Is it really ~648.69 km/s Delta-V to “Land” on the Surface of the Sun?Are Delta-V requirements for leaving the surface of a planet proportional to gravity?Calculating solar system escape and and sun-dive delta V from lower Earth orbitDelta-V required for lift-off from a planet/ asteroidNuking the Sun?Do any current ICBM's have the delta-V to target the sun?Is this really an image of the sun, or an “artist's conception”?How much delta v does it take to get to the Sun-Earth Lagrange 3 point?Is oxygen present in the sun?Could a spacecraft be propelled by the deflection of a very high number of charged particles?Delta V to get to the Sun-Earth Lagrange Point 1?
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Is it really ~648.69 km/s Delta-V to “Land” on the Surface of the Sun?
Are Delta-V requirements for leaving the surface of a planet proportional to gravity?Calculating solar system escape and and sun-dive delta V from lower Earth orbitDelta-V required for lift-off from a planet/ asteroidNuking the Sun?Do any current ICBM's have the delta-V to target the sun?Is this really an image of the sun, or an “artist's conception”?How much delta v does it take to get to the Sun-Earth Lagrange 3 point?Is oxygen present in the sun?Could a spacecraft be propelled by the deflection of a very high number of charged particles?Delta V to get to the Sun-Earth Lagrange Point 1?
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$begingroup$
According to the following diagram it would require ~648.69 km/s to do all of the intermediate transfers that would land you at or around the sun's surface at perigee. Is this a real number? How would they have calculated this number, is the sun's mass and other quantities known well enough for this to be absolutely accurate? It seems like a massive number, am I reading this wrong?
I understand partially the vis-viva equations, but not in terms of the sun. It seems odd to use it around the body you're currently orbiting, do I have to think of it in terms of a different frame of reference? I was thinking this is the delta-v for a non-circularized orbit with a lowest point being at the suns surface and the highest at Low Sun Orbit (that's my thought on the 440 number)... but I'm not sure.
delta-v the-sun
asked 8 hours ago
Magic Octopus UrnMagic Octopus Urn
4,2901 gold badge18 silver badges59 bronze badges
$endgroup$
|
show 2 more comments
$begingroup$
According to the following diagram it would require ~648.69 km/s to do all of the intermediate transfers that would land you at or around the sun's surface at perigee. Is this a real number? How would they have calculated this number, is the sun's mass and other quantities known well enough for this to be absolutely accurate? It seems like a massive number, am I reading this wrong?
I understand partially the vis-viva equations, but not in terms of the sun. It seems odd to use it around the body you're currently orbiting, do I have to think of it in terms of a different frame of reference? I was thinking this is the delta-v for a non-circularized orbit with a lowest point being at the suns surface and the highest at Low Sun Orbit (that's my thought on the 440 number)... but I'm not sure.
delta-v the-sun
asked 8 hours ago
Magic Octopus UrnMagic Octopus Urn
4,2901 gold badge18 silver badges59 bronze badges
$endgroup$
5
$begingroup$
You can only do it at night.
$endgroup$
– Organic Marble
8 hours ago
1
$begingroup$
@OrganicMarble that's what North Korea did.
$endgroup$
– Magic Octopus Urn
8 hours ago
1
$begingroup$
Seems plausible. The delta v for landing is mostly cancellation of orbital speed, and solar orbit speed at Earth’s altitude is 30km/s — from low solar orbit it would be much higher. WP gives the sun’s mass to four decimal places; I don’t know if it’s better known than that.
$endgroup$
– Russell Borogove
7 hours ago
$begingroup$
@RussellBorogove WP? Also that makes more sense, the numbers are so high because these are calculations of transfers to circular orbits, not transfers to have one side of the orbit at that altitude.
$endgroup$
– Magic Octopus Urn
7 hours ago
$begingroup$
WP = Wikipedia en.m.wikipedia.org/wiki/Sun
$endgroup$
– Russell Borogove
7 hours ago
|
show 2 more comments
$begingroup$
According to the following diagram it would require ~648.69 km/s to do all of the intermediate transfers that would land you at or around the sun's surface at perigee. Is this a real number? How would they have calculated this number, is the sun's mass and other quantities known well enough for this to be absolutely accurate? It seems like a massive number, am I reading this wrong?
