Orbital Velocities

[bknight]

OK, be gentle on me as I haven't done this completed algebra solutions for about 20 years.
Looking up the velocity at apoapsis (apogee) and periapsis (perigee) from
https://en.wikipedia.org/wiki/Vis-viva_equation

Solving for V_a and V_p
I get
V_a² = 2*GM(R_a)/(R_p(R_a+R_p))
V_p² = 2*GM(R_p)/(R_a(R_p+R_a))

Feel free to correct any errors, see above.

What my question is if velocity is changed at either apogee or perigee, then the velocity/R of the perigee or apogee velocity will change by the above formulas, so how does one determine how the V/R is changed?
In other words which is changed preferentially?

[gwiz]

If, for example, you change the velocity at apogee, that changes the perigee altitude per your first equation, but leaves the apogee the same, so you then use the new perigee altitude to get the new perigee velocity per your second equation, and the the other way round for a velocity change at perigee.

[bknight]

Sounds good and I want to verify so to calculate for Ra given a Vp, mass, Rp I ran onto a quadratic that I don't remember how to solve
R_a² + R_a * R_p is equal to?
Any help from anyone?

EDIT:
remove R_p  so that R_a is alone on one side of the equation.

[ka9q]

The key thing to remember about the vis-viva equation is that it's just energy conservation in action. Without drag or thrust, the kinetic and potential energy in a 2-body orbit must sum up to a constant. When potential energy increases at altitude, kinetic energy must decrease by the same amount.

Specific (per kilogram) kinetic energy is 1/2 v², while specific potential energy is -GM/r. (Potential energy is never positive). The value of GM for the earth is 3.986004418(9)×10¹⁴ m³/s²

[gwiz]

Sounds good and I want to verify so to calculate for Ra given a Vp, mass, Rp I ran onto a quadratic that I don't remember how to solve

Google is your friend.
http://www.math.com/students/calculators/source/quadratic.htm

[bknight]

Sounds good and I want to verify so to calculate for Ra given a Vp, mass, Rp I ran onto a quadratic that I don't remember how to solve

Google is your friend.
http://www.math.com/students/calculators/source/quadratic.htm

Been there and it doesn't fit my case, but thanks for the effort

[bknight]

The key thing to remember about the vis-viva equation is that it's just energy conservation in action. Without drag or thrust, the kinetic and potential energy in a 2-body orbit must sum up to a constant. When potential energy increases at altitude, kinetic energy must decrease by the same amount.

Specific (per kilogram) kinetic energy is 1/2 v², while specific potential energy is -GM/r. (Potential energy is never positive). The value of GM for the earth is 3.986004418(9)×10¹⁴ m³/s²

Yes, but how would I reorganize the formula such that R_p is a function of R_aV_aV_p-GM

[ka9q]

Well, rockets only change velocity, which means they can only change kinetic energy (and momentum), not potential energy (at least not right away).

So if your pre-burn velocity is V1 and your post-burn velocity is V2, then your change in specific kinetic energy is (V2² - V1²)/2, and that is also your change in total (kinetic + potential) specific energy.

Note that while velocity is a vector, velocity squared -- energy -- is a scalar, so the magnitude of your change in velocity is not necessarily equal to the magnitude of your delta V. But we'll assume that your burns are all along your velocity vector, which is the case for an impulsive Hohmann transfer.

For a given planet, the orbital period of a satellite in a closed orbit and the straight-line distance between its apogee and perigee points are both completely determined by the specific energy. (Half the distance between the apogee and perigee points is known as the semi-major axis). So if you change your specific energy with a motor burn, you can calculate your new semi-major axis. Given that the previous apogee (if you burned there) or previous perigee (if you burned there) did not change, with your new semi-major axis you can now compute your new perigee or apogee height.

Things get more complex in the general case of a burn performed out-of-plane, somewhere between apogee and perigee, or non-parallel to your velocity vector even at apogee or perigee. Some of the delta-V will go into changing the orbital plane and/or rotating the orbit within its plane, with less effect on the specific energy, orbital period and semi-major axis. Both apogee and perigee might change. In that case you need more general formulas. While the specific energy is enough to compute period and semi-major axis, it is not enough to compute apogee and perigee unless you also know the new orbital eccentricity. Perigee radius (note: not altitude) is a(1-e), where a is the semi-major axis and e is eccentricity, and apogee radius (again, not altitude) is a(1+e). The eccentricity must range between 0 and 1, not including exactly 1, for the orbit to be closed and the apogee to be defined.

