;-)"

Nope, but maybe out of that mysterious Dark Energy.

The reason is that it will need a 'push' from somewhere... ;-)

-J

Hello Jorrie

I did a calc as follows. Took mass of Earth = 6*10²⁴ kg and radius 6400 km, not to be too precise.

Equation 1 for distance from Earth centre to centre-of-mass of the 2 bodies. Simple formula, but obviously also includes unknown satellite mass.

Equation 2 between centrifugal force on the satellite and gravitational attraction - for centrifugal force assumed the r in v²/r is distance from the satellite to the C-o-M. For the gravitational attraction assumed appropriate distance is between centres of Earth and satellite.

Solving I get mass of satellite = 7.9*10²³ kg and distance to C-o-M = 978 km. I.e. well inside Earth, reasonable for ratio of masses and distance.

At only 2000km from Earth's surface it would have to be some sort of collapsed object, but the Challenge doesn't rule that out.

Does that sound anything like?

Cheers........Codey

Hi Codey, you wrote: "Equation 1 for distance from Earth centre to centre-of-mass of the 2 bodies. Simple formula, but obviously also includes unknown satellite mass."

Not sure which equations you referred to, but that result that you got is not valid for the (implied) circular orbit of the challenge - that would require a negative mass with around that same absolute value, which is not possible.

If I plug a positive mass of that value into the 2-body equations, with your other values, it does produce a valid orbit, but one where the orbital speed varies between 6.5 km/s and 8.2 km/s, altitude between 2000 km and 290 km above the equator. This is fine for a compact body, but you see, if you allow elliptical orbits, I can plug virtually any mass in and get a valid orbit, but it will never be circular.

With the values supplied, one can only solve for a specific mass if a specific ellipticity for the orbit is also given, which is not the case here. So, no, this is not a solution to the challenge.

-J

Hello Jorrie

They're my equations, I should have made that clearer, and numbered them the other way round! More details:-

H = C-o-M distance from centre of Earth,

m_s = mass of satellite

R_e = radius of Earth 6400 km

M_e = mass of Earth 6*10²⁴ kg

h = height of satellite h from Earth surface, 2000 km

v = velocity of satellite, 6.5 km/s

Then distance from centre of Earth to centre of satellite = h + R_e

Distance from centre of satellite to C-o-M = h + R_e - H

Equation 2 is for centripetal acceleration = v²/(h + R_e – H), and gravitational field = g*(R_e/(R_e + h))² on assumptions in #49. Equating these gives H = 978 km.

Equation 1 is for C-o-M, M_e*H = m_s*(h + R_e – H)

Putting in H = 978 km gives m_s = 7.9*10²³ kg

Also putting H = 0 in Equation 1 gives m_s = 0, as expected

Putting H = (h + R_e)/2 gives m_s = 6*10²⁴ kg = M_e, as expected, 2 bodies of equal mass with C-o-M halfway between.

Using my figures for R_e and M_e velocity of satellite of mass << Me comes to 6.915 km/s.

Putting v = 6.91 in Equation 2 gives H = 12 km and m_s = 8.9*10²¹ kg - reasonable figures.

Putting in v = 6.92 gives H = -12 km and m_s = -8.4*10²¹ kg - giving as expected a non-physical situation.

I realise above checks don't prove anything, but do give some support.

Also I don't follow your comment in #10 - I'd have thought a more massive body would need lower velocity, as the gravitational field is the same, but distance to the C-o-M is smaller (on my assumptions about appropriate distances).

Cheers........Codey

Just trying this as a test to see if I can insert a link. I've not had much joy so far!

C:\Documents and Settings\Satellite.html

The link must point to a internet web page, not a file on your hard drive. You need to upload the file to some web server and then point the link to that page.

Thanks AH, I'll try that. One of these days I'll get it right!

Cheers.......Codey

Hi Codey, you wrote:

"Equation 2 is for centripetal acceleration = v²/(h + R_e – H), and gravitational field = g*(R_e/(R_e + h))² on assumptions in #49. Equating these gives H = 978 km."

