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Iranian Space program

ماهواره ملی فجر نخستین ماهواره بومی ایران است که مأموریت مانور مداری دارد. این ماهواره از زیرسامانه پیش‌رانش گاز سرد برای انتقال مداری استفاده می‌کند. علاوه بر سامانه پیش‌رانش گاز سرد ماهواره فجر از رانش‌زای پالس پلاسمایی که یک رانش‌زای الکتریکی با ضربه ویژه بسیار بالا است به‌عنوان محموله فرعی بهره‌گیری خواهد کرد. این رانش‌زا باعث دوران ماهواره حول مرکز جرم خود خواهد شد. از این ماهواره در سال 1389 رونمایی شد.

ماهواره فجر چگونه یک و نیم سال عمر می‌کند؟
ماهواره امید نخستین ماهواره بومی ایران که با یک ماهواره‌بر بومی به فضا پرتاب شد، حدود سه ماه در فضا ماند و پس از حدود ۱۲۰۰ چرخش دور زمین در این مدت، ارتفاع مداری‌اش به‌اندازه‌ای کاهش یافت که وارد جو شد و سوخت؛ اما ماهواره فجر که نسخه پیشرفته‌تر ماهواره امید است، از نظر زمانی می‌تواند پنج برابر بیشتر در فضا بماند. ارتفاع مداری اولیه هردو ماهواره امید و فجر تقریباً یکی است ولی روی ماهواره فجر رانش‌زا (موتور کوچک فضایی) نصب شده است. با هر چرخش ماهواره به دور زمین، اصطکاک آن با هوای رقیق موجود در آن ارتفاع باعث اتلاف انرژی جنبشی ماهواره شده و ارتفاع مداری آن را کاهش می‌دهد؛ اما هر بار ماهواره به نقطه اوج خود می‌رسد، رانش‌زای روی آن به حالت ضربه‌ای روشن شده و در فضای نیروی لازم برای افزایش ارتفاع مداری را فراهم می‌کند. بعد از چندین ضربه رانش‌زا در مدار، ماهواره فجر از مداری بیضوی به یک مدار دایروی با ارتفاع حدود ۴۷۰ کیلومتری از سطح زمین منتقل شده و بدین ترتیب تا حدود یک سال و نیم در مدار زمین باقی می‌ماند. این فرایند در چنین مأموریتی درواقع یک تجربه برای ساخت ماهواره‌های بعدی کشور جهت حضور در مدارهای بالاتر است.
موفقیت اولیه ماهواره فجر در ارتقای مدار


ISA agency itself says so....
rast migi ha:D

my bad bro , you're right :)
 
is it right that the apogee should go down to 450km for the thrusters to start ?
Depends on what they want the final circular radius to be. If their decision is to keep have a circular orbit 450 km above earth and only want to apply the thrust when the satellite is on the apogee, then yes, they need to wait until it reaches 450km.
 
Dude,

Ion thrusters could not even generate specific thrust to even move a satellite by a cm. This is a basic physical limitation, not of engineering one which you could overcome. Putting it on satellite is stupid.

This while assuming that Iran even has Ion thrusters, given that Iran has no interplanetary program; only place where they could be used.

You media is known for it's hyperbole. It is probably that only.


the ion thrusters cause the satellite rotate around its center of mass to stabilize its position.
 
2 Feb:

NORAD ID: 40387
Int'l Code: 2015-006A
Perigee: 231.2 km
Apogee: 476.7 km
Inclination: 55.5 °
Period: 91.5 minutes
Semi major axis: 6724 km

5 Feb:

NORAD ID: 40387
Int'l Code: 2015-006A
Perigee: 229.6 km
Apogee: 463.2 km
Inclination: 55.5 °
Period: 91.3 minutes
Semi major axis: 6717 km

6 Feb:

NORAD ID: 40387
Int'l Code: 2015-006A
Perigee: 229.4 km
Apogee: 453.1 km
Inclination: 55.5 °
Period: 91.2 minutes
Semi major axis: 6712 km
 
2 Feb:

NORAD ID: 40387
Int'l Code: 2015-006A
Perigee: 231.2 km
Apogee: 476.7 km
Inclination: 55.5 °
Period: 91.5 minutes
Semi major axis: 6724 km

5 Feb:

NORAD ID: 40387
Int'l Code: 2015-006A
Perigee: 229.6 km
Apogee: 463.2 km
Inclination: 55.5 °
Period: 91.3 minutes
Semi major axis: 6717 km

