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Little orbits?

by Glenn
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Glenn
#1
Jan28-05, 10:52 PM
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What are the smallest masses that could theoretically be put in space and set up in an orbital configuration. example 1 - equal masses , example 2- sun-pluto type relationship.

Thanks,
Glenn
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Labguy
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Jan28-05, 11:47 PM
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Quote Quote by Glenn
What are the smallest masses that could theoretically be put in space and set up in an orbital configuration. example 1 - equal masses , example 2- sun-pluto type relationship.

Thanks,
Glenn
If you mean "free space", then I think that two electrons or an electron-positron pair could orbit each other, if at the right distances and velocities.

For "Big-Little" it would be a massive black hole being orbited by photons at the "photon sphere" which is 3GM/c2.
marcus
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Jan29-05, 02:02 AM
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I am picturing a toy "solar system" with planets the size of marbles
in orbit around a bowlingball size thing----way out in space of course

you asked about the smallest------you could make the masses so small you couldnt see them, but then its no fun

1. be sure to make the masses electrically neutral, otherwise you will be seeing not pure gravity but some effect of attraction and repulsion of electric charges

2. get far away so the gravity of your own body doesnt confuse the toy planets by its own gravitational attraction--------dont leave any equipment around either, everything attracts everything else by gravity.


this is certainly not the theoretical smallest but imagine this: two solid gold spheres each 1 centimeter in radius.
place them 3 centimeters apart center to center
(so there is a 1 centimeter gap between them)

start them moving in circular orbit around each other and get everything else far away so its gravity doesnt interfere

watch them with binoculars, as they orbit.

Question: how many minutes is the orbital period? how long does it take the two solid gold balls to go around each other in one complete circuit?

Labguy often says how much he loves to calculate and he could certainly tell you how many minutes. piece of cake. you might be surprised how long it would take them to go around

Labguy
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Jan29-05, 01:26 PM
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Little orbits?

Quote Quote by Marcus
this is certainly not the theoretical smallest but imagine this: two solid gold spheres each 1 centimeter in radius.
place them 3 centimeters apart center to center
(so there is a 1 centimeter gap between them)

start them moving in circular orbit around each other and get everything else far away so its gravity doesnt interfere

watch them with binoculars, as they orbit.

Question: how many minutes is the orbital period? how long does it take the two solid gold balls to go around each other in one complete circuit?

Labguy often says how much he loves to calculate and he could certainly tell you how many minutes. piece of cake. you might be surprised how long it would take them to go around
Yeah; I love it. It must be somewhere in here:

For circular Keplerian orbits where:
Vc = velocity of a circular orbit
Vesc = escape velocity
M = Total mass of orbiting and orbited bodies
G = Gravitational constant (defined below)
( G = 9.80665 m / s^2 )
u = G * M (can be measured much more accurately than G or M)
K = -G * M / 2 / a
r = radius of orbit (measured from center of mass of system)
V = orbital velocity
P = orbital period
a = semimajor axis of orbit

Vc = sqrt(M * G / r)
Vesc = sqrt(2 * M * G / r) = sqrt(2) * Vc
V^2 = u/a
P = 2 pi/(Sqrt(u/a^3))
K = 1/2 V**2 - G * M / r (conservation of energy)

The period of an eccentric orbit is the same as the period
of a circular orbit with the same semi-major axis.
but I'm not going to punch it in. I'm too dumb to get "u" figured correctly because the "s^2" in G is losing me. Someone please do it for me. I came up with ~ 2.60 to 2.75 seconds, but that sounds way too long. (??).....
tony873004
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Jan29-05, 01:57 PM
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It would take 2 hours and 38 minutes for two gold balls, 1 cm in radius, spaced 3 cm apart, center to center, to perform one circular orbit around their common barycenter.

First, we need the mass of the balls:
volume of sphere = 4/3 pi r^3
volume of each ball = 4/3 pi 1^3 = 4/3 pi
volume of each ball = 4.189 cubic centimeters

density of gold: 19.32 g/cm^3
mass = density * volume
mass = 19.32 * 4.189
mass = 80.927 grams
mass = 0.08927 kg
mass of both balls = 0.17854 kg

Next we need the circular orbital velocity:
M = mass of both balls in kg = 0.17854
G = Gravitational constant = 6.67e-11
r = seperation between the 2 balls (center to center) = 0.03 meters
Vc = sqr (M * G / r)
Vc = sqr(0.17854 * 6.67e-11 / 0.03)
Vc = 1.99237028017719E-05

Next we need to know how far a ball travels in one orbit:
distance to travel = 2 * pi * r
d = 2 * pi * 0.03 meters
d = 0.188495559215387 meters

Finally, we can compute the time.
time = d / Vc
time = 0.188495559215387 / 1.99237028017719E-05
time = 9461 seconds
time = 2.62 hours
time = 2 hours 38 minutes
Labguy
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Jan29-05, 06:04 PM
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Quote Quote by tony873004
time = 2.62 hours
I knew that! I said 2.6 seconds, you say 2.6 hours; that means I was only off by a factor of 3,600. It must have been my rounding errors..

