I'm curious in what magnitude of mass an object would need to have its own orbit. Or is it only a question of density? There's a whole bunch of information on how to calculate escape velocities and such, and I've tried using that to derive a way to find what sort of mass you need before you start getting an orbit of other objects, but to no avail.
So basically I want to know what sort of size an object needs before it has a great enough gravitational attraction for an orbit to occur. Google has been very unhelpful too! Though I'm not really sure what to search for.. Thanks in advance, dudes.
Re: A question about orbits... [Solved]
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Re: A question about orbits... [Solved]
Last edited by Bateman on Sun Sep 25, 2011 3:24 pm UTC, edited 1 time in total.
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Re: A question about orbits...
In an otherwiseempty space (only two objects), there is no lower mass limit. Orbit around an object of any given mass and at any given radius is possible with a low enough orbital velocity.
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Last updated 6/29/108
Last updated 6/29/108
Re: A question about orbits...
Yeah, any mass will work, say for example, you have an object that's 1kg, with an object of negligible mass orbiting it at a distance of 1 meter
http://en.wikipedia.org/wiki/Orbital_period
T = 2*pi* sqrt(a^{3}/GM)
a = orbit's semimajor axis = 1m
G = gravitational constant = 6.674 * 10^{11} m^{3} kg^{1} s^{2}.
M = mass of the object = 1kg
T = 2 * pi * sqrt(1m^{3} / ( 6.674 * 10^{11} m^{3} kg^{1} s^{2} * 1kg )
T = 2 * pi * sqrt(1m^{3} / 6.674 * 10^{11} m^{3} s^{2})
T = 2 * pi * sqrt(1.498*10^{10} s^{2})
T = 2 * pi * 122407 s
T = 769107s
T = 8.9 days.
So, if you had a 1kg mass, an another object of negligible mass was orbiting it in a circular orbit of 1m, it would take about 8.9 days for it to make a complete orbit. And since it's going a distance of 1m * 2 * pi = 6.283m. It would have an orbital velocity of 6.283m/769107s, or 8.169*10^{6}m/s. Or 8.169μm/s. So. if you had an object of negligible mass, 1m from a 1kg object, and gave it a push to the side, giving it a velocity of 8.169μm/s, it would orbit the 1kg object, in a circular orbit, taking about 8.9 days to make an orbit.
If you want to take the orbitting mass into account, which you probably should for things this small, you can just add the masses together to get the orbital period. Say for example you have 2 people, each 100kg, and you want to make them orbit, in a circular orbit around their common center of gravity, at a distance of 3m (just out of reach of each other. )
a = orbit's semimajor axis = 3m
G = gravitational constant = 6.674 * 10^{11} m^{3} kg^{1} s^{2}.
M = mass of the object = 200kg
T = 2 * pi * sqrt(3m^{3} / ( 6.674 * 10^{11} m^{3} kg^{1} s^{2} * 200kg )
T = 2 * pi * sqrt(3m^{3} / 1.3348 * 10^{8} m^{3} s^{2})
T = 2 * pi * sqrt(2.247*10^{8} s^{2})
T = 2 * pi * 14990 s
T = 94185s
T = 1.09 days
3m * 2pi = 18.85m
18.85m/94185s = .0002m/s = 0.2mm/s
So, if you had 2 100kg people, 3m apart, give one of them (or both of them?) a slight tap to the side to send them floating at 0.2mm/s, and they'd end up in a circular orbit around their center of mass, returning back to their starting points every 26 hours or so (also assuming that they're spherically symmetrical )
See also http://en.wikipedia.org/wiki/Escape_velocity For the maximum speed an object can have at a certain distance, from an object of a certain mass, and still be in orbit of it, more than that, and it would just keep moving away.
http://en.wikipedia.org/wiki/Orbital_period
T = 2*pi* sqrt(a^{3}/GM)
a = orbit's semimajor axis = 1m
G = gravitational constant = 6.674 * 10^{11} m^{3} kg^{1} s^{2}.
M = mass of the object = 1kg
T = 2 * pi * sqrt(1m^{3} / ( 6.674 * 10^{11} m^{3} kg^{1} s^{2} * 1kg )
T = 2 * pi * sqrt(1m^{3} / 6.674 * 10^{11} m^{3} s^{2})
T = 2 * pi * sqrt(1.498*10^{10} s^{2})
T = 2 * pi * 122407 s
T = 769107s
T = 8.9 days.