I understand partially the vis-viva equations, but not in terms of the sun. It seems odd to use it around the body you're currently orbiting, do I have to think of it in terms of a different frame of reference? I was thinking this is the delta-v for a non-circularized orbit with a lowest point being at the suns surface and the highest at Low Sun Orbit (that's my thought on the 440 number)... but I'm not sure.
delta-v the-sun
asked 8 hours ago
Magic Octopus UrnMagic Octopus Urn
4,2901 gold badge18 silver badges59 bronze badges
$endgroup$
According to the following diagram it would require ~648.69 km/s to do all of the intermediate transfers that would land you at or around the sun's surface at perigee. Is this a real number? How would they have calculated this number, is the sun's mass and other quantities known well enough for this to be absolutely accurate? It seems like a massive number, am I reading this wrong?
I understand partially the vis-viva equations, but not in terms of the sun. It seems odd to use it around the body you're currently orbiting, do I have to think of it in terms of a different frame of reference? I was thinking this is the delta-v for a non-circularized orbit with a lowest point being at the suns surface and the highest at Low Sun Orbit (that's my thought on the 440 number)... but I'm not sure.
delta-v the-sun
delta-v the-sun
asked 8 hours ago
Magic Octopus UrnMagic Octopus Urn
4,2901 gold badge18 silver badges59 bronze badges
asked 8 hours ago
Magic Octopus UrnMagic Octopus Urn
4,2901 gold badge18 silver badges59 bronze badges
edited 8 hours ago
Magic Octopus Urn
asked 8 hours ago
Magic Octopus UrnMagic Octopus Urn
4,2901 gold badge18 silver badges59 bronze badges
asked 8 hours ago
Magic Octopus UrnMagic Octopus Urn
4,2901 gold badge18 silver badges59 bronze badges
asked 8 hours ago
Magic Octopus UrnMagic Octopus Urn
4,2901 gold badge18 silver badges59 bronze badges
4,2901 gold badge18 silver badges59 bronze badges
5
$begingroup$
You can only do it at night.
$endgroup$
– Organic Marble
8 hours ago
1
$begingroup$
@OrganicMarble that's what North Korea did.
$endgroup$
– Magic Octopus Urn
8 hours ago
1
$begingroup$
Seems plausible. The delta v for landing is mostly cancellation of orbital speed, and solar orbit speed at Earth’s altitude is 30km/s — from low solar orbit it would be much higher. WP gives the sun’s mass to four decimal places; I don’t know if it’s better known than that.
$endgroup$
– Russell Borogove
7 hours ago
$begingroup$
@RussellBorogove WP? Also that makes more sense, the numbers are so high because these are calculations of transfers to circular orbits, not transfers to have one side of the orbit at that altitude.
$endgroup$
– Magic Octopus Urn
7 hours ago
$begingroup$
WP = Wikipedia en.m.wikipedia.org/wiki/Sun
$endgroup$
– Russell Borogove
7 hours ago
|
show 2 more comments
5
$begingroup$
You can only do it at night.
$endgroup$
– Organic Marble
8 hours ago
1
$begingroup$
@OrganicMarble that's what North Korea did.
$endgroup$
– Magic Octopus Urn
8 hours ago
1
$begingroup$
Seems plausible. The delta v for landing is mostly cancellation of orbital speed, and solar orbit speed at Earth’s altitude is 30km/s — from low solar orbit it would be much higher. WP gives the sun’s mass to four decimal places; I don’t know if it’s better known than that.
$endgroup$
– Russell Borogove
7 hours ago
$begingroup$
@RussellBorogove WP? Also that makes more sense, the numbers are so high because these are calculations of transfers to circular orbits, not transfers to have one side of the orbit at that altitude.
$endgroup$
– Magic Octopus Urn
7 hours ago
$begingroup$
WP = Wikipedia en.m.wikipedia.org/wiki/Sun
$endgroup$
– Russell Borogove
7 hours ago
5
5
$begingroup$
You can only do it at night.
$endgroup$
– Organic Marble
8 hours ago
$begingroup$
You can only do it at night.
$endgroup$
– Organic Marble
8 hours ago
1
1
$begingroup$
@OrganicMarble that's what North Korea did.
