# Ratio of position error of size of shuttle

1. The problem statement, all variables and given/known data

The space shuttle tracking system predicts the position of the shuttle orbiter with an sccurscy that varies between 30 m and 100 m. Its orbital radius is slightly larger than the radius of the Earth (average orbit altitude is 340 km). What, approximately, is the ratio of the position error to the size of the shuttle’s orbit? The orbiter is approximately 37 m long and 17 m high. (REarth = 6378 km)

Given:
r = 340 + 6378 = 6718 km
length of orbiter = 37 m
height of orbiter = 17 m

Required:
ratio

2. Relevant equations
I’m not sure what relevant equations to use.

3. The attempt at a solution
I’m confused about how to go about this question. I’m not sure where to use the dimmensions of the orbiter or the range. I tried solving it, but have no idea if I’m doing this even remotely right.

I’m trying to look for the ratio of position error.
range of accuracy = 30 m → 100 m = 0.03 km → 0.1 km

+/- 0.03 km
rorbiter 1 = 6718.03
rorbiter 1 = 6717.97

______________________________
0.06 km

+/- 0.1 km
rorbiter 2 = 6718.1
rorbiter 2 = 6717.9

______________________________
0.2 km

0.2/ 0.06 = 3.333

Ratio is 3.3 : 1 ????

But I didn’t use the shuttle dimmensions!

Any help would be greatly appreciated! Thanks! :shy:
:confused:

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