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 (420 km above the Earth), but 24 km behind it.

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

___________ hr

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

Answer 1

(a) Use the expression –

V = √ (GM/r) = √ (GM) /√ r

Here -

r1 = 6378000 + 420000 = 6798000 m
r2 = r1 - 1000 = 6798000 - 1000 = 6797000 m

√ r1 = 2607.30
√ r2 = 2607.11
now -
G = 6.67 x 10^ -11
M = 5.97 × 10^24
√GM = 19,954,924.2
V1 = 19,954,924.2 / 2607.30 = 7653.48 m/s
V2 = 19,954,924.2 / 2607.11 = 7654.04 m/s
∆V = V2 - V1 = 7654.04 – 7653.48 = 0.56 m/s (increase)

Now we have -

d = vt
v = d/t
t = d/v

t = 24000/0.56 = 42857 s = 11.90 Hrs

(b) Now as per the above calculations, it tooks 11.90 hrs. to catch up with orbital radius reduced by 24000 m
What orbital radius reduction to catch up in 7.0 hrs?
We want 7.0 / 11.90 time = 0.588
Velocity must increase by 1/0.588 = 1.70 times
V3 increase = 1.70 x 0.56 m/s = 0.952 m/s
V3 = 7653.48 + 0.952 = 7654.432 m/s
V = √ (GM/r) = √ (GM) /√ r
7654.432 m/s = 19,954,924.2 / √ r

=> √ r3 = 19,954,924.2 / 7654.432 = 2606.98

=> r3 = 6796326 m

therefore -

∆r = 6798000 m - 6796326 = 1674 m

so the orbital radius must be reduced by 1674 m.