Deviation of semi-major axis
If I calculate the semi-major axis of Molniya-1T with
$a = sqrt[3]{dfrac{GM}{n^2}}$ with $n=3.18683728text{ }d^{-1}$, I get another apoapsis ($a=19505.7$ km) as noted at Heavens Above:
apogee height: 25659 km
perigee height: 595 km
which is $a=frac{25659+595}{2}=13127$ km.
This is my approach in Octave/MATLAB:
# Computes the semi major axis a from n with constant GM
# @params:
# GM constant (cubic km per square second)
# n: mean motion (revs/day)
# @return:
# a: semi major axis (km)
function a = getSemiMajorAxis(GM,n)
n = n*2*pi/(24*60*60); #conversion (revs/day) --> (rad/s)
a = (GM/(power(n,2))).^(1./3);
endfunction
getSemiMajorAxis(398600.44,3.18683728) # Test with Molniya-1T
So, what is the origin of the deviation? Isn't apogee the same as apoapsis basically?
orbital-mechanics orbit orbital-elements
edited Dec 28 '18 at 2:26
Peter Mortensen
1947
asked Dec 27 '18 at 17:03
sequoiasequoia
1416
add a comment |
If I calculate the semi-major axis of Molniya-1T with
$a = sqrt[3]{dfrac{GM}{n^2}}$ with $n=3.18683728text{ }d^{-1}$, I get another apoapsis ($a=19505.7$ km) as noted at Heavens Above:
apogee height: 25659 km
perigee height: 595 km
which is $a=frac{25659+595}{2}=13127$ km.
This is my approach in Octave/MATLAB:
# Computes the semi major axis a from n with constant GM
# @params:
# GM constant (cubic km per square second)
# n: mean motion (revs/day)
# @return:
# a: semi major axis (km)
function a = getSemiMajorAxis(GM,n)
n = n*2*pi/(24*60*60); #conversion (revs/day) --> (rad/s)
a = (GM/(power(n,2))).^(1./3);
endfunction
getSemiMajorAxis(398600.44,3.18683728) # Test with Molniya-1T
So, what is the origin of the deviation? Isn't apogee the same as apoapsis basically?
orbital-mechanics orbit orbital-elements
edited Dec 28 '18 at 2:26
Peter Mortensen
1947
asked Dec 27 '18 at 17:03
sequoiasequoia
1416
I saw your profile; here are some geoinformatics questions you might find interesting: Shape and dimensions of the Moon's reference surface for selenographic latitude/longitude? and How do we define geographical coordinates on non-spheroid celestial bodies?
– uhoh
Dec 28 '18 at 2:22
add a comment |
If I calculate the semi-major axis of Molniya-1T with
$a = sqrt[3]{dfrac{GM}{n^2}}$ with $n=3.18683728text{ }d^{-1}$, I get another apoapsis ($a=19505.7$ km) as noted at Heavens Above:
apogee height: 25659 km
perigee height: 595 km
which is $a=frac{25659+595}{2}=13127$ km.
This is my approach in Octave/MATLAB:
# Computes the semi major axis a from n with constant GM
# @params:
# GM constant (cubic km per square second)
# n: mean motion (revs/day)
# @return:
# a: semi major axis (km)
function a = getSemiMajorAxis(GM,n)
n = n*2*pi/(24*60*60); #conversion (revs/day) --> (rad/s)
a = (GM/(power(n,2))).^(1./3);
endfunction
getSemiMajorAxis(398600.44,3.18683728) # Test with Molniya-1T
So, what is the origin of the deviation? Isn't apogee the same as apoapsis basically?
orbital-mechanics orbit orbital-elements
edited Dec 28 '18 at 2:26
Peter Mortensen
1947
asked Dec 27 '18 at 17:03
sequoiasequoia
1416
If I calculate the semi-major axis of Molniya-1T with
$a = sqrt[3]{dfrac{GM}{n^2}}$ with $n=3.18683728text{ }d^{-1}$, I get another apoapsis ($a=19505.7$ km) as noted at Heavens Above:
apogee height: 25659 km
perigee height: 595 km
which is $a=frac{25659+595}{2}=13127$ km.
This is my approach in Octave/MATLAB:
# Computes the semi major axis a from n with constant GM
# @params:
# GM constant (cubic km per square second)
# n: mean motion (revs/day)
# @return:
# a: semi major axis (km)
function a = getSemiMajorAxis(GM,n)
n = n*2*pi/(24*60*60); #conversion (revs/day) --> (rad/s)
a = (GM/(power(n,2))).^(1./3);
endfunction
getSemiMajorAxis(398600.44,3.18683728) # Test with Molniya-1T
So, what is the origin of the deviation? Isn't apogee the same as apoapsis basically?
orbital-mechanics orbit orbital-elements
orbital-mechanics orbit orbital-elements
edited Dec 28 '18 at 2:26
Peter Mortensen
1947
asked Dec 27 '18 at 17:03
sequoiasequoia
1416
edited Dec 28 '18 at 2:26
Peter Mortensen
1947
asked Dec 27 '18 at 17:03
sequoiasequoia
1416
edited Dec 28 '18 at 2:26
Peter Mortensen
1947
edited Dec 28 '18 at 2:26
Peter Mortensen
1947
edited Dec 28 '18 at 2:26
Peter Mortensen
1947
1947
asked Dec 27 '18 at 17:03
sequoiasequoia
1416
asked Dec 27 '18 at 17:03
sequoiasequoia
1416
asked Dec 27 '18 at 17:03
sequoiasequoia
1416
1416
I saw your profile; here are some geoinformatics questions you might find interesting: Shape and dimensions of the Moon's reference surface for selenographic latitude/longitude? and How do we define geographical coordinates on non-spheroid celestial bodies?
