- Questions
ASTRONOMY
WHAT CAUSE AN ELECTRON TO OCCUPY A HIGHER ORBIT??
Do you really mean an electron or are you really referring to a planet or satellite since the title of your post is astronomy?
If you really meant a planet or satellite:
Consider a satellite in a circular orbit around the earth. In order to raise the satellite to a specific higher altitude orbit, a specific propulsive force must ba applied to the satellite to kick it out of orbit into an elliptical orbit with an appogee of the desired new altitude. Once the satellite has reached the new appogee altitude, another specific propulsive force must be applied to increase the velocity of the satellite to the circular velocity for that altitude.
Lets consider the scenario where the space shuttle has to rendevouz with the old MIR space station, each being in a different orbit. The Space Shuttle Orbiter typically launches into a 50 x 100 miles temporary orbit, burning out at 50 miles and ascending to 100 miles. Upon reaching the 100 mile apogee, the Orbiter fires its Orbital Maneuvering System motors to inject the Orbiter into another elliptical transfer orbit with an apogee of the desired final mission altitude. Upon reaching this new apogee, the OMS fires again to increase the velocity to circular velocity for that altitude. Rather than stepping through this somewhat involved sequence of events, I will offer you a simplified scenario that will suffice to get the point across regarding what we are trying to learn about here.
Lets assume that our target vehicle is still the MIR Space Station in a 232 miles high circular orbit. Lets assume, however, that the Space Shuttle Orbiter launched into a ~110 mile high circular orbit to deposit another satellite and, from here, will rendezvous with the MIR. The circular velocity of this orbit is ~25,587 fps (~17,442 MPH) and the period 88 minutes. The Orbiter must now change its orbit and ascend to the MIR orbit in order to execute the rendezvous. How best can this be achieved?
You might have already guessed how we are going to accomplish our rendezvous. You saw above that increasing our velocity places us on a new elliptical orbit with an apogee somewhat higher than our firing point. Isn't this what we are trying to accomplish, i.e., raise our alitude from a 110 mile circular orbit to a 232 mile circular orbit? We could execute several orbital exchanges in the hope of reaching our destination as quickly as possible and with a minimum of propellant expenditures. Might there be an optimum approach to the problem?
The successful execution of this type of orbital exchange maneuver was first recognized and described by the famous German engineer Dr. Walter Hohmann. In 1925, many years before space travel was a reality, he derived what was considered to be the most energy efficient method for transferring a vehicle between two coplanar circular orbits. Not surprisingly, the method involved the utilization of an elliptical orbit between the two circular orbits involved. The velocity of the vehicle in the lower obrit is increased by an amount that will place the vehicle on an elliptical path to a higher altitude. The firing point, the altitude of the lower orbit, becomes the perigee of the transfer orbit, with the apogee the altitude of the higher orbit, 180 degrees away from perigee. The elliptical transfer orbit is, by definition, cotangent to both of the circular orbits. Upon reaching the apogee of the transfer orbit, cotangent to the higher circular orbit, the velocity has dropped below the required circular velocity for the higher orbit. Thus, the velocity of the vehicle is again increased to bring it up to the required circular velocity, and we are safely in the higher orbit. Relatively simple and elegant, yes? This method, historically referred to as the Two Burn Hohmann Transfer Orbit, HTO, is used for just about every transfer of vehicles between differing altitude orbits.
So how do we go about raising the orbit of our Orbiter? All we have to do is fire our OMS engines, increase our velocity to 25,755 fps (17,570 MPH) and ascend to our 232 mile target altitude in the half period of 45 minutes. Boy, that was easy. But hold on a minute. Yes, we have ascended to 232 miles but our velocity has dropped to 25,026 fps (17,059 MPH) and we are about to descend on our way back down to our original 110 mile altitude. What must we do? Of course, as Dr. Hohmann proposed, we must increase our 25,026 fps. velocity to the 25,212 fps. circular velocity of the target orbit and we are home free. You might be surprised by the small velocity increments involved (typically referred to as deltaV's for delta velocities). An increase of only 188 fps to raise our altitude and an additional 186 fps. to circularize our orbit.
