1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Perigee and apogee

  1. Jun 15, 2016 #1
    why spacecraft launches are usually terminated at either perigee or apogee?
  2. jcsd
  3. Jun 15, 2016 #2


    User Avatar
    Gold Member
    2016 Award

    Who says they are? I'm not even sure what you are talking about as it doesn't seem to make sense.
  4. Jun 16, 2016 #3


    User Avatar
    Science Advisor
    Gold Member

    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 :smile:

  5. Jun 16, 2016 #4


    User Avatar

    Staff: Mentor

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

    From here: http://www.braeunig.us/space/orbmech.htm

    Geosynchronous 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.

  6. Jun 16, 2016 #5

    Vanadium 50

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor

    I hope this thread doesn't turn into the guessing game "maybe the OP means..."
  7. Jun 16, 2016 #6
  8. Jun 16, 2016 #7
    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.
  9. Jun 16, 2016 #8


    User Avatar

    Staff: Mentor

    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.
  10. Jun 16, 2016 #9
    Yep I tried but I don't understand the programming language
  11. Jun 16, 2016 #10


    User Avatar
    Science Advisor

    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]
  12. Jun 16, 2016 #11

    Filip Larsen

    User Avatar
    Gold Member

  13. Jun 16, 2016 #12
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted