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
asked 2 hours ago
sequoia
364
New contributor
sequoia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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
asked 2 hours ago
sequoia
364
New contributor
sequoia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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
asked 2 hours ago
sequoia
364
New contributor
sequoia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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
asked 2 hours ago
sequoia
364
New contributor
sequoia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
sequoia
364
New contributor
sequoia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
asked 2 hours ago
sequoia
364
New contributor
sequoia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
sequoia
364
asked 2 hours ago
sequoia
364
364
New contributor
sequoia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
sequoia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
sequoia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Ok, that's embarrassing:
You just have to add earth's radii of about 6378 to the a calculated from the apogee/perigee.
answered 1 hour ago
sequoia
364
New contributor
sequoia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "508"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
sequoia is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fspace.stackexchange.com%2fquestions%2f33126%2fdeviation-of-semi-major-axis%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Ok, that's embarrassing:
You just have to add earth's radii of about 6378 to the a calculated from the apogee/perigee.
answered 1 hour ago
sequoia
364
New contributor
sequoia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
Ok, that's embarrassing:
You just have to add earth's radii of about 6378 to the a calculated from the apogee/perigee.
answered 1 hour ago
sequoia
364
New contributor
sequoia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
Ok, that's embarrassing:
You just have to add earth's radii of about 6378 to the a calculated from the apogee/perigee.
answered 1 hour ago
sequoia
364
New contributor
sequoia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Ok, that's embarrassing:
You just have to add earth's radii of about 6378 to the a calculated from the apogee/perigee.
answered 1 hour ago
sequoia
364
New contributor
sequoia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
sequoia
364
New contributor
sequoia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
sequoia
364
answered 1 hour ago
sequoia
364
364
New contributor
sequoia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
sequoia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
sequoia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
sequoia is a new contributor. Be nice, and check out our Code of Conduct.
sequoia is a new contributor. Be nice, and check out our Code of Conduct.
sequoia is a new contributor. Be nice, and check out our Code of Conduct.
sequoia is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Space Exploration Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fspace.stackexchange.com%2fquestions%2f33126%2fdeviation-of-semi-major-axis%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
