|
|
| Author |
Message |
pouya1
science forum beginner
Joined: 30 Apr 2006
Posts: 3
|
 Posted: Tue May 16, 2006 6:41 pm Post subject:
Asteroid orbitting the sun
|
|
|
hi, please help with the following question:
An asteroid is in circular orbit around the sun with period T. If the
asteroid is instantaneously stopped in its orbit, it will fall towards the
sun. How long will it take to get there? (the heing says regard the asteroid
as ALMOST stopped, so that it goes to a highly eccentric elliptical orbit,
whose major axis is a bit greater than the radius of the original circular
orbit) |
|
| Back to top |
|
 |
Mike Yarwood
science forum Guru Wannabe
Joined: 02 May 2005
Posts: 117
|
| Posted: Tue May 16, 2006 8:11 pm Post subject:
Re: Asteroid orbitting the sun
|
|
|
"pouya" <sabouri@shaw.ca> wrote in message
news:O1pag.167598$7a.89266@pd7tw1no...
| Quote: | hi, please help with the following question:
An asteroid is in circular orbit around the sun with period T. If the
asteroid is instantaneously stopped in its orbit, it will fall towards
the
sun. How long will it take to get there? (the heing says regard the
asteroid
as ALMOST stopped, so that it goes to a highly eccentric elliptical orbit,
whose major axis is a bit greater than the radius of the original circular
orbit)
So you find the radius of your original circular orbit r= C*T^k where C and |
k are some constants.
then you plug it into your formula for the period of an elliptical orbit and
take a bit less than 1/2 that period.
Best of Luck - Mike |
|
| Back to top |
|
|
tadchem
science forum Guru
Joined: 03 May 2005
Posts: 1348
|
| Posted: Wed May 17, 2006 12:04 am Post subject:
Re: Asteroid orbitting the sun
|
|
|
pouya wrote:
| Quote: | hi, please help with the following question:
An asteroid is in circular orbit around the sun with period T. If the
asteroid is instantaneously stopped in its orbit, it will fall towards the
sun. How long will it take to get there? (the heing says regard the asteroid
as ALMOST stopped, so that it goes to a highly eccentric elliptical orbit,
whose major axis is a bit greater than the radius of the original circular
orbit)
|
The kinetic energy maintains the circular (eccentricity = 0) orbit.
Take away the kinetic energy and the asteroid stops completely.
It free-falls toward the sum.
Potential energy is converted to kinetic energy.
Can you calculate the time for free fall from its original orbit's
radius straight into the sun?
It is 1/2 the orbital period of an orbit with semi-major axis = 1/2 the
semi-major axis of the circular orbit.
Tom Davidson
Richmond, VA |
|
| Back to top |
|
|
tadchem
science forum Guru
Joined: 03 May 2005
Posts: 1348
|
| Posted: Wed May 17, 2006 8:22 pm Post subject:
Re: Asteroid orbitting the sun
|
|
|
PD wrote:
| Quote: | pouya wrote:
hi, please help with the following question:
An asteroid is in circular orbit around the sun with period T. If the
asteroid is instantaneously stopped in its orbit, it will fall towards the
sun. How long will it take to get there? (the heing says regard the asteroid
as ALMOST stopped, so that it goes to a highly eccentric elliptical orbit,
whose major axis is a bit greater than the radius of the original circular
orbit)
The semi-major axis is essentially unchanged. This should tell you
something about the period of the new orbit. The fact that falling into
the sun now happens at some time in the orbit doesn't change that, but
it does tell you when it happens.
|
Hi, PD -
The major axis of a circular orbit is the diameter; the semi-major axis
is the radius.
Given that, the 'new' orbit starts at aphelion, with perihelion
arbitrarily close to the sun on the opposite side from the aphelion.
The major axis of the 'elliptical' orbit is the line segment connecting
the aphelion to the perihelion.
That makes the major axis ever so slightly longer than the circular
orbit's radius, and the new semi-major axis half of that.
Please don't ask me to draw a picture - ASCII stinks at drawing
ellipses.
Tom Davidson
Richmond, VA |
|
| Back to top |
|
|
PD
science forum Guru
Joined: 03 May 2005
Posts: 4363
|
| Posted: Wed May 17, 2006 10:12 pm Post subject:
Re: Asteroid orbitting the sun
|
|
|
tadchem wrote:
| Quote: | PD wrote:
pouya wrote:
hi, please help with the following question:
An asteroid is in circular orbit around the sun with period T. If the
asteroid is instantaneously stopped in its orbit, it will fall towards the
sun. How long will it take to get there? (the heing says regard the asteroid
as ALMOST stopped, so that it goes to a highly eccentric elliptical orbit,
whose major axis is a bit greater than the radius of the original circular
orbit)
The semi-major axis is essentially unchanged. This should tell you
something about the period of the new orbit. The fact that falling into
the sun now happens at some time in the orbit doesn't change that, but
it does tell you when it happens.
