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⇤ ← Revision 1 as of 2018-05-07 07:42:21
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| The stopping distance is $ 2 L = {v_0/2}^2 / ( 2 a_v ) $ so $ L = {v_0}^2 / ( 16 a_v ) = {v_0}^2 / ( 80 a ) = $ 80 km. The total length and stopping distance is $ 2 L $ or 160 km. We will arbitrarily assign the datum at the center, so the entry point at t = 0 is x = -L $. The velocity of the vehicle is $ v = 4000 + 50 t $ and the acceleration to 8000 m/s takes 80 seconds. The distance is the integral of this velocity added to -80000m, so $ d = -80000 + 4000 t + 25 t^2 $ | The stopping distance is $ 2 L = {v_0/2}^2 / ( 2 a_v ) $ so $ L = {v_0}^2 / ( 16 a_v ) = {v_0}^2 / ( 80 a ) = $ 80 km. The total length and stopping distance is $ 2 L $ or 160 km. We will arbitrarily assign the datum at the center, so the entry point at t = 0 is $ x = -L $. The velocity of the vehicle is $ v = 4000 + 50 t $ and the acceleration to 8000 m/s takes 80 seconds. The distance is the integral of this velocity added to -80000m, so $ d = -80000 + 4000 t + 25 t^2 $. |
SpaceportTorque
Roger Arnold's Spaceport captures incoming vehicles at perigee and accelerates them (while it decelerates). The vehicles arrive at a fraction of Spaceport orbital velocity and speed up to orbital velocity as the Spaceport slows down. The spaceport is moving slightly faster than circular orbital perigee in the beginning, and thus has a small net-upward acceleration, which we will ignore for now.
For ease of calculation, we will assume the spaceport orbits at v_0 = 8000 m/s, the vehicle arrives at 4000 m/s, and gee at altitude is a = 10 m/s² . The vehicle acceleration a_v within the spaceport is assumed to be 5 gees or 50 m/s², or 5 a. The spaceport is assumed symmetrical around its center of mass, extending a distance -L towards the inlet and +L towards the terminus. The vehicle mass is M and the spaceport mass is 2 m L >> M , with mass evenly distributed along its length.
The stopping distance is 2 L = {v_0/2}^2 / ( 2 a_v ) so L = {v_0}^2 / ( 16 a_v ) = {v_0}^2 / ( 80 a ) = 80 km. The total length and stopping distance is 2 L or 160 km. We will arbitrarily assign the datum at the center, so the entry point at t = 0 is x = -L . The velocity of the vehicle is v = 4000 + 50 t and the acceleration to 8000 m/s takes 80 seconds. The distance is the integral of this velocity added to -80000m, so d = -80000 + 4000 t + 25 t^2 .