If you reduce your orbital RADIUS, you will acquire a different small circular orbit and will never get to the satellite. You will pass between it and the earth.
This is a poorly conceived question, in my opinion.
What the astronaut needs to do is apply reverse thrust to acquire an elliptical orbit. Its semimajor diameter will be reduced, and it will return to meet the satellite on the next orbit.
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 (450 km above the Earth), but 24 km behind it.
How long will it take to overtake the satellite if you reduce your orbital radius by 1.4 km?
By how much must you reduce your orbital radius to catch up in 7.0 h?
3 answers
Assume, for discussion purposes, a satellite and the shuttle are in the same 91 minute circular orbit, the satellite 2 minutes ahead of the space shuttle.
The shuttle fires its on-orbit thrusters to reduce its orbital velocity. In doing so, the space shuttle drops into an elliptical orbit with a period 1 minute slower than the original circular orbit.
In 90 minutes, the space shuttle returns to the firing point of the elliptical orbit, now only 1 minute away from the satellite.
Remaining in the same elliptical orbit once more, the space shuttle ultimately catches up with the satellite. It now fires its thrusters to speed up the space shuttle to the same circular orbit velocity that it had when the catch up operation started.
As for your specific problem:
Assume an earth radius of 6378km
Orbital altitude = 450km
Orbital radius = radius is 6378 + 450 = 6828km or 6,828,000meters.
Earth’s gravitational constant GM = µ = 3.986365x10^14m^3/sec^2
The orbital velocity is Vc = sqrt(µ/r) = 7641m/s.
The orbital period T = 2(Pi)sqrt(a^3/µ) = 5615sec. = 93.58minutes.
…(a = semimajor axis = r for the circular orbit)
Reducing the velocity of the space places the space shuttle into an elliptical orbit with an apogee of 6826km and a new perigee of 6828 – 1.4 = 6826.6 km and a semi-major axis of 6827.3km.
The period of this new orbit becomes 5614sec. Placing the shuttle 1 second closer to the satellite.
The circular orbit perimeter is 6828(2)3.14 = 42,901.6km.
The satellite is 24(5615)/42,901.6 = 3.14sec. ahead of the space shuttle.
Therefore, after 3 of the new elliptical orbits are completed, the two vehicles are within .14sec. of one another which can be effectively accomplished with the reaction control system.
The shuttle fires its on-orbit thrusters to reduce its orbital velocity. In doing so, the space shuttle drops into an elliptical orbit with a period 1 minute slower than the original circular orbit.
In 90 minutes, the space shuttle returns to the firing point of the elliptical orbit, now only 1 minute away from the satellite.
Remaining in the same elliptical orbit once more, the space shuttle ultimately catches up with the satellite. It now fires its thrusters to speed up the space shuttle to the same circular orbit velocity that it had when the catch up operation started.
As for your specific problem:
Assume an earth radius of 6378km
Orbital altitude = 450km
Orbital radius = radius is 6378 + 450 = 6828km or 6,828,000meters.
Earth’s gravitational constant GM = µ = 3.986365x10^14m^3/sec^2
The orbital velocity is Vc = sqrt(µ/r) = 7641m/s.
The orbital period T = 2(Pi)sqrt(a^3/µ) = 5615sec. = 93.58minutes.
…(a = semimajor axis = r for the circular orbit)
Reducing the velocity of the space places the space shuttle into an elliptical orbit with an apogee of 6826km and a new perigee of 6828 – 1.4 = 6826.6 km and a semi-major axis of 6827.3km.
The period of this new orbit becomes 5614sec. Placing the shuttle 1 second closer to the satellite.
The circular orbit perimeter is 6828(2)3.14 = 42,901.6km.
The satellite is 24(5615)/42,901.6 = 3.14sec. ahead of the space shuttle.
Therefore, after 3 of the new elliptical orbits are completed, the two vehicles are within .14sec. of one another which can be effectively accomplished with the reaction control system.
Addendum to my earlier response.
A reduction in velocity of only 1.5m/sec. will drop the space shuttle into an ever so slightly elliptical orbit with an apogee of 6828km, a perigee of 6825.6km. and a period of 5611.86 sec. or 93.53min., allowing the shuttle to catch up with the satellite in one orbit.
A 7 hour closing time is somewhat impractical.
A reduction in velocity of only 1.5m/sec. will drop the space shuttle into an ever so slightly elliptical orbit with an apogee of 6828km, a perigee of 6825.6km. and a period of 5611.86 sec. or 93.53min., allowing the shuttle to catch up with the satellite in one orbit.
A 7 hour closing time is somewhat impractical.
0 answers