Question

You are an astronaut in the space shuttle pursuing a satellite in need of repair. You are in a circular orbit of the same radius as the satellite(410 above the Earth),but 30 behind it.Part AHow long will it take to overtake the satellite if you reduce your orbital radius by 1.2 ?Express your answer using two significant figures.

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Part BBy how much must you reduce your orbital radius to catch up in 5.5 ?Express your answer using two significant figures.

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Answer 1

solved a similar question already with different numbers.. i put the question and the answer.. hope it helps!! :) kindly rate a lifesaver :)

You are an astronaut in the space shuttle pursuing a satellite in need of repair. You are in a circular orbit of the same radius as the satellite (430km above the Earth), but 23 km behind it.

(a) How long will it take to overtake the satellite if you reduce your orbital radius by 1.0 km?

(b) By how much must you reduce your orbital radius to catch up in 7.0 hours?

ANSWER:

v =velocity
μ = 398600 for Earth.
radius of the Earthis 6378.1 km
orbital radius = r = 6378.1 + altitude
θ = angle from starting position (we're going to say the shuttle starts at θ = 0°)
T = period of an orbit
t = time

everything having to do with the shuttle will have a 1 and everything having to do with the satellite will have a 2 (ex: T(1) would be the period ofthe shuttle's orbit and v(2) would be the velocity of the satellite)

ok, here we go.

a)
r(1) = 6378.1 + 429 = 6807.1 km
r(2) = 6378.1 + 430 = 6808.1 km

v = sqrt(μ/r) for acircular orbit
θ = 2π*t / T
T = 2π*r / v

ok, lets get the velocity of each orbit

v(1) = sqrt(398600/6807.1) = 7.6522224 km/s
v(2) = sqrt(398600/6808.1) = 7.6516604 km/s

From these, we can get the period of each orbit

T(1) = 2π*r(1) / v(1) = 2π*6807.1 / 7.6522224
T(1) = 5589.26 s

T(2) = 2π*6808.1 / 7.6516604
T(2) = 5590.49 s

From these, we can figure out what angle either of them are at after some time 't'
First we need the starting angle for the satellite.
Find the circumference of the satellite's orbit

C(2) = 2π * r(2) = 42776.55 km
Now, it starts 23 km ahead. Find this in degrees via ratio
23 / 42776.55 = θ(2) / 360
θ(2) = 0.19356°

ok, now we use θ = 2π*t / T
we know the periods of both orbits, as well as their radial starting locations. When the shuttle catches the satellite, the radial locations will beequal to eachother.

θ(1) = 2π*t / T(1)
θ(2) = 2π*t / T(2) + 0.19356°
um, mismatched units. Get degrees in radians.
0.19356° * π / 180 = 0.0033783 rad

θ(2) = 2π*t / T(2) + 0.0033783 rad

equate θ(2) and θ(1)
2π*t / T(1) = 2π*t / T(2) + 0.0033783

T(1) = 5589.26 s
T(2) = 5590.49 s

2π*t / 5589.26 = 2π*t / 5590.49 + 0.0033783

t = 13658.953 s = 3.794 h [ANSWER]

For the 7 hours, reverse engineer it. Use the final equation here. The difference will be that you know t, but T(1) is unknown. All the other numbersare the same. Solve for T(1).
With this, use { v = sqrt(μ/r) } &{ T = 2π*r / v } to get r with your new T(1).

Answer 2

v_sat=31577.92 m/s
v_shut=31621.43 m/s

v_shutrelative to satellite = 43.51 m/s

v=d/t => t=d/v

t = 20,000m / 43.51m/s = 459.66s = 0.13 hr

Answer 3

too tired for numbers: here are references.

http://quest.nasa.gov/space/teachers/lif…
http://www.braeunig.us/space/orbmech.htm
http://en.wikipedia.org/wiki/In-orbit_re…
Get serious
http://www.cdeagle.com/ommatlab/maneuver…