6 Feb:

NORAD ID: 40387
Int'l Code: 2015-006A
Perigee: 229.4 km
Apogee: 453.1 km
Inclination: 55.5 °
Period: 91.2 minutes
Semi major axis: 6712 km

:D
 
2 Feb:

NORAD ID: 40387
Int'l Code: 2015-006A
Perigee: 231.2 km
Apogee: 476.7 km
Inclination: 55.5 °
Period: 91.5 minutes
Semi major axis: 6724 km

5 Feb:

NORAD ID: 40387
Int'l Code: 2015-006A
Perigee: 229.6 km
Apogee: 463.2 km
Inclination: 55.5 °
Period: 91.3 minutes
Semi major axis: 6717 km

6 Feb:

NORAD ID: 40387
Int'l Code: 2015-006A
Perigee: 229.4 km
Apogee: 453.1 km
Inclination: 55.5 °
Period: 91.2 minutes
Semi major axis: 6712 km
:cray:cray beyatch cray

Lol
 
image courtesy of north korea:

pict1.jpg
 
    • Elliptical Orbits: most orbits are not perfectly circular. All orbits are ellipses (flattened circles) with a high point (apogee) and a low point (perigee).
      • At apogee, when the satellite is farthest from the earth, it is going the slowest - it's ready to fall back toward the earth.
      • As the satellite falls it gains speed, and "overshoots" the earth, swinging quickly through perigee, then gaining altitude back toward apogee.
      • The satellite doesn't stay in orbit at the apogee distance because it isn't going fast enough when it reaches that point. It doesn't stay in orbit at the perigee distance because it's picked up so much speed by that point that it starts climbing again.
    orbit1.gif
    • Transfer Orbit:
      • If we speed the satellite up while it's in low circular earth orbit it will go into elliptical orbit, heading up to apogee.
      • If we do nothing else, it will stay in this elliptical orbit, going from apogee to perigee and back again.
      • BUT, if we fire a rocket motor when the satellite's at apogee, and speed it up to the required circular orbit speed, it will stay at that altitude in circular orbit. Firing a rocket motor at apogee is called "apogee kick", and the motor is called the "apogee kick motor".
    orbit2.gif

Position in an Elliptical Orbit

Johannes Kepler was able to solve the problem of relating position in an orbit to the elapsed time, t-to, or conversely, how long it takes to go from one point in an orbit to another. To solve this, Kepler introduced the quantity M, called the mean anomaly, which is the fraction of an orbit period that has elapsed since perigee. The mean anomaly equals the true anomaly for a circular orbit. By definition,

eq4-38.gif


where Mo is the mean anomaly at time to and n is the mean motion, or the average angular velocity, determined from the semi-major axis of the orbit as follows:

eq4-39.gif


This solution will give the average position and velocity, but satellite orbits are elliptical with a radius constantly varying in orbit. Because the satellite's velocity depends on this varying radius, it changes as well. To resolve this problem we can define an intermediate variable E, called the eccentric anomaly, for elliptical orbits, which is given by

eq4-40.gif


where
nu.gif
is the true anomaly. Mean anomaly is a function of eccentric anomaly by the formula

eq4-41.gif


For small eccentricities a good approximation of true anomaly can be obtained by the following formula (the error is of the order e3):

eq4-42.gif


The preceding five equations can be used to (1) find the time it takes to go from one position in an orbit to another, or (2) find the position in an orbit after a specific period of time. When solving these equations it is important to work in radians rather than degrees, where 2
pi.gif
radians equals 360 degrees.
 
Dont cry, joon.

2 Feb:

NORAD ID: 40387
Int'l Code: 2015-006A
Perigee: 231.2 km
Apogee: 476.7 km
Inclination: 55.5 °
Period: 91.5 minutes
Semi major axis: 6724 km

5 Feb:

NORAD ID: 40387
Int'l Code: 2015-006A
Perigee: 229.6 km
Apogee: 463.2 km
Inclination: 55.5 °
Period: 91.3 minutes
Semi major axis: 6717 km

6 Feb morning:

NORAD ID: 40387
Int'l Code: 2015-006A
Perigee: 229.4 km
Apogee: 453.1 km
Inclination: 55.5 °
Period: 91.2 minutes
Semi major axis: 6712 km

6 Feb evening:

NORAD ID: 40387
Int'l Code: 2015-006A
Perigee: 228.8 km
Apogee: 447.3 km
Inclination: 55.5 °
Period: 91.1 minutes
Semi major axis: 6709 km

:smitten:
 
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