So, if you had a 1kg mass, an another object of negligible mass was orbiting it in a circular orbit of 1m, it would take about 8.9 days for it to make a complete orbit. And since it's going a distance of 1m * 2 * pi = 6.283m. It would have an orbital velocity of 6.283m/769107s, or 8.169*10^{6}m/s. Or 8.169μm/s. So. if you had an object of negligible mass, 1m from a 1kg object, and gave it a push to the side, giving it a velocity of 8.169μm/s, it would orbit the 1kg object, in a circular orbit, taking about 8.9 days to make an orbit.
If you want to take the orbitting mass into account, which you probably should for things this small, you can just add the masses together to get the orbital period. Say for example you have 2 people, each 100kg, and you want to make them orbit, in a circular orbit around their common center of gravity, at a distance of 3m (just out of reach of each other. )
a = orbit's semimajor axis = 3m
G = gravitational constant = 6.674 * 10^{11} m^{3} kg^{1} s^{2}.
M = mass of the object = 200kg
T = 2 * pi * sqrt(3m^{3} / ( 6.674 * 10^{11} m^{3} kg^{1} s^{2} * 200kg )
T = 2 * pi * sqrt(3m^{3} / 1.3348 * 10^{8} m^{3} s^{2})
T = 2 * pi * sqrt(2.247*10^{8} s^{2})
T = 2 * pi * 14990 s
T = 94185s
T = 1.09 days
3m * 2pi = 18.85m
18.85m/94185s = .0002m/s = 0.2mm/s
So, if you had 2 100kg people, 3m apart, give one of them (or both of them?) a slight tap to the side to send them floating at 0.2mm/s, and they'd end up in a circular orbit around their center of mass, returning back to their starting points every 26 hours or so (also assuming that they're spherically symmetrical )
See also http://en.wikipedia.org/wiki/Escape_velocity For the maximum speed an object can have at a certain distance, from an object of a certain mass, and still be in orbit of it, more than that, and it would just keep moving away.

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Re: A question about orbits...
And for multiple bodies, the sphere of influence is a pretty decent approximation: http://en.wikipedia.org/wiki/Sphere_of_ ... ynamics%29
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Re: A question about orbits...
Orbiting objects are simply falling bodies with horizontal motion such that they get out of the central object's way as they fall. Bodies in freefall are under the influence of no forces other than gravity. Such objects exhibit behavior that is independent of mass. In the atmosphere, air resistance is important at high velocities, but is often negligible when the falling time is short. In space, there are usually no other forces involved.
The usual illustration is a canon firing a projectile on the surface of a very small "planet" with increasing muzzle velocities. As the speed of the projectile increases, it follows the curvature of the planet around, taking longer and longer to land. At some muzzle velocity, the projectile travels horizontally far enough in a given time such that the surface is always a fixed distance away; this is a circular orbit.
You can find this velocity (for circular orbits) by setting the centripetal acceleration a=v^{2}/r equal to the acceleration due to gravity g=GM/r^{2} and doing a bit of algebra.
If there was an atmosphere, mass would be important. A heavy object would experience less deceleration from air resistance and follow the ideal orbit more closely. However, the orbit would not be stable and air resistance would gradually slow it down until the ground rushed up to it faster than it could get out of the way!
The usual illustration is a canon firing a projectile on the surface of a very small "planet" with increasing muzzle velocities. As the speed of the projectile increases, it follows the curvature of the planet around, taking longer and longer to land. At some muzzle velocity, the projectile travels horizontally far enough in a given time such that the surface is always a fixed distance away; this is a circular orbit.
You can find this velocity (for circular orbits) by setting the centripetal acceleration a=v^{2}/r equal to the acceleration due to gravity g=GM/r^{2} and doing a bit of algebra.
If there was an atmosphere, mass would be important. A heavy object would experience less deceleration from air resistance and follow the ideal orbit more closely. However, the orbit would not be stable and air resistance would gradually slow it down until the ground rushed up to it faster than it could get out of the way!
Re: A question about orbits... [Solved]
Thanks guys, you were all fantastic help!