Well, we have now discovered how to raise the orbit of our Orbiter, but we have yet to address the primary objective of our mission, rendezvousing with the MIR. Clearly, what we really want to do is raise our orbit, but simultaneously, meet up with the MIR, or at worst, reach a point a few minutes ahead of, or behind, MIR as discussed ealier. So what are we faced with? A very unique game of space tag. MIR is in a 232 mile orbit with a 92 minute period. We are in a 110 mile orbit with an 88 minute period. We have two options depending on our actual angular relationship to one another.
Lets examine the most desireable approach first. Ideally, we want to place our Orbiter on a HTO that will ascend to, and reach, a point in the MIR orbit at exactly the moment that MIR reaches the same point. Knowing that our HTO ascent will take 45 minutes, we therefore, must fire our engines at a time when MIR is 45 minutes away from our designated meeting point in their orbit. Since MIR's orbital half period is 46 minutes, we must fire our engines when MIR is one minute ahead of us in their orbit or ~4 degrees ahead of us. At the point of firing, both MIR and our Orbiter have 45 minutes to travel before meeting. Upon reaching the meeting point, the Orbiter fires its engines again to circularize the orbit and the final rendezvous maneuvers can commense.
The other option available to us is to ascend on a HTO that will place the Orbiter a few minutes ahead of, or behind, the MIR, from which point they can execute the catch up maneuvers described earlier. It is obvious that the key orbital maneuver that must be executed in the first place is the placement of the Orbiter in the proper angular relationship to MIR in the initial lower circular orbit, which, in the final analysis, is what determines launch time. The Shuttle must lift off, reach the 50 x 100 mile parking orbit, and ultimately ascend to the first, low altitude, 110 mile circular orbit for depositing the satellite payload, and be in the proper relationship to MIR for the ultimate HTO and rendezvous.
Lets now consider the scenario where the two vehicles are in totally different orbits. The Space Shuttle Orbiter typically launches into a 50 x 100 miles temporary orbit, burning out at 50 miles and ascending to 100 miles. Upon reaching the 100 mile apogee, the Orbiter fires its Orbital Maneuvering System motors to inject the Orbiter into another elliptical transfer orbit with an apogee of the desired final mission altitude. Upon reaching this new apogee, the OMS fires again to increase the velocity to circular velocity for that altitude. Rather than stepping through this somewhat involved sequence of events, I will offer you a simplified scenario that will suffice to get the point across regarding what we are trying to learn about here.
Lets assume that our target vehicle is still the MIR Space Station in a 232 miles high circular orbit. Lets assume, however, that the Space Shuttle Orbiter launched into a ~110 mile high circular orbit to deposit another satellite and, from here, will rendezvous with the MIR. The circular velocity of this orbit is ~25,587 fps (~17,442 MPH) and the period 88 minutes. The Orbiter must now change its orbit and ascend to the MIR orbit in order to execute the rendezvous. How best can this be achieved?
You might have already guessed how we are going to accomplish our rendezvous. You saw above that increasing our velocity places us on a new elliptical orbit with an apogee somewhat higher than our firing point. Isn't this what we are trying to accomplish, i.e., raise our alitude from a 110 mile circular orbit to a 232 mile circular orbit? We could execute several orbital exchanges in the hope of reaching our destination as quickly as possible and with a minimum of propellant expenditures. Might there be an optimum approach to the problem?