$endgroup$
– Magic Octopus Urn
8 hours ago
$begingroup$
@OrganicMarble that's what North Korea did.
$endgroup$
– Magic Octopus Urn
8 hours ago
1
1
$begingroup$
Seems plausible. The delta v for landing is mostly cancellation of orbital speed, and solar orbit speed at Earth’s altitude is 30km/s — from low solar orbit it would be much higher. WP gives the sun’s mass to four decimal places; I don’t know if it’s better known than that.
$endgroup$
– Russell Borogove
7 hours ago
$begingroup$
Seems plausible. The delta v for landing is mostly cancellation of orbital speed, and solar orbit speed at Earth’s altitude is 30km/s — from low solar orbit it would be much higher. WP gives the sun’s mass to four decimal places; I don’t know if it’s better known than that.
$endgroup$
– Russell Borogove
7 hours ago
$begingroup$
@RussellBorogove WP? Also that makes more sense, the numbers are so high because these are calculations of transfers to circular orbits, not transfers to have one side of the orbit at that altitude.
$endgroup$
– Magic Octopus Urn
7 hours ago
$begingroup$
@RussellBorogove WP? Also that makes more sense, the numbers are so high because these are calculations of transfers to circular orbits, not transfers to have one side of the orbit at that altitude.
$endgroup$
– Magic Octopus Urn
7 hours ago
$begingroup$
WP = Wikipedia en.m.wikipedia.org/wiki/Sun
$endgroup$
– Russell Borogove
7 hours ago
$begingroup$
WP = Wikipedia en.m.wikipedia.org/wiki/Sun
$endgroup$
– Russell Borogove
7 hours ago
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
The Vis-viva equation is
$$ v = sqrt GM left(frac2r - frac1a right) , $$
The $GM$ product for the Sun is 1.327E+20 m^3/s^2.
If 1 AU is 150E+09 meters, then when you are in a circular ($r=a$) Heliocentric orbit at Earth escape/capture point your velocity is 29.7 km/s.
If you then change to an ellipse with aphelion still at 1 AU and perihelion at the surface of the Sun, then $a = (1 AU +695700 km)/2 = 75.35 $ million km and your velocity at aphelion ($r= 1AU$) is now only 2.86 km/s, and at perihelion (grazing the Sun) it's 616.2 km/s. You've had to loose $Delta v$ of 26.9 km/s to get into that orbit, and you'll have to loose another 616.2 km/s to stop on the Sun the next time you graze it.
That's 643.1 km/s, plus the 12.6 km/s to get from Earth's surface to Earth escape/capture, so the total I get is 655.7 km/s.
That's pretty close to your number. I've treated the Earth's orbit as a circle at 150 million km exactly, which it isn't.
answered 7 hours ago
uhohuhoh
49.1k22 gold badges196 silver badges637 bronze badges
$endgroup$
1
$begingroup$
Wow, when explained like that it's much easier to understand the math behind those charts. I was definitely overthinking the application of that equation... Thanks for expanding on the actual calculations.
$endgroup$
– Magic Octopus Urn
5 hours ago
add a comment |
$begingroup$
It's a "real number", a delta-v chart isn't really concerned with the realism of various missions. It's quite simply a table of velocity changes, how you would go about achieving these velocity changes is outside its scope.
As to how to calculate it, the parts of the journey in space can be calculated like any other part of the diagram, using the patched conic approximation to split it up into two body systems, and from then the Vis-viva equation (written on the chart).
The other lading/ascension values on the chart appears to consider atmospheric drag, but they skipped it for the Sun since any realistic drag model for a spacecraft there doesn't exist. It's just the speed you would crash into the Sun with from low orbit. (or more accurately, the way you have described it in your question, as a small elliptical transfer). Using the Vis-viva equation for the body you are currently orbiting isn't odd at all, it's the normal use case.
Are the relevant quantities for the Sun known well enough to give a reliable number? Yes and no. The Sun's mass is known very accurately due to accurate measurements of the planets orbiting it, but the other relevant number, the radius is more so-so. The Sun doesn't really have a solid surface, so the surface radius is more a question of definition than accuracy of measurements.
