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View Full Version : What do orbits depend on?

Plat
2005-Mar-09, 04:08 AM

Kristophe
2005-Mar-09, 04:25 AM

Of...? The gravitational force is the gradiant of the gravitational potential, both of which are functions of mass and radius. So, orbits depend on mass and radius.

lti
2005-Mar-09, 09:25 AM
u mean the orbital period depends on mass and radius?

Kepler's laws of planetary motion are excellent at describing orbits.

the velocity and mass of an object will define the radius of the resulting orbit.

Nergal
2005-Mar-09, 02:17 PM
What do orbits depend on?

ToSeek
2005-Mar-09, 03:09 PM

If you want to call it that. If the relevant attracting force (gravity) observes an inverse square law, then the orbits form the familiar conic sections. For an inverse cube law, orbits spiral inwards. I don't recall what happens with a straight inverse law, but I don't think it's stable.

Glom
2005-Mar-09, 03:24 PM
Lot of people who eat sugary foods but want to keep their teeth healthy so opt for a sugar free gum but don't like the other brands.

2005-Mar-09, 04:49 PM
Orbit also depends on if there are any other objects that would cause a gravitational pull in the vicinity

Grey
2005-Mar-09, 07:31 PM
If you want to call it that. If the relevant attracting force (gravity) observes an inverse square law, then the orbits form the familiar conic sections. For an inverse cube law, orbits spiral inwards. I don't recall what happens with a straight inverse law, but I don't think it's stable.
The only stable ones are a force that goes like the inverse square or one that increases directly with distance (F = -kx, like a spring).

Evan
2005-Mar-09, 08:39 PM
Here's a fun little toy:

http://www.ionaphysics.org/lab/Explore/dswmedia/orbit.htm

jfribrg
2005-Mar-10, 03:03 AM
What do orbits depend on?

Since orbits are elliptical, I don't think of the radius as determining the orbit. If you measure the mass, velocity, and location of every object in the system (no Heisenberg nitpicks please), then you can determine the orbits (or parabolic or hyperbolic trajectories as well).

Evan
2005-Mar-10, 07:09 AM
If you measure the mass, velocity, and location of every object in the system (no Heisenberg nitpicks please), then you can determine the orbits (or parabolic or hyperbolic trajectories as well).

Uhh, no. You need to know the velocity vector of every object as well.

A Thousand Pardons
2005-Mar-10, 09:14 AM
If you measure the mass, velocity, and location of every object in the system (no Heisenberg nitpicks please), then you can determine the orbits (or parabolic or hyperbolic trajectories as well).

Uhh, no. You need to know the velocity vector of every object as well.
Velocity is a vector. Just ask Sam5 :)

Evan
2005-Mar-10, 03:47 PM
Yes it is. But the vector must be specified in 3-space plus time. How often have you heard that such and such is moving at x kps, only?

A Thousand Pardons
2005-Mar-10, 04:20 PM
Just once (http://www.squidge.org/~peja/startrekenterprise/DoctorsOrders.htm)

Saluki
2005-Mar-10, 04:20 PM
Yes it is. But the vector must be specified in 3-space plus time. How often have you heard that such and such is moving at x kps, only?

Just because lay people sometimes misuse the term, it does not change the fact that "velocity" is a vector quantity by definition. "Speed" is a scalar term that is equal to the magnitude of velocity. If it is reported in km/s with no direction, it is speed, not velocity.

BTW, "kbps" is a computer term and has nothing to do with speed or velocity. :wink:

jfribrg
2005-Mar-10, 05:10 PM
If you measure the mass, velocity, and location of every object in the system (no Heisenberg nitpicks please), then you can determine the orbits (or parabolic or hyperbolic trajectories as well).

Uhh, no. You need to know the velocity vector of every object as well.

&lt;soapbox>
I thought I had all the bases covered with my post. I get beat up enough on this site for letting a little BA slip into my posts from time to time. This time however, I make a correct statement, using the correct term (in this case velocity, which we all have known since college is always a vector), and I still get beat up because other people supposedly use the term incorrectly. Nowhere in my post did I imply that velocity was anything other than the vector quantity that it is. I think I will become a GLP'er instead of a BAbbler. That way I can vent my frustration without fear of getting banned.
&lt;/soapbox>

Evan
2005-Mar-10, 06:10 PM
BTW, "kbps" is a computer term and has nothing to do with speed or velocity

So? I wrote "kps" as in kilometers per second.

pghnative
2005-Mar-10, 06:27 PM
If you measure the mass, velocity, and location of every object in the system (no Heisenberg nitpicks please), then you can determine the orbits (or parabolic or hyperbolic trajectories as well).

