BA in JPL press release on Galileo?

[kucharek]

http://galileo.jpl.nasa.gov/news/rel...ess030917.html

Its entry point into the giant planet's atmosphere is about 1/4 degree south of Jupiter's equator. If there were observers floating along at the cloud tops, they would see Galileo streaming in from a point about 22 degrees above the local horizon.

Where is this "22 degrees" coming from?

[ToSeek]

I interpret it as meaning that Galileo is coming in at a 22-degree angle with respect to the "surface" of Jupiter, if "surface" is taken as the top of the visible atmosphere.

[RichField]

I'm not sure they actually provide enough information to independently come up with the 22 degree figure, but it's not too difficult to show that the angular separation from the equator will not equal the observer's angular separation from the zenith.

I started with a circle drawn on a sheet of paper and a concentric arc outside the circle. The circle represents the planet, the equator is vertical and there is a little stick figure person standing on the edge. A horizontal line defines his local horizon. The arc represents the boundary of the atmosphere.

Next, I drew a line, an angle theta off the equator from the center of the planet. The point where it crosses the arc represents the satellite's impact point with the atmosphere. Then a line from the observer to the impact point is drawn. This is at an angle Beta to the equator. The angle phi is the angle of the impact point up from the horizon and is easily shown to be 90-Beta. On a paper sized sketch it is easy to see that theta is not equal to Beta and that Beta is in fact a larger angle. A horizontal line from the impact point to the equator line forms all the triangles we should need. That horizontal distance is h, the radius of the planet is r_p and the thickness of the atmosphere is d_A.

Assume r_p = 142900 km
d_A = 31 km (Earth's atmosphere, we'll change this later)
theta = 0.25 degrees

Find: phi - angle above the horizon.

phi = 90-Beta

h/d_A = tan(Beta)

h/(d_A+r_p) = tan(theta)

solving for phi gives:
phi = 90 - arctan((1+r_p/d_A)*tan(theta)
phi = 2.85 degrees

That's actually really low on the horizon, so to get up to 22 degrees you'd need to change the atmospheric thickness values. You can do this by solving the above equations for d_A, I've done it both as presented and with the modified assumption that d_A + r_p = 142900, it doesn't really make that much of a difference. For the latter case, it comes out to a person standing 142650 km from the center of the planet with 250 km of atmosphere overhead. Galileo will then be seen to enter at 22 degrees above the horizon.

Things neglected:
Atmospheric transparency
All objects are in a plane (third dimension ignored)

[kilopi]

And what if they were "standing" right under it? Then the angle above the horizon would be 90 degrees, right?

[kucharek]

I know think, they want to say that if you stand at the point where it will make the entry, it will come in at an elevation of 22 degrees. Just to express that is not simply dropping down vertically. IIRC, the Apollo spacecraft had a re-entry angle of 6.5 degrees when coming from the moon.

[Tobin Dax]

I agree with kucharek. It sounds like they're saying the angle of impact will be 22 deg up from the surface (or tangent to the surface at the point of impact). No, they don't tell us enough to figure this out for ourselves, but press releases/articles hardly ever do (a pet peeve of mine). And while kilopi is technically correct, you wouldn't be under it for very long and couldn't be under it if you were at the "crash site."