Spacecraft Launches: Perigee & Apogee - Why Terminate?

[jose medina pedraza]

why spacecraft launches are usually terminated at either perigee or apogee?

 

[phinds]

why spacecraft launches are usually terminated at either perigee or apogee?

Who says they are? I'm not even sure what you are talking about as it doesn't seem to make sense.

 

[davenn]

why spacecraft launches are usually terminated at either perigee or apogee?

As Phinds said ... that didn't make a lot of sense

do you understand what the definitions perigee or apogee are ?
Describe what you think they are and we can work from there D

 

[Drakkith]

Perhaps the OP is referring to the practice of launching satellites into Geosynch.

From here: http://www.braeunig.us/space/orbmech.htmGeosynchronous orbits (GEO) are circular orbits around the Earth having a period of 24 hours. A geosynchronous orbit with an inclination of zero degrees is called a geostationary orbit. A spacecraft in a geostationary orbit appears to hang motionless above one position on the Earth's equator. For this reason, they are ideal for some types of communication and meteorological satellites. A spacecraft in an inclined geosynchronous orbit will appear to follow a regular figure-8 pattern in the sky once every orbit. To attain geosynchronous orbit, a spacecraft is first launched into an elliptical orbit with an apogee of 35,786 km (22,236 miles) called a geosynchronous transfer orbit (GTO). The orbit is then circularized by firing the spacecraft 's engine at apogee.

 

[Vanadium 50]

Perhaps the OP is referring

I hope this thread doesn't turn into the guessing game "maybe the OP means..."

 

[jose medina pedraza]

How do I calculate the radius of orbit from the period of orbit
http://hyperphysics.phy-astr.gsu.edu/hbase/orbv3.html#ov

 

[jose medina pedraza]

Yes Drakkith, that's what I meant, but which formulas do I use to calculate the radius 35,786 km (22,236 miles), by knowing that its period is 24 H.

 

[Drakkith]

I can't help you there at the moment. I'd have to dig around to see if it's just basic mechanics stuff, or if it's something more complicated.

 

[jose medina pedraza]

Yep I tried but I don't understand the programming language

 

[jbriggs444]

Yes Drakkith, that's what I meant, but which formulas do I use to calculate the radius 35,786 km (22,236 miles), by knowing that its period is 24 H.

The radius of the orbit is not 22,236 miles. Take a trip to Google and look up what reference point that 22,236 figure is measured from.

Now, to determine the radius of the orbit from the period, you need to know two formulas:

The centripetal force required to keep a body in circular motion: $f=\frac{mv^2}{r}$ where m is the mass of the object, v is its velocity and r is its distance from the center of the earth.

The force of gravity: $f=\frac{GmM}{r^2}$ where G is Newton's universal gravitational constant, M is the mass of the Earth, m is the mass of the object and r is the distance of the object from the center of the earth.

Solve these two equations for the radius at which gravity is exactly equal to the required centripetal force.
[Hint: the orbital velocity is easily found from the orbital radius and period.]
[Hint: pick a consistent set of units (e.g. meters, kilograms, seconds) and use that]
[Hint: keep everything symbolic until you have a formula for the quantity you are after in terms of things that you know]

 

[Filip Larsen]

Perhaps reading about orbital period [1] and geo-synchronous orbits [2] may help.

[1] https://en.wikipedia.org/wiki/Orbital_period
[2] https://en.wikipedia.org/wiki/Geostationary_orbit

 

[jose medina pedraza]

Thanks