The successful execution of this type of orbital exchange maneuver was first recognized and described by the famous German engineer Dr. Walter Hohmann. In 1925, many years before space travel was a reality, he derived what was considered to be the most energy efficient method for transferring a vehicle between two coplanar circular orbits. Not surprisingly, the method involved the utilization of an elliptical orbit between the two circular orbits involved. The velocity of the vehicle in the lower obrit is increased by an amount that will place the vehicle on an elliptical path to a higher altitude. The firing point, the altitude of the lower orbit, becomes the perigee of the transfer orbit, with the apogee the altitude of the higher orbit, 180 degrees away from perigee. The elliptical transfer orbit is, by definition, cotangent to both of the circular orbits. Upon reaching the apogee of the transfer orbit, cotangent to the higher circular orbit, the velocity has dropped below the required circular velocity for the higher orbit. Thus, the velocity of the vehicle is again increased to bring it up to the required circular velocity, and we are safely in the higher orbit. Relatively simple and elegant, yes? This method, historically referred to as the Two Burn Hohmann Transfer Orbit, HTO, is used for just about every transfer of vehicles between differing altitude orbits.
So how do we go about raising the orbit of our Orbiter? All we have to do is fire our OMS engines, increase our velocity to 25,755 fps (17,570 MPH) and ascend to our 232 mile target altitude in the half period of 45 minutes. Boy, that was easy. But hold on a minute. Yes, we have ascended to 232 miles but our velocity has dropped to 25,026 fps (17,059 MPH) and we are about to descend on our way back down to our original 110 mile altitude. What must we do? Of course, as Dr. Hohmann proposed, we must increase our 25,026 fps. velocity to the 25,212 fps. circular velocity of the target orbit and we are home free. You might be surprised by the small velocity increments involved (typically referred to as deltaV's for delta velocities). An increase of only 188 fps to raise our altitude and an additional 186 fps. to circularize our orbit.
Well, we have now discovered how to raise the orbit of our Orbiter, but we have yet to address the primary objective of our mission, rendezvousing with the MIR. Clearly, what we really want to do is raise our orbit, but simultaneously, meet up with the MIR, or at worst, reach a point a few minutes ahead of, or behind, MIR as discussed ealier. So what are we faced with? A very unique game of space tag. MIR is in a 232 mile orbit with a 92 minute period. We are in a 110 mile orbit with an 88 minute period. We have two options depending on our actual angular relationship to one another.
Lets examine the most desireable approach first. Ideally, we want to place our Orbiter on a HTO that will ascend to, and reach, a point in the MIR orbit at exactly the moment that MIR reaches the same point. Knowing that our HTO ascent will take 45 minutes, we therefore, must fire our engines at a time when MIR is 45 minutes away from our designated meeting point in their orbit. Since MIR's orbital half period is 46 minutes, we must fire our engines when MIR is one minute ahead of us in their orbit or ~4 degrees ahead of us. At the point of firing, both MIR and our Orbiter have 45 minutes to travel before meeting. Upon reaching the meeting point, the Orbiter fires its engines again to circularize the orbit and the final rendezvous maneuvers can commense.
The other option available to us is to ascend on a HTO that will place the Orbiter a few minutes ahead of, or behind, the MIR, from which point they can execute the catch up maneuvers described earlier. It is obvious that the key orbital maneuver that must be executed in the first place is the placement of the Orbiter in the proper angular relationship to MIR in the initial lower circular orbit, which, in the final analysis, is what determines launch time. The Shuttle must lift off, reach the 50 x 100 mile parking orbit, and ultimately ascend to the first, low altitude, 110 mile circular orbit for depositing the satellite payload, and be in the proper relationship to MIR for the ultimate HTO and rendezvous.
http://library.thinkquest.org/15567/lessons/2.html
If you mean an electron in position around a positive nucleus, then the "orbital" it occupies is dependent on the energy state it has. It can change to a higher energy state if energy is absorbed from a photon.
Despite the word orbitals used to describe the position of electrons about the nucleus, electrons do not orbit the nucleus in the manner of small planets about tiny sun. Instead, the quantum model of the atom helps us understand that the electrons are placed in probability fields of designated shapes away from the nucleus.