NASASpaceFlight.com Forum
General Discussion => Q&A Section => Topic started by: cube on 04/24/2022 12:52 pm
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Hello, I have always heard that it is more efficient to launch a satellite into a polar orbit from a launch site as far north as possible than from a site on the equator. But then what would be the difference in delta v necessary for a polar orbit between a launch site at the equator and a site at the pole for example? Is the difference negligible?
Thank you !
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No, it is not negligible. If trying to launch into a true polar orbit (90 degree inclination) from the equator, the launch vehicle would have to fly at an azimuth greater 180 degrees because it has to expend energy to negate the velocity contributed by the earth’s rotation. Launching further north or south decreases the velocity from the earth’s rotation
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I saw a message from someone explaining that the difference in delta v between the launch site at the equator and at the pole would be 13 m/s (√7800²+463²≈ 7813). Does this seem right? because I have a doubt.
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That doesn't seem right. Earth's rotation speed at the equator is 463 m/s, so that's what you'd have to shed in the equatorial direction.
Your equation calculates what your final speed would be if you launch towards the pole from the equator (463 m/s) with a rocket that can add 7800 m/s. This doesn't cancel out the 463 m/s.
That triangle (7800 m/s and 463 m/s) gives an angle of your orbit of 86º.
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1/sin(86º) = 1.0024
1.0024 * 7800 = 7819
This agrees with Pythagoras to the accuracy of the 86º.
Engineering is done with numbers. Ignore the opinions.
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But then the difference in delta v between the launch site at the equator and at the pole would be 19 m/s?
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You have initial velocity (463, 0) (Eastward). You want velocity (0, 7800). You need to apply a delta v of (-463,7800). This has a magnitude of sqrt(7800^2+463^2) = 7814 m/s and a direction of atan2(7800, -463) = 93.0 degrees. This angle is the direction you aim the rocket--not the orbit it ends up in. It should make a lot of sense that you need to aim the rocket slightly west of due north in order to cancel out the equatorial velocity.
From the pole, of course, you only need 7800 m/s, so you save 14 m/s by launching into a polar orbit from the pole rather than the equator. Obviously this is negligible.
By comparison, to get into an equatorial orbit from the equator, you need a delta v of (7337,0) or 477 m/s less. This is a very substantial saving.
And to get into an equatorial orbit from the pole is really expensive, since you have to first get into orbit and then change planes. If you try to do the plane change in one burn (assuming a circular orbit), it'll cost you an extra 11,030 m/s, which is coincidentally the same amount to reach escape velocity. (v times sqrt(2)).
Upshot: put your launch site as close to the equator as possible. If you need more-inclined orbits, it'll be cheap to angle the launch for them, but if you need less-inclined orbits, you'll need at least two burns, and it'll be pretty expensive to get there. (2*v*sin(delta_i/2))
Apologies if there are any errors in this. If so, I'm sure someone will point them out. :-)
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Thank you all for your explanations !
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Its possible for equatorial launches to go east to west into retrograde orbit quite DV hit. Israel does this as it can't fly east over its neighbors.
Because satellite and earth are rotating in opposite directs, orbits are lot quicker relative to ground than prograde for same altitude. Can be plus or minus depends on what satellite's job is.
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