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Orbit Impact Angle

by guss
Tags: angle, impact, orbit
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guss
#1
Feb8-13, 03:04 PM
P: 248
Let's say a craft is in an orbit at a height h above a planet or moon. A force acts directly against the craft's direction of motion so as the force vector is always parallel to the surface of the central body. So, the craft will slowly decelerate and impact the surface. What would be the best way to compute the velocity vector upon impact? I don't have much experience at all with orbital trajectories, and it seems like an interesting question with simple parameters.
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mfb
#2
Feb8-13, 05:10 PM
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P: 11,575
If the force acts for a short time: Calculate the position+velocity afterwards, derive parameters of a new Kepler orbit, evaluate this at the radius of the planet/moon.
If the force is significant and acts in a complicated way for a longer time: Numerical simulation?
If the force is like air drag and small everywhere and the initial orbit is roughly circular: Assume a nearly circular orbit at impact, calculate the corresponding speed.
guss
#3
Feb8-13, 10:45 PM
P: 248
I wanted a more exact result for that third option, and I couldn't derive a formula, so I did a simulation. It was actually much easier than I thought it was gonna be, pretty cool as well. I got an impact speed of .3116 m/s in the straight down direction. Craft mass = 1.3kg (cubesat), F = .0001 N in the direction opposite of velocity, starting from 50km orbit. Very strange to thing about an object coming in for landing this way, nearly in orbit extremely close to the surface.

mfb
#4
Feb9-13, 04:29 AM
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P: 11,575
Orbit Impact Angle

In 50km, there is nothing which I would call orbit, there is too much air drag.
guss
#5
Feb9-13, 08:05 PM
P: 248
Quote Quote by mfb View Post
In 50km, there is nothing which I would call orbit, there is too much air drag.
The moon is a vacuum for all practical purposes, especially at 50km altitude.
mfb
#6
Feb10-13, 07:10 AM
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P: 11,575
But how do you get air drag there? Or do you provide that force via thrusters?


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