View Full Version : A question about objects in orbit

Astronot

2003-Jun-26, 06:14 PM

If I were working outside the International Space Station and threw a ball toward space, precisely opposite the pull of gravity at say 5 MPH what would happen to the ball? Lets ignore the drag from the thin atmosphere that slows the station down requiring a reboost. I assume it would just go into a higher orbit but how much higher would that orbit be? How about if I threw it straight down at the earth at 5 mph, would that be enough to cause it to fall until it burned up? What about forward or backwards?

Glom

2003-Jun-26, 06:55 PM

Well, let's assume a circular orbit or radius 6780000m. At the point of throwage, the station occupies the point with position vector 6780000Im. It's velocity is in the fundamental plane, of the coordinate system, with vector, 7670Jm/s, circular velocity at that point.

You impart a velocity of 2.22Im/s, so now, the new velocity is (2.22I + 7670J)m/s. with r and v, we can now describe the orbit fully.

First, the fundamental vectors.

Specific angular momentum, h = r ×v = 5.2×10^10Km²/s.

Ascending node vector will be zero because we've defined the fundamental plane to be coincident with the plane of the orbit, although the orbit of ISS is about 30° to the equatorial plane.

Eccentricity vector, e = 1/µ ((v²-µ/r)r + (r.v)v) = 6.49×10^-4I + 2.90×10-4J

So, eccentricity of the orbit is 7.1×10-4, or not a lot.

Next, we can use the equation for specific mechanical energy to calculate the semi-major axis:

E=½v²-µ/r=-µ/2a => a=6784404m, or not a change as far as the accuracy I've been using is concerned.

By using the dot product of r and e, I work out that the ball is 24° ahead of perigee.

That's pretty much how it works.

The point is that throwing the ball is like doing a small orbital burn. It slightly changes the orbit. If you look at the eccentricity, you'll see that it didn't increase much so the ball will remain in an ever so slightly elliptical orbit if you throw it out. Looking at the negligable change to the semi-major axis, throwing it downwards will also produce a negligable change and will not send it towards Earth, not with that speed. Similarly, throwing it forwards or backwards will only slightly change the orbit.

ToSeek

2003-Jun-26, 08:41 PM

What about forward or backwards?

If you are at apogee or perigee and throw the ball forwards or backwards, the ball will have an orbit with a higher or lower (respectively) opposite position than the space station you threw it from (e.g., if at perigee, you will affect apogee, and vice versa). This is a very common maneuver.

To look at what Glom is saying qualitatively, understand that you're already traveling at 18,000 miles per hour. If the ball is thrown upwards, then the ball is now traveling at something like 18,001 miles per hour on a slightly higher vector than the space station (or a lower vector if you throw it downwards). So it's not going to have much of an effect, just a slightly different orbit.

[Edited to correct spelling]

Peter B

2003-Jun-27, 04:37 AM

Astronot

Glom and ToSeek have given you specific answers to your question, but I thought I might add a general rule to help explain things.

If we ignore atmospheres and other miscellaneous effects on orbits, orbits are stable. As a general rule, orbits are elliptical, but one special type of ellipse is the circle. That’s Kepler’s First Law. His Second Law is that orbits trace out equal areas in equal times. While that mightn’t make much sense at first, what it means in practice is that when an orbit is an ellipse, the satellite travels fastest when it’s closest to the planet (or whatever).

Imagine the ISS is in a circular orbit. That means it doesn’t alter its height above the Earth, and its speed doesn’t change either. You’re floating outside the ISS, and chuck a cricket ball (or, if you must, a baseball!) in some direction. The ball is going to enter an elliptical orbit. Depending on the direction you chuck the ball, the point in space where you chuck the ball could be the high point of the orbit (apogee) or the low point (perigee). If you chuck the ball forwards, that point will be the perigee, and the ball will climb to a high point on the opposite side of the Earth. If you chuck the ball backwards, that point will be the apogee, and the ball will descend to a low point on the opposite side of the Earth.

Now that you have the ball in an elliptical orbit, it’ll remain in that orbit until you apply another force. If you want to circularise the ball’s orbit, you’ll need to apply that force on the opposite side of the Earth to where you boosted it the first time.

This is the same process used to get satellites into geosynchronous orbit. This orbit is about 35000 kilometres above the Earth (someone’s bound to provide the exact figure). Firstly, the satellite is launched into a circular low Earth orbit (say, 400 kilometres above the Earth). Secondly, it’s boosted into an elliptical orbit with an apogee of 35000 kilometres and a perigee of 400 kilometres. Thirdly, at apogee, it gets another boost which circularises the orbit at 35000 kilometres.

Astronot

2003-Jun-27, 01:11 PM

Thanks to all for the answers. There is something about orbits and angular momentum that I find rather difficult. One of the reasons I like this board is because of the talented people here that contribute to increase the level of knowledge of those of us that are in different fields.

ToSeek

2003-Jun-27, 02:19 PM

I was watching a program last night about the ill-fated Columbia mission, and it was pointed out that to re-enter the atmosphere the shuttle only needs to be slowed by a few hundred miles per hour - apparently that's enough to put perigee sufficiently inside the atmosphere for drag to take over.

daver

2003-Jun-27, 06:34 PM

I was watching a program last night about the ill-fated Columbia mission, and it was pointed out that to re-enter the atmosphere the shuttle only needs to be slowed by a few hundred miles per hour - apparently that's enough to put perigee sufficiently inside the atmosphere for drag to take over.

Reentry burns tend to be on the order of 250 feet/second, which is about the speed of a good arrow. So you might not be able to hit the earth with a baseball from the shuttle, but you could with an arrow.

Irishman

2003-Jun-29, 10:20 PM

Remember: out takes you up, up takes you back, back takes you down, and down takes you out. Orbital mechanics - you gotta love it.

Irishman

2003-Jun-29, 10:23 PM

That probably makes more sense if you replace "out" with "forward". I think I scrambled something. (I think "out" is supposed to mean "up", not forward, which screws up what I said.)

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