Lastly, it seems like a massive number, can it be right? Going from the surface of a body, and then navigating some in space, you would expect spending about the same delta-v as the escape velocity of the body. Given the Sun's escape velocity of 620km/s, this seems correct.
answered 7 hours ago
Hohmannfan♦Hohmannfan
11.8k1 gold badge43 silver badges97 bronze badges
$endgroup$
$begingroup$
Just want to clarify, that the 440 delta-v is not for a "circular orbit at the surface of the sun" (I know that doesn't make sense) but an orbit at which Apogee is LSO and Perigee is Sun Surface? However, the 178 delta-v is a transfer from Earth orbit to a circular Low Sun Orbit both probably approximated with 0 eccentricity? +1 anyway, for the links and the factoids.
$endgroup$
– Magic Octopus Urn
7 hours ago
1
$begingroup$
1. Yes, that's correct. I can edit that in to better address that part of the question. 2. Not quite. The node at 178 distance is the Earth-LSO transfer, at which point you are already in orbit around the Sun in an ellipse touching LSO and Earth's orbit.
$endgroup$
– Hohmannfan♦
7 hours ago
$begingroup$
Ah, so literally none of these are transfers to circular orbits, just the delta-v for the first burn (assumed instantaneous) at the apogee (being LEO) to lower from LEO to LSO on only one side of the orbit. Thanks!
$endgroup$
– Magic Octopus Urn
5 hours ago
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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votes
$begingroup$
The Vis-viva equation is
$$ v = sqrt GM left(frac2r - frac1a right) , $$
The $GM$ product for the Sun is 1.327E+20 m^3/s^2.
If 1 AU is 150E+09 meters, then when you are in a circular ($r=a$) Heliocentric orbit at Earth escape/capture point your velocity is 29.7 km/s.
If you then change to an ellipse with aphelion still at 1 AU and perihelion at the surface of the Sun, then $a = (1 AU +695700 km)/2 = 75.35 $ million km and your velocity at aphelion ($r= 1AU$) is now only 2.86 km/s, and at perihelion (grazing the Sun) it's 616.2 km/s. You've had to loose $Delta v$ of 26.9 km/s to get into that orbit, and you'll have to loose another 616.2 km/s to stop on the Sun the next time you graze it.
That's 643.1 km/s, plus the 12.6 km/s to get from Earth's surface to Earth escape/capture, so the total I get is 655.7 km/s.
That's pretty close to your number. I've treated the Earth's orbit as a circle at 150 million km exactly, which it isn't.
answered 7 hours ago
uhohuhoh
49.1k22 gold badges196 silver badges637 bronze badges
$endgroup$
1
$begingroup$
Wow, when explained like that it's much easier to understand the math behind those charts. I was definitely overthinking the application of that equation... Thanks for expanding on the actual calculations.
$endgroup$
– Magic Octopus Urn
5 hours ago
add a comment |
$begingroup$
The Vis-viva equation is
$$ v = sqrt GM left(frac2r - frac1a right) , $$
The $GM$ product for the Sun is 1.327E+20 m^3/s^2.
If 1 AU is 150E+09 meters, then when you are in a circular ($r=a$) Heliocentric orbit at Earth escape/capture point your velocity is 29.7 km/s.
If you then change to an ellipse with aphelion still at 1 AU and perihelion at the surface of the Sun, then $a = (1 AU +695700 km)/2 = 75.35 $ million km and your velocity at aphelion ($r= 1AU$) is now only 2.86 km/s, and at perihelion (grazing the Sun) it's 616.2 km/s. You've had to loose $Delta v$ of 26.9 km/s to get into that orbit, and you'll have to loose another 616.2 km/s to stop on the Sun the next time you graze it.
That's 643.1 km/s, plus the 12.6 km/s to get from Earth's surface to Earth escape/capture, so the total I get is 655.7 km/s.
That's pretty close to your number. I've treated the Earth's orbit as a circle at 150 million km exactly, which it isn't.
answered 7 hours ago
uhohuhoh
49.1k22 gold badges196 silver badges637 bronze badges
$endgroup$
1
$begingroup$
Wow, when explained like that it's much easier to understand the math behind those charts. I was definitely overthinking the application of that equation... Thanks for expanding on the actual calculations.
$endgroup$
– Magic Octopus Urn
5 hours ago
add a comment |
$begingroup$
The Vis-viva equation is
$$ v = sqrt GM left(frac2r - frac1a right) , $$
The $GM$ product for the Sun is 1.327E+20 m^3/s^2.