Hi, PD -
The major axis of a circular orbit is the diameter; the semi-major axis
is the radius.
Given that, the 'new' orbit starts at aphelion, with perihelion
arbitrarily close to the sun on the opposite side from the aphelion.
The major axis of the 'elliptical' orbit is the line segment connecting
the aphelion to the perihelion.
That makes the major axis ever so slightly longer than the circular
orbit's radius, and the new semi-major axis half of that.
|
Yes, indeed. I misspoke.
| Quote: |
Please don't ask me to draw a picture - ASCII stinks at drawing
ellipses.
Tom Davidson
Richmond, VA |
|
|
| Back to top |
|
|
Gordon
science forum Guru Wannabe
Joined: 12 May 2005
Posts: 115
|
| Posted: Tue Jul 11, 2006 4:27 pm Post subject:
Re: Asteroid orbitting the sun
|
|
|
On Tue, 16 May 2006 18:41:50 GMT, "pouya" <sabouri@shaw.ca>
wrote:
| Quote: | hi, please help with the following question:
An asteroid is in circular orbit around the sun with period T. If the
asteroid is instantaneously stopped in its orbit, it will fall towards the
sun. How long will it take to get there? (the heing says regard the asteroid
as ALMOST stopped, so that it goes to a highly eccentric elliptical orbit,
whose major axis is a bit greater than the radius of the original circular
orbit)
Another question along this same line...if an asteroid is in a |
circular, or elliptical orbit around the sun, or even around one
of the planets, would it be possible for another asteroid to
pass near the first one, engaging in a Hohmann Transfer of energy
by mutual gravitational interaction, from the first to the second
asteroid, leaving the first one with little or no remaining
kinetic energy?
Gordon |
|
| Back to top |
|
|
Greg Neill
science forum Guru Wannabe
Joined: 31 May 2005
Posts: 180
|
| Posted: Wed Jul 12, 2006 1:22 pm Post subject:
Re: Asteroid orbitting the sun
|
|
|
"Gordon" <gordonlr@DELETEswbell.net> wrote in message
news:gfj7b2dve94p3o9oens55dbji5otbiurfu@4ax.com...
| Quote: | On Tue, 16 May 2006 18:41:50 GMT, "pouya" <sabouri@shaw.ca
wrote:
hi, please help with the following question:
An asteroid is in circular orbit around the sun with period T. If the
asteroid is instantaneously stopped in its orbit, it will fall towards the
sun. How long will it take to get there? (the heing says regard the asteroid
as ALMOST stopped, so that it goes to a highly eccentric elliptical orbit,
whose major axis is a bit greater than the radius of the original circular
orbit)
Another question along this same line...if an asteroid is in a
circular, or elliptical orbit around the sun, or even around one
of the planets, would it be possible for another asteroid to
pass near the first one, engaging in a Hohmann Transfer of energy
by mutual gravitational interaction, from the first to the second
asteroid, leaving the first one with little or no remaining
kinetic energy?
|
Terminology point: A Hohmann Transfer is the free-falling
(unpowered) trajectory of least energy between two closed
orbits (circular, elliptical). It doesn't describe a type
of energy transfer, but the orbit trajectory itself.
If two equally massive bodies were orbiting in opposite
directions, then their collision could result in
debris that's essentially stationary. Otherwise, you
might want to take a look at the formulae that
describe gravity assist manoeuvres to see what the mass
ratio would have to be to leave one of the bodies
without forward momentum in the Sun-centered inertial
frame of reference.
For Pouya'a question, if the orbit is narrow enough and
the perihelion close enough to the center of the Sun,
given that Kepler's Laws still apply to the orbit regardless
of the eccentricity, you should be able to determine the
period of such an orbit. What portion of this orbit would
correspond to a straight in fall? |
|
| Back to top |
|
|
Gordon
science forum Guru Wannabe
Joined: 12 May 2005
Posts: 115
|
| Posted: Wed Jul 12, 2006 2:52 pm Post subject:
Re: Asteroid orbitting the sun
|
|
|
On Wed, 12 Jul 2006 09:22:58 -0400, "Greg Neill"
<gneillREM@OVE.THIS.netcom.ca> wrote:
| Quote: | "Gordon" <gordonlr@DELETEswbell.net> wrote in message
news:gfj7b2dve94p3o9oens55dbji5otbiurfu@4ax.com...
On Tue, 16 May 2006 18:41:50 GMT, "pouya" <sabouri@shaw.ca
wrote:
hi, please help with the following question:
An asteroid is in circular orbit around the sun with period T. If the
asteroid is instantaneously stopped in its orbit, it will fall towards the
sun. How long will it take to get there? (the heing says regard the asteroid
as ALMOST stopped, so that it goes to a highly eccentric elliptical orbit,
whose major axis is a bit greater than the radius of the original circular
orbit)
Another question along this same line...if an asteroid is in a
circular, or elliptical orbit around the sun, or even around one
of the planets, would it be possible for another asteroid to
pass near the first one, engaging in a Hohmann Transfer of energy
by mutual gravitational interaction, from the first to the second
asteroid, leaving the first one with little or no remaining
kinetic energy?
Terminology point: A Hohmann Transfer is the free-falling
(unpowered) trajectory of least energy between two closed
orbits (circular, elliptical). It doesn't describe a type
of energy transfer, but the orbit trajectory itself.
If two equally massive bodies were orbiting in opposite
directions, then their collision could result in
debris that's essentially stationary. Otherwise, you
might want to take a look at the formulae that
describe gravity assist manoeuvres to see what the mass
ratio would have to be to leave one of the bodies
without forward momentum in the Sun-centered inertial
frame of reference.
For Pouya'a question, if the orbit is narrow enough and
the perihelion close enough to the center of the Sun,
given that Kepler's Laws still apply to the orbit regardless
of the eccentricity, you should be able to determine the
period of such an orbit. What portion of this orbit would
correspond to a straight in fall?
Thanks, Greg. This clears up some issues. I had gotten some |
information from Wikipedia
http://en.wikipedia.org/wiki/Hohmann_transfer
http://en.wikipedia.org/wiki/Bi-elliptic_transfer
In any case, this talks about orbital transfers that are the
result of kinetic energy being added to or removed from the
object.(thrusters, or reverse thrusters turned on briefly)
resulting in the object's moving into an elliptical orbit. Then,
at the other end of the ellipse the thrusters are turned on again
adding or removing enough kinetic energy to allow the object to
remain in the new orbit.
This process can be applied either way. That is it can be used to
move the object to a larger radius or to a smaller radius orbit.
It is just a matter of applying two brief bursts of thrust
(kinetic energy) to the object at the opposite ends of the
elliptical transfer path.
My thinking is that a near miss between a comparatively large
asteroid and a much smaller one, moving in the opposite
direction, could result in most of the momentum (velocity) of the
smaller asteroid being deleted. That is, nearly equal but
opposite momentum would be transferred to the smaller object,
leaving it with nearly zero velocity relative to the system.
A deep space probe can be directed to pass near a planet, and in
the process, pick up additional momentum which will reduce the
fuel requirements, and provide the necessary energy to put the
probe deeper into space. This momentum picked up by the probe is
"robbed" from the planet as the probe passes near the planet.
Isn't this a form of Hohmann transfer?
Gordon |
|
| Back to top |
|
|
Greg Neill
science forum Guru Wannabe
Joined: 31 May 2005
Posts: 180
|
| Posted: Wed Jul 12, 2006 3:27 pm Post subject:
Re: Asteroid orbitting the sun
|
|
|
"Gordon" <gordonlr@DELETEswbell.net> wrote in message
news:mb1ab25jvb92kd8evc68ee38ka952dqdes@4ax.com...
| Quote: | A deep space probe can be directed to pass near a planet, and in
the process, pick up additional momentum which will reduce the
fuel requirements, and provide the necessary energy to put the
probe deeper into space. This momentum picked up by the probe is
"robbed" from the planet as the probe passes near the planet.
Isn't this a form of Hohmann transfer?
|
It's generally referred to as a gravity assist manoeuvre.
A Hohmann Transfer includes the manoeuvres at both ends
of the trajectory that take the body from a condition of
being in one elliptical orbit to another, different
elliptical orbit. The essential characteristic of a Hohmann
Transfer is that the trajectory is unpowered except at
the end points. |
|
| Back to top |
|
|
Google
|
|
| Back to top |
|
|
|
|
The time now is Thu Feb 21, 2019 2:31 pm | All times are GMT
|
|
Copyright © 2004-2005 DeniX Solutions SRL
|
|
Other DeniX Solutions sites:
Electronics forum |
Medicine forum |
Unix/Linux blog |
Unix/Linux documentation |
Unix/Linux forums |
send newsletters
|
| |
|
Powered by phpBB © 2001, 2005 phpBB Group
|
|
FAQ
Search
Memberlist
Usergroups
Profile
Log in to check your private messages
Log in
Forum index
»
Science and Technology
»
Physics
