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As liftoff proceeds, the desired inclination of your rocket equals the lattitude from which you are lifting off.


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6 minutes ago, MetricKerbalist said:

Hi @K^2 or anyone else,

Wow!  I am getting overwhelmed.  I didn't realize how much I didn't know.

May I please ask you this?  Let's say that a rocket were lifting off from Kennedy Space Center, which is at roughly 28 degrees North latitude.  Could you kindly show on a map or on Google Earth or on something what one complete orbit around Earth would look like?

Thank you for your consideration.

Stanley

nHzoKSz.png

(Don't worry about the weird twistiness to the curve, it's in Earth reference so it tracks the spinning surface. The orbit itself is just an ellipse, but you can pretty much see it here.)

The spaceship launches from the Cape and heads due east. It enters orbit where the red curve finishes. The plane it's in can be defined by two nearby points along the rocket's trajectory, and the center of the Earth. You can see that this plane is inclined relative to the equator.

I used flightclub.io to make this crude plot.

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Hi @cubinator,

Thank you so very much!  That picture was so helpful, you have no idea.

If you don't mind, I am going to press my luck and hope that I don't overtax your good will.  Would you kindly produce one more picture, this time of a rocket's orbital trajectory as it lifts off from somewhere well south of the equator -- let's say in South America, or Cape Town, or Australia.  (I assume that it will head towards the northeast.)

Incidentally, the following  paragraph and its representation on your picture was additionally helpful

43 minutes ago, cubinator said:

The spaceship launches from the Cape and heads due east. It enters orbit where the red curve finishes.... You can see that this plane is inclined relative to the equator.

Thank you for your consideration.

Stanley

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1 hour ago, MetricKerbalist said:

Hi @K^2 or anyone else,

Wow!  I am getting overwhelmed.  I didn't realize how much I didn't know.

May I please ask you this?  Let's say that a rocket were lifting off from Kennedy Space Center, which is at roughly 28 degrees North latitude.  Could you kindly show on a map or on Google Earth or on something what one complete orbit around Earth would look like?

Thank you for your consideration.

Stanley

The thing is, once the rocket leaves the ground it is no longer connected to the ground. So its orbit is no longer connected to the surface of the earth.

All this is actually demonstrated better by playing KSP than by talking about it here!

orbital_mechanics.png

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2 hours ago, MetricKerbalist said:

Hi @Hannu2,

I don't understand the sentence that I quoted.  Would you mind explaining further.

Hi @K^2

Would it be burdensome for you to show this trig?

I thank both of you for your posts and your consideration.

Stanley

Assuming you want a circular orbit in LEO, orbital velocity is ~7800m/s.

Ground speed is cos(latitude of launch site)*460m/s due east.

If your intended orbit is at inclination theta, then your heading (from North) should be at angle phi. 

 

With two sides and the angle between them (the angle between red and orange is theta), the green DV can be found from the cosine rule: a^2 =b^2 + c^2 -2bc*cos(A).

The missing angle can then be found from the sin rule, a/sin(A) = b/sin(B) = c/sin(C).

 

Note that the calculated DV will not be inclusive of gravity and atmospheric losses, so a little extra will be required which is highly dependent on how efficiently a particular rocket is at getting to orbit.

 

Edited by RCgothic
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1 hour ago, MetricKerbalist said:

Hi @K^2 or anyone else,

Wow!  I am getting overwhelmed.  I didn't realize how much I didn't know.

May I please ask you this?  Let's say that a rocket were lifting off from Kennedy Space Center, which is at roughly 28 degrees North latitude.  Could you kindly show on a map or on Google Earth or on something what one complete orbit around Earth would look like?

Thank you for your consideration.

Stanley

Not really no, it would depend so much on the orbit chosen, there are an infinite number of orbits it could take.  all we could say for certain is that none of them would have an inclination below 28 degrees or above 152 (assuming the rocket did not make the Dogleg maneuver @YNMdescribed earlier)

If the planet was not rotating we could also say for certain that the ground trace would describe a great circle around the globe that would cross the equator twice on opposite sides of the globe, and would pass over Kennedy again every orbit.  on a flat map projection it would look most probably like a sine wave but things get fuzzy if it gets close to the poles, and if it crosses them then it just looks like a vertical line, when you also have a rotating planet the ground trace can look like all manner of things.

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2 hours ago, MetricKerbalist said:

If you don't mind, I am going to press my luck and hope that I don't overtax your good will.  Would you kindly produce one more picture, this time of a rocket's orbital trajectory as it lifts off from somewhere well south of the equator -- let's say in South America, or Cape Town, or Australia.  (I assume that it will head towards the northeast.)

Glad I could help! This time I just drew it on the same picture in MS Paint. Here the new rocket is launching from São Paulo, Brazil and heading due east.

czlf7Zt.png

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5 hours ago, MetricKerbalist said:

Would it be burdensome for you to show this trig?

There might be a shortcut I'm missing, but I've converted both coordinate systems into Cartesian and worked from there. The Z axis will run through the poles, and without loss of generality we can take the X axis to be passing through ascending node. Also, we really don't care what the longitude's going to be, so we can set 0° at ascending node as well just for simplicity. We will also be working with true anomaly, so we don't care about the actual radius, so I'm going to take R to be 1 for both planet and target orbit. This is correct to within oblateness, and that's a separate can of worms. For the same reason, we can take argument of periapsis to be 0°, so we'll be counting true anomaly from the ascending node.

So the spherical coordinates are latitude and longitude: θ, φ.  The orbit coordinates are true anomaly and inclination: ν, i.

Conversion from polar coordinates to Cartesian:

x = cos(θ) cos(φ)

y = cos(θ) sin(φ)

z = sin(θ)

Conversion from orbital elements to Cartesian:

x = cos(ν)

y = sin(ν) cos(i)

z = sin(ν) sin(i)

So then you just have these 3 equations to work with. You really only need two of them, but it might be convenient to switch around which one you're using depending on what you're solving for.

cos(θ) cos(φ) = cos(ν)

cos(θ) sin(φ) = sin(ν) cos(i)

sin(θ) = sin(ν) sin(i)

If we are looking at situation where we know the desired inclination i, and the starting latitude θ, then you know cos/sin of these angles and you can work out what the ν and φ are going to be. Now, you shouldn't treat φ as actual longitude of the launch - rather, it tells you what the launch window should be to match your desired longitude of the ascending node.

Finally, if you want the angle for the heading, you have to do a bit of calculus. If you treat orbit as a parametric curve parametrized by ν, you know that the tangent in polar coordinates will be given by ∂θ/∂ν and ∂φ/∂ν.

∂θ/∂ν = cos(ν) sin(i) / cos(θ)

∂φ/∂ν = cos(ν) cos(i) / (cos(θ) cos(φ))

So the heading angle will be given by arctan((∂φ/∂ν) / (∂θ/∂ν)). Keep in mind that this is true heading, and you need to do considerably more math to get ground heading from it.

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