2006-09-29 13:32:27 · 11 answers · asked by bill n 2 in Arts & Humanities ➔ Other - Arts & Humanities
if you were ON the shuttle, it'd travel as far as the wall...
if you were out on your little tether dealie, it'd keep traveling until it met with an obect large enough to change it's path or stop it all together.
arrows to aerospace, baby...
2006-09-29 13:34:52 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Depends on which way you shot it. It would likely go into orbit along with the shuttle.
2006-09-29 20:40:34 · answer #2 · answered by Paul P 5 · 0⤊ 0⤋
Unless you could launch the arrow at earth's escape velocity, it would continue in orbit until the orbit decays. It would travel quite a distance in orbit but I can't do the physics.
2006-09-29 20:42:43 · answer #3 · answered by williegod 6 · 0⤊ 0⤋
Do you mean a space shuttle, or the shuttle from the airport?
2006-09-29 20:45:43 · answer #4 · answered by The Gadfly 5 · 2⤊ 0⤋
It would go on until it hit something, or until it was attracted by the magnetic pull of a planet, i.e. Earth and would then burn up in the atmosphere. Along with , one hopes all the other discarded rubbish we have put up there.
2006-09-29 20:47:13 · answer #5 · answered by Social Science Lady 7 · 0⤊ 0⤋
Depends on the strength of the bow you used.
With a 50 lb. compound bow and aluminum shaft arrows it would go really, really far.
2006-09-29 20:35:48 · answer #6 · answered by Anonymous · 0⤊ 0⤋
I believe it will travel forever unless gravity from a planet or it crashes into something makes it stop or change course.
2006-09-29 20:40:19 · answer #7 · answered by leckscheid 3 · 1⤊ 0⤋
if nuthing stops it it will keep going and going thats called inurtia didnt ur teacher teach u that beacuse im purty sure ur biger than me and im in 5 th grade
2006-09-29 23:50:03 · answer #8 · answered by samantha o 1 · 0⤊ 0⤋
I don't know but I am sure you can slowly see it float away
2006-09-29 20:40:23 · answer #9 · answered by Bullz_ eye 6 · 0⤊ 0⤋
hmmm who knows
2006-09-29 20:34:17 · answer #10 · answered by cherdwum4 3 · 0⤊ 0⤋