If 1 AU is 150E+09 meters, then when you are in a circular ($r=a$) Heliocentric orbit at Earth escape/capture point your velocity is 29.7 km/s.
If you then change to an ellipse with aphelion still at 1 AU and perihelion at the surface of the Sun, then $a = (1 AU +695700 km)/2 = 75.35 $ million km and your velocity at aphelion ($r= 1AU$) is now only 2.86 km/s, and at perihelion (grazing the Sun) it's 616.2 km/s. You've had to loose $Delta v$ of 26.9 km/s to get into that orbit, and you'll have to loose another 616.2 km/s to stop on the Sun the next time you graze it.
That's 643.1 km/s, plus the 12.6 km/s to get from Earth's surface to Earth escape/capture, so the total I get is 655.7 km/s.
That's pretty close to your number. I've treated the Earth's orbit as a circle at 150 million km exactly, which it isn't.
answered 7 hours ago
uhohuhoh
49.1k22 gold badges196 silver badges637 bronze badges
$endgroup$
The Vis-viva equation is
$$ v = sqrt GM left(frac2r - frac1a right) , $$
The $GM$ product for the Sun is 1.327E+20 m^3/s^2.
If 1 AU is 150E+09 meters, then when you are in a circular ($r=a$) Heliocentric orbit at Earth escape/capture point your velocity is 29.7 km/s.
If you then change to an ellipse with aphelion still at 1 AU and perihelion at the surface of the Sun, then $a = (1 AU +695700 km)/2 = 75.35 $ million km and your velocity at aphelion ($r= 1AU$) is now only 2.86 km/s, and at perihelion (grazing the Sun) it's 616.2 km/s. You've had to loose $Delta v$ of 26.9 km/s to get into that orbit, and you'll have to loose another 616.2 km/s to stop on the Sun the next time you graze it.
That's 643.1 km/s, plus the 12.6 km/s to get from Earth's surface to Earth escape/capture, so the total I get is 655.7 km/s.
That's pretty close to your number. I've treated the Earth's orbit as a circle at 150 million km exactly, which it isn't.
answered 7 hours ago
uhohuhoh
49.1k22 gold badges196 silver badges637 bronze badges
answered 7 hours ago
uhohuhoh
49.1k22 gold badges196 silver badges637 bronze badges
answered 7 hours ago
uhohuhoh
49.1k22 gold badges196 silver badges637 bronze badges
answered 7 hours ago
uhohuhoh
49.1k22 gold badges196 silver badges637 bronze badges
49.1k22 gold badges196 silver badges637 bronze badges
1
$begingroup$
Wow, when explained like that it's much easier to understand the math behind those charts. I was definitely overthinking the application of that equation... Thanks for expanding on the actual calculations.
$endgroup$
– Magic Octopus Urn
5 hours ago
add a comment |
1
$begingroup$
Wow, when explained like that it's much easier to understand the math behind those charts. I was definitely overthinking the application of that equation... Thanks for expanding on the actual calculations.
$endgroup$
– Magic Octopus Urn
5 hours ago
1
1
$begingroup$
Wow, when explained like that it's much easier to understand the math behind those charts. I was definitely overthinking the application of that equation... Thanks for expanding on the actual calculations.
$endgroup$
– Magic Octopus Urn
5 hours ago
$begingroup$
Wow, when explained like that it's much easier to understand the math behind those charts. I was definitely overthinking the application of that equation... Thanks for expanding on the actual calculations.
$endgroup$
– Magic Octopus Urn
5 hours ago
add a comment |
$begingroup$
It's a "real number", a delta-v chart isn't really concerned with the realism of various missions. It's quite simply a table of velocity changes, how you would go about achieving these velocity changes is outside its scope.
As to how to calculate it, the parts of the journey in space can be calculated like any other part of the diagram, using the patched conic approximation to split it up into two body systems, and from then the Vis-viva equation (written on the chart).
The other lading/ascension values on the chart appears to consider atmospheric drag, but they skipped it for the Sun since any realistic drag model for a spacecraft there doesn't exist. It's just the speed you would crash into the Sun with from low orbit. (or more accurately, the way you have described it in your question, as a small elliptical transfer). Using the Vis-viva equation for the body you are currently orbiting isn't odd at all, it's the normal use case.
