Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Earth and the Itokawa asteroid

  1. Oct 4, 2009 #1
    Next year the Hyabusa spacecraft will return to Earth from the asteroid Itokawa. Inspired by this, you are required to report on your individual study of the following:

    The free trajectory options, associated Δv requirements and mission times to take a spacecraft from a low Earth orbit (LEO) to a low orbit around Itokawa, if Itokawa is at:

    a. perihelion, (147,098,074 Km)
    b. apohelion. (152,097,701 Km)
    What are the pros and cons of each mission?

    LEO is at 8378 Km, How ever I have no idea how to caluclate the oribit of Itokawa, to get the distance between it and LEO. Also having problem geting the change in v.

    About Itokawa : http://en.wikipedia.org/wiki/25143_Itokawa

    Hope some one can help. Thanks in advance.
     
  2. jcsd
  3. Oct 6, 2009 #2
    Can anyone help ? Still cant figure it out. Thank you.
     
  4. Oct 6, 2009 #3
    Free trajectory involves using the mass of the destination to "bend" your flightpath so that you return to your origin without using any fuel for course corrections:
    http://en.wikipedia.org/wiki/Free_return_trajectory
    Your link gives the escape velocity of Itokawa as approximately .2m/s. I'm guessing that if your delta v were near that (and your trajectory correct), then your spacecraft would do a half orbit and then return. You would still need to know where the Earth is when Itokawa is at points a. and b. to plot a timeline.
     
  5. Oct 6, 2009 #4

    tony873004

    User Avatar
    Science Advisor
    Gold Member

    Normally, you're supposed to put homework questions in the homework help section ( https://www.physicsforums.com/forumdisplay.php?f=152 ). But you're new so I'll bend a little.

    I'm not sure what he means by free trajectory. I was guessing what Arch2008 said, except your question wants it to orbit the asteroid, rather than loop around and come home.

    Compute the amount of delta V to get from the orbit of Earth (1 AU) to the the distance of the asteriod at perihelion (147,098.074 km). This is interior to Earth, so you're making Earth the aphelion of this transfer orbit. Formulas 4.18 and 4.19 on BobB's page at http://www.braeunig.us/space/ tell you how to get this velocity. Compute the difference between this velocity and Earth's circular velocity. That's how much delta V you'll need to get to the asteroid at perihelion. You must escape the Earth with enough deltaV that your excess velocity (aka velocity at infinity) equals this amount. Escape velocity with hyperbolic excess is sqrt((2*GM/r)+vinfinity^2). Once you get to the asteroid, compute your encounter velocity (you can probably safely ignore the asteroid's weak gravity). How much higher is this than the velocity for circular orbit around the asteroid? (look up the asteroid's mass, and radius and use the circular velocity formula). You'll need that much deltaV to brake into orbit. As for the time, compute the semi-major axis of your transfer orbit. From that you can compute the period. You're doing half an orbit to get there, so divide this by 2.

    Repeat above for the asteroid's aphelion. As for pros and cons, one way takes more fuel (deltaV) than the other. One is also faster than the other. So draw your own conclusions after you've parts a and b.
     
  6. Oct 9, 2009 #5
    hahahah.... it's due monday so you better get a move on.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Earth and the Itokawa asteroid
  1. Asteroid hitting earth (Replies: 5)

Loading...