– uhoh
Dec 28 '18 at 2:22
add a comment |
I saw your profile; here are some geoinformatics questions you might find interesting: Shape and dimensions of the Moon's reference surface for selenographic latitude/longitude? and How do we define geographical coordinates on non-spheroid celestial bodies?
– uhoh
Dec 28 '18 at 2:22
I saw your profile; here are some geoinformatics questions you might find interesting: Shape and dimensions of the Moon's reference surface for selenographic latitude/longitude? and How do we define geographical coordinates on non-spheroid celestial bodies?
– uhoh
Dec 28 '18 at 2:22
I saw your profile; here are some geoinformatics questions you might find interesting: Shape and dimensions of the Moon's reference surface for selenographic latitude/longitude? and How do we define geographical coordinates on non-spheroid celestial bodies?
– uhoh
Dec 28 '18 at 2:22
add a comment |
1 Answer
1
active
oldest
votes
Ok, that's embarrassing:
You just have to add earth's radius (traditionally the equatorial radius) of about 6378 or 6378.137 km to apogee or perigee heights to get distances to the center.
edited Dec 28 '18 at 0:50
uhoh
35.1k18123439
answered Dec 27 '18 at 17:48
sequoiasequoia
1416
3
An endless source of confusion. Get used to it :)
– SF.
Dec 27 '18 at 21:17
4
It's good that you wrote an answer, others may well have the same issue!
– Organic Marble
Dec 27 '18 at 21:50
The clock in San Dimas, is always running... you have to dial one number higher.
– Mazura
Dec 28 '18 at 7:23
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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votes
Ok, that's embarrassing:
You just have to add earth's radius (traditionally the equatorial radius) of about 6378 or 6378.137 km to apogee or perigee heights to get distances to the center.
edited Dec 28 '18 at 0:50
uhoh
35.1k18123439
answered Dec 27 '18 at 17:48
sequoiasequoia
1416
3
An endless source of confusion. Get used to it :)
– SF.
Dec 27 '18 at 21:17
4
It's good that you wrote an answer, others may well have the same issue!
– Organic Marble
Dec 27 '18 at 21:50
The clock in San Dimas, is always running... you have to dial one number higher.
– Mazura
Dec 28 '18 at 7:23
add a comment |
Ok, that's embarrassing:
You just have to add earth's radius (traditionally the equatorial radius) of about 6378 or 6378.137 km to apogee or perigee heights to get distances to the center.
edited Dec 28 '18 at 0:50
uhoh
35.1k18123439
answered Dec 27 '18 at 17:48
sequoiasequoia
1416
3
An endless source of confusion. Get used to it :)
– SF.
Dec 27 '18 at 21:17
4
It's good that you wrote an answer, others may well have the same issue!
– Organic Marble
Dec 27 '18 at 21:50
The clock in San Dimas, is always running... you have to dial one number higher.
– Mazura
Dec 28 '18 at 7:23
add a comment |
Ok, that's embarrassing:
You just have to add earth's radius (traditionally the equatorial radius) of about 6378 or 6378.137 km to apogee or perigee heights to get distances to the center.
edited Dec 28 '18 at 0:50
uhoh
35.1k18123439
answered Dec 27 '18 at 17:48
sequoiasequoia
1416
Ok, that's embarrassing:
You just have to add earth's radius (traditionally the equatorial radius) of about 6378 or 6378.137 km to apogee or perigee heights to get distances to the center.
edited Dec 28 '18 at 0:50
uhoh
35.1k18123439
answered Dec 27 '18 at 17:48
sequoiasequoia
1416
edited Dec 28 '18 at 0:50
uhoh
35.1k18123439
edited Dec 28 '18 at 0:50
uhoh
35.1k18123439
edited Dec 28 '18 at 0:50
uhoh
35.1k18123439
35.1k18123439
answered Dec 27 '18 at 17:48
sequoiasequoia
1416
answered Dec 27 '18 at 17:48
sequoiasequoia
1416
answered Dec 27 '18 at 17:48
sequoiasequoia
1416
1416
3
An endless source of confusion. Get used to it :)
– SF.
Dec 27 '18 at 21:17
4
It's good that you wrote an answer, others may well have the same issue!
– Organic Marble
Dec 27 '18 at 21:50
The clock in San Dimas, is always running... you have to dial one number higher.
– Mazura
Dec 28 '18 at 7:23
add a comment |
3
An endless source of confusion. Get used to it :)
– SF.
Dec 27 '18 at 21:17
4
It's good that you wrote an answer, others may well have the same issue!
– Organic Marble
Dec 27 '18 at 21:50
The clock in San Dimas, is always running... you have to dial one number higher.
– Mazura
Dec 28 '18 at 7:23
3
3
An endless source of confusion. Get used to it :)
– SF.
Dec 27 '18 at 21:17
An endless source of confusion. Get used to it :)
– SF.
Dec 27 '18 at 21:17
4
4
It's good that you wrote an answer, others may well have the same issue!
– Organic Marble
Dec 27 '18 at 21:50
It's good that you wrote an answer, others may well have the same issue!
– Organic Marble
Dec 27 '18 at 21:50
The clock in San Dimas, is always running... you have to dial one number higher.
– Mazura
Dec 28 '18 at 7:23
The clock in San Dimas, is always running... you have to dial one number higher.
– Mazura
Dec 28 '18 at 7:23
add a comment |
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I saw your profile; here are some geoinformatics questions you might find interesting: Shape and dimensions of the Moon's reference surface for selenographic latitude/longitude? and How do we define geographical coordinates on non-spheroid celestial bodies?
– uhoh
Dec 28 '18 at 2:22