I do not agree with your gravitational field equation, as it is only valid for a satellite with negligible mass. As an example, with your final mass (~13% of Earth's mass, which does not qualify as negligible), your "g" would be ~ 11 m/s^2, if such a point mass could be dropped from near Earth's surface. It is given by G(M_earth+M_sat)/Re², so it depends on the unknown mass, which you do not know up front. The problem is, you cannot first make a coarse approximation (negligible mass) and later use a previous result to calculate a significant mass as the 'answer'.

In any case, if we cannot show how it will produce a circular orbit at 2000 km altitude, I don't think it solves the challenge, not so?

-J

Hi Codey,

I wrote: "I do not agree with your gravitational field equation, as it is only valid for a satellite with negligible mass. As an example, with your final mass (~13% of Earth's mass, which does not qualify as negligible), your "g" would be ~ 11 m/s^2, if such a point mass could be dropped from near Earth's surface. It is given by G(M_earth+M_sat)/Re², so it depends on the unknown mass, which you do not know up front."

Sorry, I was a bit hasty - what I wrote there is nonsense! Relative to Earth's CoM the acceleration remains 9.8 m/s^2.

Your gravitational field extrapolation is correct. However, as I said in my prior posts, you do not have a solution to the puzzle, because it requires an orbit at 2000 km altitude circular with 6.5 km/s constant speed relative to Earth. Anything else would need some arbitrary extra assumptions.

-J

Hi again Codey, something I haven't responded to in your post #58:

You wrote: "Also I don't follow your comment in #10 - I'd have thought a more massive body would need lower velocity, as the gravitational field is the same, but distance to the C-o-M is smaller (on my assumptions about appropriate distances)."

This is true for orbital velocity relative to the CoM, but not relative to Earth (geocentric). For any significant mass, Earth also moves relative to the CoM, but in the opposite direction (in x,y,z coordinates). The geocentric orbital speed goes up when the mass of the orbiting body increases.

I think part of the confusion (also my own) on this is due to Wikipedia's http://en.wikipedia.org/wiki/Orbital_speed equation. It uses the distance (r) between the two bodies m₁ and m₂ (geocentric), but calculates the speed (v_o) relative to the CoM (barycentric). The geocentric orbital velocity is simply: v_geo = √[G(m₁+m₂)/r)], so with your values, the geocentric orbit speed needs to be ~7.343 km/s for a circular orbit at r = 8400 km.

This also places a question mark into my mind about the reference frames when we say that a massive object also falls at g=-9.8 m/s^2 near Earth's surface. Relative to what frame? I think it must also be true in the CoM frame, so in the geocentric frame a massive body should fall at a faster acceleration, not so? Maybe my 11 m/s^2 was not wrong, just ill defined.

-J

Hello Jorrie

To try to answer all your posts -

1. I must admit I didn't take too much notice of "geocentric" in the Challenge question. If it were simply a low-mass satellite the centre of orbit would be Earth centre, but this clearly isn't the case as the 6.5 km/s and 2000 km don't agree.

I assume both bodies orbit round the C-o-M, which is stationary (in absolute space, as Newton would have said). The Challenge doesn't say 6.5 km/s relative to Earth. It's possible to imagine centre of Earth stationary with the C-o-M and satellite going round, but I'm not sure how it could happen physically.

Not sure what you mean by "Relative to Earth's CoM the acceleration remains 9.8 m/s^2" in #69.

2. For simplicity, when considering Earth rather than a more general problem, I usually calculate gravity field from g at Earth surface, to avoid bringing G into it (which I have to look up!). But with my approximate figures g, G, M_e and R_e don't correlate exactly, so I've altered it in line with Wikipedia. Also put in more precise figures for M_e and R_e taken from #2. Recalculation gives

H = 844 km, m_s = 6.61*10²³ kg

3. Opened the Wikipedia link, but can't see formula v_geo ~ √[G(m₁+m₂)/r)] (not sure why it's ~). I saw v_geo ~ √[G*m₂²(m₁+m₂)/r)] which using my nomenclature is √[G*M_e²(M_e+m_s)/(R_e + h)].

Putting my figure for m_s into this gives 6.5 km/s (gratifyingly ). Of course could use the Wiki formula up front to find m_s = G*M_e²/[v²*(R_e + h)] which gives same answer. Wiki doesn't say how to find H but the Challenge didn't ask for that.

Also can find velocity of Earth = √[G*m_s²(M_e+m_s)/(R_e + h)] = 0.728 km/s, which agrees with 6.5*H/(R_e + h - H) as expected.

So I'm not yet convinced my answer isn't right!

Cheers......Codey

Hi Codey,

I think before we go into more technicalities, we must first try to agree what the challenge is that we are trying to solve. They gave us just two values and one indication: a single altitude, a single orbital speed and they indicated "geocentrically". I think it is reasonable to assume that since the altitude implies the distance from Earth's center (Re+h), the speed is also relative to that frame, i.e., geocentric. In this case, a circular orbit is not possible, as many respondents has also indicated.

If, on the other hand, the challenge intended to use the Wikipedia convention and the orbital speed (6.5 km/s) is indeed given in barycentric coordinates, I found from the the Wiki formula that the mass of the satellite is ~ 7.5*10²³ kg for a circular orbit. This is not that far from your m_s = 6.61*10²³ kg, perhaps just due to rounding of input numbers. I also get your H = 832 km, not very different to yours. So, if this interpretation of the challenge is valid, I think you can take the accolades for the first correct answer.

-J

P.S. I think there's a typo in your v_geo ~ √[G*m₂²(m₁+m₂)/r)]. Should rather be v_o ~ √[G*m₂²/((m₁+m₂)r)] (note, also not v_geo). It is relatively easy to prove that this cannot be a geocentric orbit speed, but must be a barycentric speed, because the centrifugal force is caused by the motion around the barycenter, not the geocenter. The gravitational force, on the other hand, must be geocentric. This mixing of coordinates systems is why I dislike the Wiki formula...

The equation v_geo = √[G(m₁+m₂)/r)] is not in the Wiki article, but it is also quite easily derived. Because of all the issues and the lack of a single, easy references, I'm busy writing up all the relevant equations and considerations. Will put it on my CR4 Blog for comments sometime in the near future.

-J

Hello again Jorrie

You're right, I missed out a / and a pair of brackets. Also I misunderstood what you were saying about v_o and v_geo and thought your formula was meant to be straight from Wiki.

Not only that but I made a mistake in my revised figures. Should have been

H = 938 km, m_s = 7.53*10²³ kg. Not sure how that happened, I'm inclined to blame Mathcad but it wouldn't be the first time I thought the computer was wrong but it turned out to be my fault .

Be interesting to see what the "official" answer is!

Cheers.........Codey

the radius of the orbiting satellite has to be less than 2000km!

BIG!

this means the volume of the satellite is less than a 30th of the volume of the earth's volume, the same density for earth and the satellite results in a masss less than 2e23kg for the satellite!

The question did not mention either the density or the size of the satellite.

if you know the density and the volume of the satellite you know the mass too

Orbital Speed is something like 24319km/h or 6.7752777777km/s; nearly 6.5km/s

5kg.

I correct myself: 5 Newtons. Mass was requested, not weight.

I correct myself again: 500 dynes. (.005 Newtons) Decimal error.

N (the Newton) is a unit of force, kg (the kilogram) is a unit of mass, both in SI units.

dyn (the dyne) is a unit of force in the (defunct) CGS system of units.

I correct myself again: my answer is 5 grams. You are quite right. I had the units right the first time but not the magnitude. It is amazing how one can forget such a basic fact.

As they say at school - "you only get full marks if you show how you worked it out".

Easy. F_centripetal=mrω²=mr(2∏/t)²

r=(2000km+6378km_eq)=8378km

t=2∏r/v=8089 seconds

F=m(8378km)(6.02E-7sec^-2)=.005N

(This is where I failed to complete the problem initially)

So: .005N/(1N/kg)=m=.005kg=5 grams.

where did you fetch the .005N?

