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Newly Discovered Asteroid To Pass Within Geostationary Orbit Sunday
theshowmecanuck writes: A newly found asteroid the size of a house will give earth a close flyby this weekend. It will pass just below satellites in geostationary orbit, and above New Zealand around 14:18 EDT / 18:18 GMT / 06:18 NZST this coming Sunday (Monday morning in NZ). "Asteroid 2014 RC was initially discovered on the night of August 31 by the Catalina Sky Survey near Tucson, Arizona, and independently detected the next night by the Pan-STARRS 1 telescope, located on the summit of Haleakal on Maui, Hawaii," NASA officials said in a statement.
You can expect a magnitude of +11.5 according to some sources. So no, definitely not visible to the naked eye. Should be easy with a good motorised telescope.
I'll worry about more likely concerns for local-scale damage.
Like say, a tornado. Today. That's more likely than a house sized asteroid hitting anywhere in my region in my lifetime.
Asteroids are primarily a concern due to the civilization terminating potential. And intrasystem asteroids the size of houses don't pose that threat.
From TFA
At its close approach, the 60-foot (20 meters) asteroid will fly about 25,000 miles (40,000 km) from the center of Earth. The average radius of the Earth (the distance from the center of the planet to its surface) is about 3,959 miles (6,371 km).
Geostationary orbit is ~42,164 km from the center of Earth, so TFS is correct based on this info.
An asteroid the size of a house would have to be going extraordinarily fast to pose much of a threat to the planet as a whole.
It's about the same size as the Chelyanbinsk meteor:
http://en.wikipedia.org/wiki/C...
Which hit the earth with the force of about 500kilitons of TNT
Here's some video footage in case you're not terrified yet:
https://www.youtube.com/watch?...
From what I'm reading, this asteroid is going even faster, but it's hard to tell how fast it will be going if it actually hit us.
Your estimates of the impact are way off for the scale of this rock, which is only 15-25 m in diameter. Even if it was quite dense rock and managed to hit at 90 degrees, it would still mostly break up in the air and you would get a spray of fragments over a couple hundred meters not strong enough to create any large crater. Even the 90 degree case in both shallow and deep water will not create tsunami more than a meter high.
The total kinetic energy of the thing in space is a couple of megatons, a lot of which is lost upon hitting the atmosphere before even breaking up. You're not going to get devastation orders of magnitude larger than a large nuclear weapon under worst case scenario. And if it comes in at something less than a 90 degree angle, you could end up with something like the Chelyabinsk meteor, since this is nearly the same size and a bit faster.
to answer GP (who I assume is an AC): geostationary is by no means arbitrary.
A geostationary orbit is one in which the orbiting body does not move relative to a point on the surface of its parent (in the context case, specifically Earth). This requires a specific orbital distance (22,236 miles*) at a specific inclination (0 to the equator), to maintain a sidereal orbital period of 23 hours 56 minutes 4 seconds (approximately). which is equal to the sidereal rotation period of Earth - how about that? In a two-body problem this would be simple, but we have this thing called the Moon, and this thing called the Sun, and to a lesser extent every other body with mass in the Universe, to deal with in maintaining a geostationary orbit. NBody physics introduces a certain degree of chaos to orbital predictions.
*this number is known by calculation using: cube root mu over omega squared. Refer to the Wikipedia.
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