Are the relevant quantities for the Sun known well enough to give a reliable number? Yes and no. The Sun's mass is known very accurately due to accurate measurements of the planets orbiting it, but the other relevant number, the radius is more so-so. The Sun doesn't really have a solid surface, so the surface radius is more a question of definition than accuracy of measurements.
Lastly, it seems like a massive number, can it be right? Going from the surface of a body, and then navigating some in space, you would expect spending about the same delta-v as the escape velocity of the body. Given the Sun's escape velocity of 620km/s, this seems correct.
answered 7 hours ago
Hohmannfan♦Hohmannfan
11.8k1 gold badge43 silver badges97 bronze badges
$endgroup$
$begingroup$
Just want to clarify, that the 440 delta-v is not for a "circular orbit at the surface of the sun" (I know that doesn't make sense) but an orbit at which Apogee is LSO and Perigee is Sun Surface? However, the 178 delta-v is a transfer from Earth orbit to a circular Low Sun Orbit both probably approximated with 0 eccentricity? +1 anyway, for the links and the factoids.
$endgroup$
– Magic Octopus Urn
7 hours ago
1
$begingroup$
1. Yes, that's correct. I can edit that in to better address that part of the question. 2. Not quite. The node at 178 distance is the Earth-LSO transfer, at which point you are already in orbit around the Sun in an ellipse touching LSO and Earth's orbit.
$endgroup$
– Hohmannfan♦
7 hours ago
$begingroup$
Ah, so literally none of these are transfers to circular orbits, just the delta-v for the first burn (assumed instantaneous) at the apogee (being LEO) to lower from LEO to LSO on only one side of the orbit. Thanks!
$endgroup$
– Magic Octopus Urn
5 hours ago
add a comment |
$begingroup$
It's a "real number", a delta-v chart isn't really concerned with the realism of various missions. It's quite simply a table of velocity changes, how you would go about achieving these velocity changes is outside its scope.
As to how to calculate it, the parts of the journey in space can be calculated like any other part of the diagram, using the patched conic approximation to split it up into two body systems, and from then the Vis-viva equation (written on the chart).
The other lading/ascension values on the chart appears to consider atmospheric drag, but they skipped it for the Sun since any realistic drag model for a spacecraft there doesn't exist. It's just the speed you would crash into the Sun with from low orbit. (or more accurately, the way you have described it in your question, as a small elliptical transfer). Using the Vis-viva equation for the body you are currently orbiting isn't odd at all, it's the normal use case.
Are the relevant quantities for the Sun known well enough to give a reliable number? Yes and no. The Sun's mass is known very accurately due to accurate measurements of the planets orbiting it, but the other relevant number, the radius is more so-so. The Sun doesn't really have a solid surface, so the surface radius is more a question of definition than accuracy of measurements.
Lastly, it seems like a massive number, can it be right? Going from the surface of a body, and then navigating some in space, you would expect spending about the same delta-v as the escape velocity of the body. Given the Sun's escape velocity of 620km/s, this seems correct.
answered 7 hours ago
Hohmannfan♦Hohmannfan
11.8k1 gold badge43 silver badges97 bronze badges
$endgroup$
$begingroup$
Just want to clarify, that the 440 delta-v is not for a "circular orbit at the surface of the sun" (I know that doesn't make sense) but an orbit at which Apogee is LSO and Perigee is Sun Surface? However, the 178 delta-v is a transfer from Earth orbit to a circular Low Sun Orbit both probably approximated with 0 eccentricity? +1 anyway, for the links and the factoids.
$endgroup$
– Magic Octopus Urn
7 hours ago
1
$begingroup$
1. Yes, that's correct. I can edit that in to better address that part of the question. 2. Not quite. The node at 178 distance is the Earth-LSO transfer, at which point you are already in orbit around the Sun in an ellipse touching LSO and Earth's orbit.
$endgroup$
– Hohmannfan♦
7 hours ago
$begingroup$
Ah, so literally none of these are transfers to circular orbits, just the delta-v for the first burn (assumed instantaneous) at the apogee (being LEO) to lower from LEO to LSO on only one side of the orbit. Thanks!