Thank you for that. I see that my first calculation was correct (#14) and I lost the conversion to meters from kilograms in subsequent "corrections."

F=m(8378kmx1000m/km)(6.02E-7sec^-2)=5N

Scientific notation: F=m(8378000m)x(.00000602sec^-2)=5N

As I was reminded, F centripetal is unit of force, in Newtons from meters and seconds units, which opposes a rotating mass of 5.0 kilograms, at two significant figures.

As they also say in school - check your work!

Ignore the given velocity in the challenge question.

Now, for the given altitude above Earth, and a circular orbit, calculate the satellite orbital velocity.

Compare that with the given orbital velocity and explain the discrepancy.

Why would I ignore the given??

Formula for centripetal force is perfectly adequate for this problem:

F=ma, a_centripetal=rω^2, F_centripetal=mrω² , ω=2∏/t, t=∏D/v,

F=m in SI units in this case,

m=v²/r, v in m/sec, r in meters, m in kilograms.

m=(6.5E3m/sec)²/8378137m=5.0 kg. (answered in #14)

QED.

Do this simple test in your head. Take the satellite at the mass you have calculated in orbit and cut it in two without imparting any additional force/velocity to either half. Now do the two halves continue in the same orbit? If not what sort of force was one half applying to the other to keep them "in sync.".
OK now take away one of the halves......

What are you asserting the two 2.5 kg halves do? If you are asserting that they continue in orbit, please tell my why?

They must continue in orbit for the same reason as you thought they were already in orbit.

Think just about half "A" for a while. If it does something different because it's attached to "B": it can only do so because "B" is applying some force to it, but, then "A" would be applying an equal and opposite force to "B". So what would it do?

If it's too difficult to think about the "forces" which might be involved in "orbit". Just, in your head drop a brick off a building in a vacuum: does it do anything different to two half bricks. Now remove the earth and replace it with a point mass at the centre of the earth: and you'll see that the brick was in orbit after all.

I did not think they were already in orbit any more than the condition of the problem. It was clear from the beginning using the orbital velocity equation that it was in a degrading orbit. I appreciate your visualizations, however they are not germain to my thought processes. It is easy to default to the point of view, when one is bright, that the person you are dealing with is probably not as bright as yourself. One misses the opportunity of having an intelligent discourse that way. I have made that mistake too many times myself, and have missed out on the fun that this stuff should be. Cheers.

Wrong! ;-) Someone is a bad student. ;-) I will help...

Formula for velocity for a circular orbit when the mass of the orbiting body is << Earth:

v = √[(G * M)/R]

Where:

G = 6.67x10^-11 (Gravitational Constant)

M = 5.97×10^24 KG

R = 8.38x10^6 meters

v = √[(6.67x10^-11 * 5.97×10^24) / 8.38x10^6] = 6,893 m/s or 6.89 km/s

Check my math, please...

Question: Why does 6.89 km/s not equal the expected 6.5 km/s?

In other words, why does the check fail?

If your original statement was right is should pass the above check, but it does not. Do you now see the contradiction? Any ideas?

Thank you for that. Was enjoying playing with the concepts as if they were again new. It has been decades since I was in engineering school and find I must now look up the formulae. Now that Codemaster has apparently solved the problem, it has lost it's appeal. The only thing of interest to me now is the origin of the question. Cheers.

"F=m in SI units in this case"

Where did that come from? With F in Newtons and m in kilograms, acceleration = 1 ms^-2 is the only possibility; so why? Which of your equations gives this result?

Completely indeterminate. Orbital velocity is independent of mass.

No, it is not. However, it is inconsequential when one mass is significantly larger than another, then the second mass can be ignored.

If you want more information, search for Two Body Orbit Problem.

If you run the math for a normal satellite at that orbital distance above Earth the orbital velocity comes out to about 6.8 km/s, which does not match the given geocenter velocity of 6.5 km/s.

This should be a clue that the object's mass is significant, at least I think that was the intent.

Now there's an idea; what if the distance measurement is to to bottom of the object?

