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2. The shuttle problem.
Solution:
Consider these two equations:
1) Pa = 2πa3/2 / (GM)½
2) v = [(2GM/r) - (GM/a)] ½
A rough approximation for the time follows this logic. Imagine two satellites following highly eccentric elliptical orbits about the Earth with semi-major axis' of A and a. The orbital period in equation 1 is independent of the eccentricity of the orbit. In other words, even if the ellipse is almost completely flat the equation is still only dependent on the length of the semi-major axis.
In this case 2A = R and 2a = r (roughly). A and a are the lengths of the semi-major axis' of the two ellipses.
Satellite a has roughly 0 velocity at distance r and satellite A covers the total distance traveled by satellite a at point r. But since it has an initial velocity, the amount of time required for A to travel satellite a's orbit is roughly:
Time = Pa2r / (vPa + 2r)
If we subtract that time from the period of A's orbit and divide by 2 we roughly have the time required to travel from R to r. Simplifying the final equation yields:
t = A - B
where
A = πR3/2
÷ (8GM)½
B = πr3/2
4(R)½ ÷ (8GM)½[π(R - r) ½ +4(R)½]
Finally the time T it will take for the shuttle to travel the distance R - r with initial velocity V' is equal to (when plugging in the value for t):
T = (R - r) / [V' + (R-r)/t]
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