$endgroup$
– Magic Octopus Urn
5 hours ago
add a comment |
$begingroup$
It's a "real number", a delta-v chart isn't really concerned with the realism of various missions. It's quite simply a table of velocity changes, how you would go about achieving these velocity changes is outside its scope.
As to how to calculate it, the parts of the journey in space can be calculated like any other part of the diagram, using the patched conic approximation to split it up into two body systems, and from then the Vis-viva equation (written on the chart).
The other lading/ascension values on the chart appears to consider atmospheric drag, but they skipped it for the Sun since any realistic drag model for a spacecraft there doesn't exist. It's just the speed you would crash into the Sun with from low orbit. (or more accurately, the way you have described it in your question, as a small elliptical transfer). Using the Vis-viva equation for the body you are currently orbiting isn't odd at all, it's the normal use case.
Are the relevant quantities for the Sun known well enough to give a reliable number? Yes and no. The Sun's mass is known very accurately due to accurate measurements of the planets orbiting it, but the other relevant number, the radius is more so-so. The Sun doesn't really have a solid surface, so the surface radius is more a question of definition than accuracy of measurements.
Lastly, it seems like a massive number, can it be right? Going from the surface of a body, and then navigating some in space, you would expect spending about the same delta-v as the escape velocity of the body. Given the Sun's escape velocity of 620km/s, this seems correct.
answered 7 hours ago
Hohmannfan♦Hohmannfan
11.8k1 gold badge43 silver badges97 bronze badges
$endgroup$
It's a "real number", a delta-v chart isn't really concerned with the realism of various missions. It's quite simply a table of velocity changes, how you would go about achieving these velocity changes is outside its scope.
As to how to calculate it, the parts of the journey in space can be calculated like any other part of the diagram, using the patched conic approximation to split it up into two body systems, and from then the Vis-viva equation (written on the chart).
The other lading/ascension values on the chart appears to consider atmospheric drag, but they skipped it for the Sun since any realistic drag model for a spacecraft there doesn't exist. It's just the speed you would crash into the Sun with from low orbit. (or more accurately, the way you have described it in your question, as a small elliptical transfer). Using the Vis-viva equation for the body you are currently orbiting isn't odd at all, it's the normal use case.
Are the relevant quantities for the Sun known well enough to give a reliable number? Yes and no. The Sun's mass is known very accurately due to accurate measurements of the planets orbiting it, but the other relevant number, the radius is more so-so. The Sun doesn't really have a solid surface, so the surface radius is more a question of definition than accuracy of measurements.
Lastly, it seems like a massive number, can it be right? Going from the surface of a body, and then navigating some in space, you would expect spending about the same delta-v as the escape velocity of the body. Given the Sun's escape velocity of 620km/s, this seems correct.
answered 7 hours ago
Hohmannfan♦Hohmannfan
11.8k1 gold badge43 silver badges97 bronze badges
edited 7 hours ago
answered 7 hours ago
Hohmannfan♦Hohmannfan
11.8k1 gold badge43 silver badges97 bronze badges
answered 7 hours ago
Hohmannfan♦Hohmannfan
11.8k1 gold badge43 silver badges97 bronze badges
answered 7 hours ago
Hohmannfan♦Hohmannfan
11.8k1 gold badge43 silver badges97 bronze badges
11.8k1 gold badge43 silver badges97 bronze badges
$begingroup$
Just want to clarify, that the 440 delta-v is not for a "circular orbit at the surface of the sun" (I know that doesn't make sense) but an orbit at which Apogee is LSO and Perigee is Sun Surface? However, the 178 delta-v is a transfer from Earth orbit to a circular Low Sun Orbit both probably approximated with 0 eccentricity? +1 anyway, for the links and the factoids.
$endgroup$
– Magic Octopus Urn
7 hours ago
1
$begingroup$
1. Yes, that's correct. I can edit that in to better address that part of the question. 2. Not quite. The node at 178 distance is the Earth-LSO transfer, at which point you are already in orbit around the Sun in an ellipse touching LSO and Earth's orbit.
$endgroup$
– Hohmannfan♦
7 hours ago
$begingroup$
Ah, so literally none of these are transfers to circular orbits, just the delta-v for the first burn (assumed instantaneous) at the apogee (being LEO) to lower from LEO to LSO on only one side of the orbit. Thanks!