Isn't there an easier way, fewer steps, since both masses, sizes and densities are identical? The most math I can do is with the capabilities of my give-away calculator that says "Thank you for shopping at WalMart" on it.
tony873004
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Jan29-05, 10:21 PM
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Quote Quote by LabguyIsn't there an easier way, fewer steps, since both masses, sizes and densities are identical? The most math I can do is with the capabilities of my give-away calculator that says "[I
Thank you for shopping at WalMart[/I]" on it.
Mass, distance and G are the only values you need. If one of the balls were lead instead of gold, but it still had the same mass, the results would be the same even though the lead ball would be smaller and denser. Size and density are unimportant, except in this example, mass was not given, so they were needed to compute the mass.

Changing the mass changes the result. So I don't think there's any shortcut introduced by the balls' similarities.
Janus
#8
Jan30-05, 12:43 PM
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Quote Quote by Labguy
I knew that! I said 2.6 seconds, you say 2.6 hours; that means I was only off by a factor of 3,600. It must have been my rounding errors..

Isn't there an easier way, fewer steps, since both masses, sizes and densities are identical? The most math I can do is with the capabilities of my give-away calculator that says "Thank you for shopping at WalMart" on it.
If the masses are equal you can use the formula:
[tex]T = \pi \sqrt{\frac{d^3}{2GM}}[/tex]

d is the distance center to center and M is the mass of either object.

This formula is the formula
[tex]T =2 \pi M \sqrt{\frac{d^3}{G(M+m)^3}} [/tex]

adjusted for M=m.

As for your calculator, you'll just have to convert to scientific notation, and use the calculator for the significant digits. You'll probably need a scratch pad to keep track of what you're doing, but at least its not the bad old days when you would have had to use a slip-stick.
tony873004
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Jan30-05, 02:00 PM
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Quote Quote by Janus
If the masses are equal you can use the formula:
[tex]T = \pi \sqrt{\frac{d^3}{2GM}}[/tex]

d is the distance center to center and M is the mass of either object.

This formula is the formula
[tex]T =2 \pi M \sqrt{\frac{d^3}{G(M+m)^3}} [/tex]

adjusted for M=m.
These formulas give an answer exactly half of the answer I got. Changing it to
[tex]T = 2\pi \sqrt{\frac{d^3}{2GM}}[/tex]
or
[tex]T =4 \pi M \sqrt{\frac{d^3}{G(M+m)^3}} [/tex]
makes them agree with what I got.
Could it be that pi only represents half the orbit, so pi needs to become 2pi to give the full orbit?
Janus
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Jan30-05, 03:03 PM
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Quote Quote by tony873004
These formulas give an answer exactly half of the answer I got. Changing it to
[tex]T = 2\pi \sqrt{\frac{d^3}{2GM}}[/tex]
or
[tex]T =4 \pi M \sqrt{\frac{d^3}{G(M+m)^3}} [/tex]
makes them agree with what I got.
Could it be that pi only represents half the orbit, so pi needs to become 2pi to give the full orbit?
The second formula must approach

[tex]T =2 \pi \sqrt{\frac{d^3}{GM}} [/tex]

as m becomes very small compared to M, Which my form does and yours doesn't.

Going back over your post, I notice that you took 0.03m as the radius of the masses' orbits when calculating the period, but this is the center to center distance between the two masses. They will orbit the barycenter which, in this case, is located halfway between the two masses. This makes the radius of the orbit 0.015m, cutting the period to half that which you got.
tony873004
#11
Jan30-05, 04:49 PM
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Quote Quote by Janus
Going back over your post, I notice that you took 0.03m as the radius of the masses' orbits when calculating the period, but this is the center to center distance between the two masses. They will orbit the barycenter which, in this case, is located halfway between the two masses. This makes the radius of the orbit 0.015m, cutting the period to half that which you got.
Cutting the radius in half would not cut the period in half. Time is not proportional to radius, but time is proportional to [tex]radius^{3/2} [/tex] by Kepler's 3rd law.