This would mean a radius of ~6.8 - 6.5 = ~.6 diameter! (or radial length, if say it's that 2001 thing )

Still not much help with the mass though

Am I missing something here?

The orbit of a satelite is not dependant on its mass, only on its velocity - for a circular orbit v = sqrt(GM/R). G = gravitaional constant, M = Mass of earth, R = radius of orbit.

For an orbital height of 2000km (R = ± 8371km) the orbital velocity would be about 6.9km/s. Not far off from what is given but without requiring the satelite mass.

Guest wrote: "Am I missing something here?"

Yep, check out http://en.wikipedia.org/wiki/Orbital_speed

-J

Jorrie

"A satellite is geocentrically orbiting 2,000 km above the surface of the Earth ..."

We know, or at least are lead to believe, that there is nothing within about 360 000km from the earth with a signifacant mass. I think it would be safe to assume that the mass of the satelite is therefore negligible compared to that of the earth, therefore should apply.

Hi Guest, but what would you say is the point of the challenge then?

BTW, nothing that we know of in Earth orbit has circular orbits either, so

does not quite apply either...

-J

Try solving your equation for R, with G = gravitational constant, M = Mass of earth, v=6.5ms^-1

A ring?

You mean like Larry Niven's 'Ringworld'? That's an interesting thought.

Yes, kind of. I confess I had to google "Ringworld".

Barycenter = Geocenter for a circular ring. And no Moon interference.

In response #4 AH showed that at an altitude of 2000 km, the orbital speed ought to be 6.89 km/sec. So if the ring is rotating at 6.5 km/sec the individual parts of the ring would want to fall inward to a lower orbit. This differential in speed would tend to cause a compression along the circumference of the ring, making it stronger. It might also create a form of artificial gravity within the ring (if the ring was hollow, like a torus shape), since a 'free floating' object inside the ring would tend to drift toward the inner wall nearest the Earth.

Yeah, compression, you know how much steel costs at those heights. BTW, we really regret having used up the Moon for gravel, but failing to do so would have jeopardised the cash flow right at the beginning. And now the orbit-by-wire system has to deal with more bodies only.

take gold, plumbum is to heavy!

[Sorry, editing problems] Yeah, compression, you know how much steel costs at those heights. BTW, we really regret having used up the Moon for gravel, but failing to do so would have jeopardised the cash flow right at the beginning. And now the orbit-by-wire system has to deal with distant bodies only.

It seems like the mass would have to be zero. Otherwise the orbit could not be truly geocentric.

Can't be zero mass or it wouldn't be "falling" - so couldn't be orbiting

the orbit is geocentrigal - the satellite is circling!

each satellite is independent of its mass orbiting around the earth!

Hi Wihelm, you wrote: "each satellite is independent of its mass orbiting around the earth!"

Only when the satellite mass is negligible compared to Earth. Check http://en.wikipedia.org/wiki/Orbital_speed.

The values given by this challenge actually requires a mass of -5.16E+23 kg, which is nonsensical (even for anti-matter). A positive mass of 5.16E+23 kg at 2000 km altitude needs an orbital speed of about 7.1 km/s to be in circular orbit, while a negligible mass needs about 6.8 km/s.

-J

OK!

A small mistake to my calculations: The Mass of the Earth is 5.973e27g (not kg)

m*v²/r=f*M*m/r²

M=r*v²/f is the Mass of the Earth!

if v is equal lightspeed, M is the Mass of the Universe for the (given) Radius r (13.7 Lightyears) of the Universe!

But the Mass of the satrellite is not to calculate (calculateable?)!

radius of earth is 6370km!

13.7 billion LY

is there an additional force at the satellite?

where are the little green man on the satellite?

Wilhelm asked: "where are the little green man on the satellite?"

Vanished, because it will most likely be a mini black hole. For that mass to fit into a radius of 2,000 km, the density must be at least 7 times the density of our moon, or 22,452 kg/m³ (about density of Iridium?), which is perhaps less likely than a mini black hole. Could perhaps be a neutron star, but AFAIK, they do not come in such 'lightweight' varieties.