$endgroup$
– Magic Octopus Urn
5 hours ago
add a comment |
$begingroup$
Just want to clarify, that the 440 delta-v is not for a "circular orbit at the surface of the sun" (I know that doesn't make sense) but an orbit at which Apogee is LSO and Perigee is Sun Surface? However, the 178 delta-v is a transfer from Earth orbit to a circular Low Sun Orbit both probably approximated with 0 eccentricity? +1 anyway, for the links and the factoids.
$endgroup$
– Magic Octopus Urn
7 hours ago
1
$begingroup$
1. Yes, that's correct. I can edit that in to better address that part of the question. 2. Not quite. The node at 178 distance is the Earth-LSO transfer, at which point you are already in orbit around the Sun in an ellipse touching LSO and Earth's orbit.
$endgroup$
– Hohmannfan♦
7 hours ago
$begingroup$
Ah, so literally none of these are transfers to circular orbits, just the delta-v for the first burn (assumed instantaneous) at the apogee (being LEO) to lower from LEO to LSO on only one side of the orbit. Thanks!
$endgroup$
– Magic Octopus Urn
5 hours ago
$begingroup$
Just want to clarify, that the 440 delta-v is not for a "circular orbit at the surface of the sun" (I know that doesn't make sense) but an orbit at which Apogee is LSO and Perigee is Sun Surface? However, the 178 delta-v is a transfer from Earth orbit to a circular Low Sun Orbit both probably approximated with 0 eccentricity? +1 anyway, for the links and the factoids.
$endgroup$
– Magic Octopus Urn
7 hours ago
$begingroup$
Just want to clarify, that the 440 delta-v is not for a "circular orbit at the surface of the sun" (I know that doesn't make sense) but an orbit at which Apogee is LSO and Perigee is Sun Surface? However, the 178 delta-v is a transfer from Earth orbit to a circular Low Sun Orbit both probably approximated with 0 eccentricity? +1 anyway, for the links and the factoids.
$endgroup$
– Magic Octopus Urn
7 hours ago
1
1
$begingroup$
1. Yes, that's correct. I can edit that in to better address that part of the question. 2. Not quite. The node at 178 distance is the Earth-LSO transfer, at which point you are already in orbit around the Sun in an ellipse touching LSO and Earth's orbit.
$endgroup$
– Hohmannfan♦
7 hours ago
$begingroup$
1. Yes, that's correct. I can edit that in to better address that part of the question. 2. Not quite. The node at 178 distance is the Earth-LSO transfer, at which point you are already in orbit around the Sun in an ellipse touching LSO and Earth's orbit.
$endgroup$
– Hohmannfan♦
7 hours ago
$begingroup$
Ah, so literally none of these are transfers to circular orbits, just the delta-v for the first burn (assumed instantaneous) at the apogee (being LEO) to lower from LEO to LSO on only one side of the orbit. Thanks!
$endgroup$
– Magic Octopus Urn
5 hours ago
$begingroup$
Ah, so literally none of these are transfers to circular orbits, just the delta-v for the first burn (assumed instantaneous) at the apogee (being LEO) to lower from LEO to LSO on only one side of the orbit. Thanks!
$endgroup$
– Magic Octopus Urn
5 hours ago
add a comment |
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5
$begingroup$
You can only do it at night.
$endgroup$
– Organic Marble
8 hours ago
1
$begingroup$
@OrganicMarble that's what North Korea did.
$endgroup$
– Magic Octopus Urn
8 hours ago
1
$begingroup$
Seems plausible. The delta v for landing is mostly cancellation of orbital speed, and solar orbit speed at Earth’s altitude is 30km/s — from low solar orbit it would be much higher. WP gives the sun’s mass to four decimal places; I don’t know if it’s better known than that.
$endgroup$
– Russell Borogove
7 hours ago
$begingroup$
@RussellBorogove WP? Also that makes more sense, the numbers are so high because these are calculations of transfers to circular orbits, not transfers to have one side of the orbit at that altitude.
$endgroup$
– Magic Octopus Urn
7 hours ago
$begingroup$
WP = Wikipedia en.m.wikipedia.org/wiki/Sun
$endgroup$
– Russell Borogove
7 hours ago