Using 0.03 is what you would want though, as orbits are measured center to center, and not barycenter to center. Two objects 1 kg each orbiting 1 meter from each other (0.5 m to barycenter) would use the same formulas as 1 massless test particle orbiting a 2 kg object from 1 meter.

Using Gravity Simulator, I set this up. Two identical objects (except color), 1 cm in radius, and 0.08927 kg orbit each other in circular orbits, 0.03 m, center to center. They take 1 hour 19 minutes to complete one half an orbit, and 2 hours 38 minutes to complete a full orbit.
http://orbitsimulator.com/orbiter/gold1.GIF (the red object is on top)
http://orbitsimulator.com/orbiter/gold2.GIF (1/2 orbit later, the green object is on top, 1h 19 m later)
http://orbitsimulator.com/orbiter/gold3.GIF (after a full orbit, the red object is back on top, 2h 38 m later)
Janus
#12
Jan30-05, 11:48 PM
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Going back over my derivation, I noted that I did make a mistake along the way. I inverted a term early on (serves me right for not double checking my work)). The correct equations should be:

[tex]T= \pi \sqrt{\frac{2d^3}{GM}}[/tex]

equal masses and

[tex]T = 2 \pi \sqrt{\frac{d^3}{G(M+m)}}[/tex]

in the General case.

While these give the same answer as tony873004 got for the period, I still have problems with how he got his answer.

In the first place, he calculates the orbital velocity as

[tex]V_{o} = \sqrt{\frac{G{M+m}}{d}}[/tex]

which is incorrect, the proper expression is

[tex]V_{o} = M \sqrt{\frac{G}{d(M+m)}}[/tex]

Where the orbital velocity is for mass m.

This is because the bodies do orbit the barycenter. An example is the Earth and Moon, the Earth makes a small orbit around the Barycenter and the moon makes a larger orbit. The period is the same for both, but since the Earth travels a much smaller circle, its orbital velocity is much smaller than the Moon's around that same barycenter. tony873004's equation does not take this into account, and apparently gives the same orbital velocity for both bodies regardless of their relative mass. He also gets a value that is too high by a factor of 2.

He then calculates the period by dividing this velocity into the circumference of an orbit with a radius of .03 m. Again, since the objects orbit the Barycenter, the radius of each orbit is only .015 m, he is high by a factor of 2 once more. Since these two errors are divided by each other, they end up canceling each other out when calculating the period.

Going back to our example of the Earth, the radius of the moon's orbit around the common barycenter is:

[tex]D_{Earth-moon}\frac{M_{earth}}{M_{Earth}+M{moon}} = 379330 km[/tex]

Which it travels at 1.012 km/sec in about 27.3 days

The Earth's orbit has a radius of

[tex]D_{Earth-moon}\frac{M_{moon}}{M_{Earth}+M{moon}} = 4670 km[/tex]

which it travels at .01246 km/sec in the same 27.3 days

When calculating orbital velocity and the size of the orbits, the relative size of the masses do matter.
tony873004
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Jan31-05, 04:03 AM
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My formula is the one commonly listed as the formula for circular velocity in Physics texts and websites such as in this website:
http://ceres.hsc.edu/homepages/class...ics/sec10.html
But it is used for spacecraft, where m << M. This formula doesn't give barycentric speeds. But in the special circumstance where masses are equal, you can add the masses together, assign them to one object and consider the other object massless. So a formula that does not provide barycentric results will work in this particular instance.

btw...
my formula needs parenthesis around M+m
[tex]V_{o} = \sqrt{\frac{G(M+m)}{d}}[/tex]
I think you meant to do that, but made a tex typo. Your tex code had {{ }} instead of {( )}

Your formula needs to be changed slightly to work, and broken down into seperate formulas for the Earth and Moon:

Where
M is the mass of the Earth
m is the mass of the Moon
For the Moon:
[tex]V_{o} = m \sqrt{\frac{G}{d(M)}}[/tex]


For the Earth:
[tex]V_{o} = m \sqrt{\frac{G}{d(m)}}[/tex]


When calculating orbital velocity and the size of the orbits, the relative size of the masses do matter.
Are you sure? I don't see why the size of the object would matter at all. Neither your formula or my formula use the objects' sizes.
Labguy
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Jan31-05, 10:47 AM
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Quote Quote by tony873004
Are you sure? I don't see why the size of the object would matter at all. Neither your formula or my formula use the objects' sizes.
Another Labguy "no-math" post:

Unless I'm totally lost, I can't see how "size" would matter in any of the calculations above. We need masses and distance and G, but sizes couldn't matter unless two objects were in contact, a rare case except in "contact binaries".

