Is the Circular to Elliptical Orbit Calculator Accurate?

[chasrob]

I have an object in a 500,000 km circular orbit around earth. I want to change it into an elliptical orbit with a perigee of 500 km. I found this cool site and put those numbers in the calculator. The ”Injection into elliptical transfer orbit delta V” answer is -741.97 m/s. Does that mean to kill that amount of the big orbit velocity (887.2 m/s)? If so, that will put me in an elliptical orbit with a perigee of 500km and apogee of 500,000 km, correct? I don’t need to circularize the final orbit.

Unfortunately, can’t find an explanation of the figures on that web page.

 

[Bandersnatch]

correct?

Yes.

 

[chasrob]

Would it be possible to determine the period of that final orbit and whether its pro or retrograde? Also, its velocity at perigee and apogee?

 

[Janus]

Unfortunately, can’t find an explanation of the figures on that web page.

There is an equation known as as the Vis Viva equation that gives you the orbital velocity of your object if you have the mass of the object it is orbiting, its present orbital distance and it average orbital distance (also known as the semi-major axis of the orbit)
it is
$$v = \sqrt{GM \left ( \frac{2}{r}- \frac{1}{a}\right ) }$$

where r is the present orbital distance and a the semi-major axis. note that for a circular orbit, r=a and you get
$$ v= \sqrt{\frac{GM}{r}}$$

If you just have the perigee and apogee of the orbit of an elliptical orbit, you can get a from taking the average : (r_p+r_a)/2
Now you can get the orbital velocity at perigee of your elliptical orbit. To go from a circular orbit of equal radius to the the apogee, you have to change velocity from circular orbit speed to elliptical orbit apogee speed. Your required velocity change will be the difference between the two.
After all this you can arrive a the following equation for orbital transfer:
$$ \Delta v = \sqrt{\frac{GM}{r_1}} \left ( \sqrt{\frac{2r_2}{r_1+r_2}}-1 \right )$$
Where r₁ is the radius of the circular orbit you are starting from and r₂ is the perigee of the new elliptical orbit. ( it also works if your starting orbit (r₁) is at the perigee of the new orbit and r₂ is the apogee of the new orbit.)

From the numbers you gave and the answer you got, I assume the apogee and perigee numbers you give of your orbit is measured from the surface of the Earth rather than the center (a 500 km perigee as measured from the center is below the Earth's surface), in which case you have to add the radius of the Earth to your numbers before plugging them into the equation since this equation assumes that r₁ and r₂ are measured from the focus of the orbit.

Would it be possible to determine the period of that final orbit and whether its pro or retrograde? Also, its velocity at perigee and apogee?

The period is found by
$$ T = 2 \pi \sqrt{\frac{a^3}{GM}}$$

With 'a' found as above. Since you a just subtracting enough velocity to drop the perigee, which will be less than the starting orbital velocity, the new orbit will be in the same direction as the starting one. The Vis Viva equation I gave earlier will give you perigee and apogee velocities.

 

[Bandersnatch]

Or, you can go the lazy way with velocities, and use the values the calculator gives you. For apogee, take the initial circular velocity and add the elliptical insertion deltaV (remember the minus sign). For perigee take final circular velocity and add the injection into circular. Janus' post explains the why of it.

 

[chasrob]

Or, you can go the lazy way with velocities, and use the values the calculator gives you. For apogee, take the initial circular velocity and add the elliptical insertion deltaV (remember the minus sign). For perigee take final circular velocity and add the injection into circular. Janus' post explains the why of it.

I actually considered that but was certain it couldn't be that simple .

 

[chasrob]

I actually considered that but was certain it couldn't be that simple .

To expand my thinking about using that site's calculator, the initial braking maneuver makes sense to me--you brake to start your infall towards the earth... but that resultant elliptical transfer orbit means your velocity at perigee is ~10700 m/s, correct? It says the final circular orbit has a 7612 m/s velocity, so wouldn't you have to brake down to that speed, instead of adding ~3080 m/s that the calculator says to get into the circular orbit at 500 km?
What am I doing wrong? Wouldn't adding 3080 to 10700 put you into a big solar orbit since the Earth's escape velocity is only 11,200 m/s? Could it be that calculator is for transfers from LEO to GEO only?

 

[Bandersnatch]

The velocity at perigee, providing you want to retain the elliptical orbit, should be on the order of 10 km/s. Circularizing the orbit at this point would require braking to match the ~7 km/s orbital velocity for the circular orbit. Otherwise you have too much speed, and you end up back where you started.

 

[Janus]

To expand my thinking about using that site's calculator, the initial braking maneuver makes sense to me--you brake to start your infall towards the earth... but that resultant elliptical transfer orbit means your velocity at perigee is ~10700 m/s, correct? It says the final circular orbit has a 7612 m/s velocity, so wouldn't you have to brake down to that speed, instead of adding ~3080 m/s that the calculator says to get into the circular orbit at 500 km?
What am I doing wrong? Wouldn't adding 3080 to 10700 put you into a big solar orbit since the Earth's escape velocity is only 11,200 m/s? Could it be that calculator is for transfers from LEO to GEO only?

To determine what delta v you would need to enter the circular orbit from the transfer orbit, you can use a variation of the equation I gave above. In this case:

$$ \Delta v = \sqrt{\frac{GM}{r_2}} \left( 1- \sqrt{\frac{2r_1}{r_1+r_2}} \right )$$

with r₁ being the radius of the circular orbit you started at and r₂ being the radius of the circular orbit you want to end up in.

with either equation a negative answer equates to a braking maneuver. So yes, you would have to brake at perigee in order to insert yourself into a circular orbit. This means that you will have reduced your velocity twice, but still end up with a higher orbital velocity than what you started with (887 m/sec at the 500,000 km circular orbit, and 7612m/s at the 500 km orbit (I get closer to 7614, but we may be using slightly different estimates for Earth's mass or radius.)

 

[chasrob]

To determine what delta v you would need to enter the circular orbit from the transfer orbit, you can use a variation of the equation I gave above. In this case:

$$ \Delta v = \sqrt{\frac{GM}{r_2}} \left( 1- \sqrt{\frac{2r_1}{r_1+r_2}} \right )$$

I used this equation above and came up with that website's result (~3080m/s), only negative 3080; which is what you-all have been saying--you have to brake down at perigee to get to the circular final LEO. Since their result is positive and I guess if it was used at perigee... the result would be ~13,800 m/s and some wild solar orbit.

 

[Bandersnatch]

Yeah, I was actually wondering why the site doesn't have a minus there, given how the box is described as 'Injection into circular orbit delta V'. It would make sense if it were a transfer from low circular to high orbit, where the rocket must add energy at both apses - after checking, the sign in this particular box doesn't change after switching the setup from low to high orbit.
Must be a mistake.