Uhh, no. You need to know the velocity vector of every object as well.

&lt;soapbox>
I thought I had all the bases covered with my post. I get beat up enough on this site for letting a little BA slip into my posts from time to time. This time however, I make a correct statement, using the correct term (in this case velocity, which we all have known since college is always a vector), and I still get beat up because other people supposedly use the term incorrectly. Nowhere in my post did I imply that velocity was anything other than the vector quantity that it is. I think I will become a GLP'er instead of a BAbbler. That way I can vent my frustration without fear of getting banned.
&lt;/soapbox>
Well, you did have ATP and Saluki stand up for you. :D

(of course, I still don't get the kbps comment...)

edited to add: I also think ATP has lost his mind with that link

Saluki
2005-Mar-10, 06:32 PM
The kbps comment was a brain freeze on my part. My mind filled the "b" in. You would not see an engineer (not sure about scientists) use "kps", but I suppose it is technically correct, if counter-productive to unit analysis.

The more correct lable would be "km/s".

A Thousand Pardons
2005-Mar-10, 07:00 PM
edited to add: I also think ATP has lost his mind with that link
Long, long, before that. :)

Anyway, you google on "moving at x kps" and tell me what pops up for you.

Saluki
2005-Mar-10, 08:16 PM
edited to add: I also think ATP has lost his mind with that link
Long, long, before that. :)

Anyway, you google on "moving at x kps" and tell me what pops up for you.

A hodgepodge of things, just like I expected. Note that the first return is a reference (incorrect) to data transmission. Perhaps that author had the same brain freeze I did.

As a contrast, look at the returns for "moving at x km/s". It is much more coherent, and on-topic.

Evan
2005-Mar-10, 09:59 PM
I figure if mph, kph and mps are all correct (try Google using "define:kph") then why isn't kps?

http://www.cogsci.princeton.edu/cgi-bin/webwn?stage=1&amp;word=kph

A Thousand Pardons
2005-Mar-11, 09:31 AM
A hodgepodge of things, just like I expected.
include the quote marks

Gsquare
2005-Mar-12, 05:18 AM
u mean the orbital period depends on mass and radius?

Kepler's laws of planetary motion are excellent at describing orbits.

the velocity and mass of an object will define the radius of the resulting orbit.

That's correct lti; and you are correct in relating 'stability' to Kepler's laws. :D
However, velocity and radius are not usually the parameters used since both are constantly varying in an elliptical orbit; instead, angular momentum L is used as the invariant.

So one definition of a 'stable' orbit is one in which the angular momentum L (and the mass) is constant, which is simply Kepler's 2nd law:

L/2m = constant

And of course, this means the time rate of change of area, A, swept out by the radius vector is also equal to the same constant:

dA/dt = constant = L/2m

It may interest those here also to know that this 'stability' is because gravity is a central force (along the radius vector), irrespective of the inverse square law. In fact, Kepler's 2nd law makes no mention of it.

It is the central force (not its variation with distance) that keeps angular momentum constant, i.e., torque = 0 (assuming other perturbations from other masses are minimal), and believe it or not, Kepler's 2nd law will work equally well even if gravity varies as 1/r or 1/r^3. :wink:

However, having said that, the inverse square law IS evident in Kepler's 1st law, and it can be shown that the elliptical orbits of the planets are a result of the inverse square law which requires the sun to be at one of the foci. :wink:

G^2

JHotz
2005-Mar-12, 12:35 PM
If you measure the mass, velocity, and location of every object in the system (no Heisenberg nitpicks please), then you can determine the orbits (or parabolic or hyperbolic trajectories as well).

Uhh, no. You need to know the velocity vector of every object as well.

&lt;soapbox>
I thought I had all the bases covered with my post. I get beat up enough on this site for letting a little BA slip into my posts from time to time. This time however, I make a correct statement, using the correct term (in this case velocity, which we all have known since college is always a vector), and I still get beat up because other people supposedly use the term incorrectly. Nowhere in my post did I imply that velocity was anything other than the vector quantity that it is. I think I will become a GLP'er instead of a BAbbler. That way I can vent my frustration without fear of getting banned.
&lt;/soapbox>

Just do what I do put maybe, perhaps, I believe or I think before every assertion.