-J

Hi !

I've just read in diagonal the various comments. Of course, there must be a trick.

It seems that, assuming a small (artificial) satellite that the speed is to small for the satellite to stay there, so the *satellite has its propeller on*, to fight against gravity.

So a possible answer could be : "the satellite mass is... monotonically decreasing ! "

The mass shouldn't even be a factor under normal conditions. The acceleration towards earth is constant (because distance is) no matter how much matter, and thus the velocity needed to overcome this centrifugal force/acceleration is the same no matter what mass we have.

In addition I think the person who wrote this miscalculated speed, perhaps by showing speed relative to the earths surface speed rather then absolute speed. Thus why we're getting negative masses/dark energy/propulsion.

So at the end of the day the answer is any object, no matter what the mass, would fall out of the sky. Without propulsion or something off-topic, that is.

In addition, if the mass were enough to pull the earth towards it, that's going to require an increase in distance or velocity. So if using the m1 and m2 equation instead of the negligible mass formula only makes the issues worse.

With what we have been given, all we know is that the object appears to be falling. The object may be accelerating back into orbit under conditions we have not been given.

But the centrifugal force decreases with the radius and is proportional to mass and square of the velocity. if the mass increases, so must the centrifugal force. At the given height of 2000000m, and at the given velocity of 6500m/s, there is a centrifugal force proportional only to the mass.

Let me correct my original post. I should not have used the word force at all in reference to centrifugal. I should have only used the term centrifugal acceleration.

Have you guys read some of the replies above?

As from post #58 and on, Codey and myself slowly started to converge on a credible solution to the challenge. One simply has to interpret the 6.5 km/s orbital speed of the satellite as relative to the barycenter (center of mass) of the two masses, giving you m_sat ~ 7.52*10²³ kg, a whopping 12.5% of Earth's mass. The orbital speed relative to Earth is then ~ 7.32 km/s, because Earth move in the opposite direction relative to the barycenter at ~ 818 m/s.

This is a bit of mixing of coordinate systems, but is common practice in astronomy, e.g. for binary stars.

That would make sense if you substitute barycenter for.geocenter.

That's right, but if it were geocentric, ie satellite and C-o-M both rotating round a stationary Earth, the satellite mass cancels from the force equation. So the situation is same as satellite of mass << Earth. Then the distance determines the velocity (or vice versa), you can't specify both. And the values in the Challenge don't agree on that basis.

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

Question of frames of reference?

Yes. This is from Wikipedia. The red cross is the barycenter and the geocenter being the center of the larger body (Earth) in this example.

The relative orbital distance is different based on which center you call it from.

Note that the radius between bodies remains constant in distance as does the distance for the barycenter for a true circular orbit.

Yep - OP shouldn't have used the term "geocentric" - but it has led to 90 odd more posts of interesting fun after that became obvious.

"OP shouldn't have used the term "geocentric" - but it has led to 90 odd more posts of interesting fun after that became obvious."

Problem is that the distance was effectively given in geocentric coordinates, with the speed barycentric, without saying so. Anyway, I agree that this made it an interesting puzzle to ponder.

-J

The mass of the satellite is it's weight at the Earth's surface divided by the acceleration due to Gravity at the same place. I.e. m(kg)= W(Newtons)/g(m/s**2).

All objects in the same (stable) orbit will have the same velocity, v= (GM/R)**1/2, irrespective of the mass of the orbital body. Where M is the mass of the orbited body. Thus, it is not possible to determine the mass of the satellite using the velocity and orbital radius.

It is possible however, to weigh the Earth, given the orbital parameters: using the 'centripetal equation', mv**2/R+h = GMm/(R+h)**2. Hence, M = ((R+h)*v**2)/G. Using a value for G = 6.67*10**-11 Nm**2.kg**2, and (r+h) = 8.37*10**6 m yields M = 5.3*10**24 kg.

"Sometimes I think the surest sign that intelligent life exists elsewhere in the universe is that none of it has tried to contact us." Calvin. (B. Watterson)