When one body is much more massive than the other, G and distance should do it, even without mass of the smaller object(s). For example, the many-ton space shuttle is orbiting at, say, 280 miles. When they do a waste dump, very small chunks of "stuff" is released. So, we have a X-ton shuttle orbiting at 280 miles, and a small piece of 3-ounce junk following in the same orbit. Just because the shuttle is large (= more mass) and the junk is small (= less mass) neither one will suddenly need to "jump" to a different orbital height (distance) would it?
Janus
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Jan31-05, 06:11 PM
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Quote Quote by tony873004
My formula is the one commonly listed as the formula for circular velocity in Physics texts and websites such as in this website:
http://ceres.hsc.edu/homepages/class...ics/sec10.html
But it is used for spacecraft, where m << M. This formula doesn't give barycentric speeds. But in the special circumstance where masses are equal, you can add the masses together, assign them to one object and consider the other object massless. So a formula that does not provide barycentric results will work in this particular instance.

btw...
my formula needs parenthesis around M+m
[tex]V_{o} = \sqrt{\frac{G(M+m)}{d}}[/tex]
I think you meant to do that, but made a tex typo. Your tex code had {{ }} instead of {( )}

Your formula needs to be changed slightly to work, and broken down into seperate formulas for the Earth and Moon:

Where
M is the mass of the Earth
m is the mass of the Moon
For the Moon:
[tex]V_{o} = m \sqrt{\frac{G}{d(M)}}[/tex]


For the Earth:
[tex]V_{o} = m \sqrt{\frac{G}{d(m)}}[/tex]
????.

Your first formula gives the the Moon an orbital velocity of 125 m/s or .125 km/s, with an orbital radius of 379330 km, this works out to a period of over 220 days.

The second formula gives the Earth an orbital velocity of 1.130 km/sec, causing it to circle the barycenter (at 4670 km distant from the Earth's GoG) in 7.2 hrs.

Even you transposed the formulas, the period for the moon would be 24.4 days ( almost three days short of it actual period), and the Earth would have a period of 2.7 days.

Both these periods have to be the same.

My formula does give the correct answers, and you really don't have to use two formulas as long as you designate m as the mass of the object for which you are calculating the velocity for (rather then any particular planet or moon).


Are you sure? I don't see why the size of the object would matter at all. Neither your formula or my formula use the objects' sizes.
In "size" I meant the magnitude of the masses. I realise that it might not have been the best choice of words, but I assumed it meaning would be obvious in the context of the text, where I made no mention of differences in physical dimensions of the bodies in question, either before or after this statement.
Sorry for the confusion.
tony873004
#16
Jan31-05, 06:59 PM
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You're right. I didn't realize little m would represent Earth on the Earth velocity formula.

I swear my 2 formulas worked last night. But they don't work today! I entered them into Visual Basic but I probably made a typo.
Janus
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Jan31-05, 07:49 PM
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Quote Quote by tony873004
You're right. I didn't realize little m would represent Earth on the Earth velocity formula.

I swear my 2 formulas worked last night. But they don't work today! I entered them into Visual Basic but I probably made a typo.
That's okay, I just realised that I misplaced a decimal point while entering the Earth moon distance into the formulas

This makes the first equation give an answer of .0125 km/sec and the
second .133 km per sec. in the first case it means the moon would have a period of 2220 days and the Earth a period of 72 hrs.

If you reverse it and make the first equation equal the Earth's oribital velocity then you get a period of 27.16 days, close to the proper value.

the second equation, now for the moon gives a period of 244 days. still way off.

However, if you change the second equation to

[tex]V_{o}(moon) = M\sqrt{\frac{G}{dM}}[/tex]


you get an orbital velocity for the moon of 1.0185 km/s and a period of 27.1 days. considering rounding, this comes close again.

But note that the difference between these new corrected equations and mine is that I have (M+m) in the demoninator and the new ones have just M.

Since M in the corrected equations is the Mass of the Earth, 81 times that o fthe moon, it dominates and is why the corrected equationss give close to the correct answer. However, as the two masses become more even in magnitude, the equation I gave will still give the correct answer and the ones with just M in the denominator will drift off in accuracy.


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