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 agelessdrifter
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Re: A question about orbits... [Solved]
I have a question about orbits as well. It's a homework question from classical mechanics I've come close to the solution a few times, but something is off with my calculations.
Two planets of mass m_{1} and m_{2} are orbiting each other in circular orbits with a period T. Show that if the planets are suddenly stopped in their orbits, the time it would take them to collide would be T/2^(3/2)
So I'm using the reduced mass approach and calculating the time it would take a body of mass M=(m_{1}m_{2}/(m_{1}+m_{2}) to collide with a body of mass (m_{1}+m_{2}). Since the orbits are circular, I have from the text that the period is going to be 4pi^{2}r^{3}/G(m_{1}+m_{2}). The energy of the system is given by
E= (1/2)Mv^{2}=(1/2)Mv^{2}GM(m1+m2)/r
and I've tried every substitution I can think of to get T into the expression, then integrate v to get r as a function of t, then t as a function of r. But I keep winding up with answers that are slightly off or completely nonsensical. The last attempt I made ended up with t(r) being an expression involving arctanh, and the time for collision being 0infinity.
So I dunno if I've got the wrong expression for T, the wrong equation for E, if I'm doing poorly chosen substitutions or just flat out integrating incorrectly (actually I've been checking that with mathematica, so I know that's not it). Can anyone offer a suggestion?
Two planets of mass m_{1} and m_{2} are orbiting each other in circular orbits with a period T. Show that if the planets are suddenly stopped in their orbits, the time it would take them to collide would be T/2^(3/2)
So I'm using the reduced mass approach and calculating the time it would take a body of mass M=(m_{1}m_{2}/(m_{1}+m_{2}) to collide with a body of mass (m_{1}+m_{2}). Since the orbits are circular, I have from the text that the period is going to be 4pi^{2}r^{3}/G(m_{1}+m_{2}). The energy of the system is given by
E= (1/2)Mv^{2}=(1/2)Mv^{2}GM(m1+m2)/r
and I've tried every substitution I can think of to get T into the expression, then integrate v to get r as a function of t, then t as a function of r. But I keep winding up with answers that are slightly off or completely nonsensical. The last attempt I made ended up with t(r) being an expression involving arctanh, and the time for collision being 0infinity.
So I dunno if I've got the wrong expression for T, the wrong equation for E, if I'm doing poorly chosen substitutions or just flat out integrating incorrectly (actually I've been checking that with mathematica, so I know that's not it). Can anyone offer a suggestion?

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Re: A question about orbits... [Solved]
What relationship is there between v, r and T?

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Re: A question about orbits... [Solved]
agelessdrifter wrote:I have a question about orbits as well. It's a homework question from classical mechanics I've come close to the solution a few times, but something is off with my calculations.
Two planets of mass m_{1} and m_{2} are orbiting each other in circular orbits with a period T. Show that if the planets are suddenly stopped in their orbits, the time it would take them to collide would be T/2^(3/2)
So I'm using the reduced mass approach and calculating the time it would take a body of mass M=(m_{1}m_{2}/(m_{1}+m_{2}) to collide with a body of mass (m_{1}+m_{2}). Since the orbits are circular, I have from the text that the period is going to be 4pi^{2}r^{3}/G(m_{1}+m_{2}). The energy of the system is given by
E= (1/2)Mv^{2}=(1/2)Mv^{2}GM(m1+m2)/r
and I've tried every substitution I can think of to get T into the expression, then integrate v to get r as a function of t, then t as a function of r. But I keep winding up with answers that are slightly off or completely nonsensical. The last attempt I made ended up with t(r) being an expression involving arctanh, and the time for collision being 0infinity.
So I dunno if I've got the wrong expression for T, the wrong equation for E, if I'm doing poorly chosen substitutions or just flat out integrating incorrectly (actually I've been checking that with mathematica, so I know that's not it). Can anyone offer a suggestion?
I think the approach you want to take here is to 1) figure out how far apart the two objects are when they're orbiting, then 2) figure out how long it takes two objects of mass m1 and m2, at a distance r, to collide if they start at rest. 1) should be fairly straightforward, and 2) sounds like something that I did when I was a budding young physics major, and shouldn't be too terribly difficult.
Ten is approximately infinity (It's very large)
Ten is approximately zero (It's very small)
Ten is approximately zero (It's very small)
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