[bknight]

What I was attempting to accomplish is reverse engineer the orbital data for Gemini VIII
https://www.hq.nasa.gov/alsj/43455667-Gemini-Program-Mission-Report-Gemini-Viii.pdf
and calculate the velocity at apogee
The table on pages 4-22,25 give the following information
Weight 8351.31 Pounds(3788.09 Kg) 3-38
At separation 25738 fps at 86.3 N Miles 4-23,29 Perigee 146.7 N Miles Apogee
So what was the Velocity at Apogee?
Thus the vis-viva equation, seemed simple until I got the quadratic that I have not been able to find a solution.
Then there were two height adjustments, a plane adjust and a plane change, coelliptical bring the Apogee/Perigee 146.7/143.9 N Miles  I wasn't actually looking to try any calculations, except the one that increased the Perigee from 86.3-143.9.  I found that phase changes increased the height slightly, half way to maximum, and the coelliptical changed the altitude by the other half.  So I'm not sure this was the best example of what change in velocity changed the Perigee(Apogee).
Confused?

[ka9q]

Assuming perigee was at separation (which is not necessarily the case), then proceed as follows.

Convert everything to SI. You'll thank me later. (You're welcome.)

Convert the apogee and perigee altitudes to radius distances by adding the radius of the earth to each one. The earth is not a perfect sphere, but good enough.

Compute the (specific) orbital energy by adding the (specific) kinetic and (specific) potential energies at perigee. Note: the potential energy and the orbital energy will both be negative. Potential energy is never positive, and orbital energy is positive only for an escape trajectory.

Compute the potential energy at apogee and subtract from the orbital energy you just determined to find the kinetic energy at apogee. (Be careful with multiple negative signs).

Convert this kinetic energy back to velocity.

Make sense?

[bknight]

Perigee   Apogee              Vel      Mass                          g m/s^2   R      
86.3           145.7   N Miles   25738   f/s   8351.31   Pounds   9.8   6371000      
159827.6   269836.4   m   7844.9424   m/s   3788.087406   Kg            
6530827.6   6640836.4   R+r   11850.44381   m/s                  
                              
                              

                                SPE (m*g*h)   SKE (.5mv^2)           Pe+Ke                     
Perigee                       -5.34501E+11   2.56983E+11   -2.77519E+11                     
Apogee                       -5.43505E+11   2.65986E+11                        

Ok see the above table and the Vel. at Apogee is approx. 11850 m/s given all the parameters are correct.  Now the next step would be the original post question to find a given change with a delta v.  Conceptually there should be a change in both v and r on the other side of the orbit (for this exercise assume at Apogee) increasing the Perigee  v and r.
EDIT:
Thanks ka9q

[ka9q]

Right. If you can make certain assumptions, the formulas for orbits are very simple. That includes:

Pure 2-body motion. No perturbations from the sun, moon or other planets. This is a reasonable assumption for low earth orbit.

No drag. This is not a good approximation for many manned LEO missions because they're so low in altitude, but it'll work for an orbit or two.

The earth is round and has spherically symmetric gravity field. This is also not a reasonable assumption for the earth except for quick calculations. The earth is actually an oblate spheroid, and its equatorial bulge causes orbits to precess.

The only way to precisely predict the trajectory of a satellite is with full-blown numerical integration that takes all the forces into account. You start with a "state vector", the 3-D vectors that specify the position and velocity of the satellite, and update them from there.

[gwiz]

Sounds good and I want to verify so to calculate for Ra given a Vp, mass, Rp I ran onto a quadratic that I don't remember how to solve

Google is your friend.
http://www.math.com/students/calculators/source/quadratic.htm

Been there and it doesn't fit my case, but thanks for the effort

Just needs a little light algebra to get the equation into the right order and away you go.

[bknight]

Just needs a little light algebra to get the equation into the right order and away you go.

Yes, you are correct, but I don't remember how to compute
R_a² + R_a * R_p in terms of R_a

[gwiz]

Just needs a little light algebra to get the equation into the right order and away you go.

Yes, you are correct, but I don't remember how to compute
R_a² + R_a * R_p in terms of R_a

In this case, R_a is the unknown x, so a=1, b=R_p and c=-2*GM(R_p)/V_p²