Impact Velocity of a bus-sized asteroid....

Ke = mv^2

Where:

Ke = Kinetic Energy

m = The mass of the object

v = The relative velocity of the object

^2 = The power of 2. In other words, the square of the velocity or v * v.

To solve for velocity you need to know the impact energy of the object and its mass. Doing basic algebra yields:

Velocity = the square root of ( Kinetic Energy / mass )

All of his formula are correct, but all that depends on how fast it is traveling and its mass. I have a piece of meteorite in my hand, it's about the size of my fist and weights about 5 lbs. Rocky meteorites weight a lot less. So the two things you have to come up with are mass and speed.

Google for "Vredefort Dome" it is believed that a 10km object hit the earth. a crater 300km in diameter was formed. A big object of 5km hit Australia and formed a 120km crater.

A bus size would really be relatively small.

Someone is teasing you...there is no answer to this question...

It depends how fast it is going initially, the shape, the angle if incidence to the atmosphere and a thousand other variables.

(I reserve the right to be no help whatsoever)

I believed, that he meant a strainght on hit scenario.

In that case see #9.!

Using that information and by applying some high level statistics we can calculate the velicty to be c/2 .

(joke folks don't shoot)

Try this link

http://www.hartrao.ac.za/other/tswaing/tswaing.html

300 000 ton. slightly bigger than a bus. They show some velocities etc.

As Del pointed out, there is no answer to the question. The incoming velocity depends on where it came from (eg. the asteroid belt, farther out, outside the solar system...) and what other massive objects it passed near on its way here ( passing by a massive object may either speed it up, make no difference, or slow it down, depending on the relative motions of the two).

The impact velocity for a given incoming velocity depends mostly on the angle of approach. If it just grazes the atmosphere, it will be slowed by friction, but will pass on by and not strike the Earth. At a little steeper angle, much of it will burn up during passage through the atmosphere (the Challenger was a bit larger than a bus, and somewhat slower than most asteroids). If it comes in vertically to the point of impact, the atmosphere is very thin and will only slow it a little.

Now on the other hand, the escape velocity from Earth is about 25,000 mph. That is the outward velocity an object must have if it is to leave the vicinity of the Earth's gravitational field. It is also the velocity acquired by an object originally at rest at a large distance from Earth, during its fall to Earth. Since most asteroids already had a significant velocity before they began falling to Earth, most of them will be traveling significantly more that 25,000 mph at the time they enter the Earth's atmosphere.

Earth's orbital velocity is ~ 30 km/s, asteroid probably doing similar speed so could take say 50 km/s impact velocity for order-of-magnitude estimates (of what I don't know).

Codey

Estimated impact veloctiy somewhere between 0 and c .

You could say that

Earth's orbital speed would be an slightly eliptical circle with a radius (a.u.) of 149,597,870 Km, traveled once a year. Actual radii are 1.0167 a.u. at aphelion and 0.9833 a.u. at Perihelion. As for the actual meteor/meteorite, it depends on the angle of approach to the Earth (opposite or with the orbital direction of the Earth, angles, etc), plus would the meteor break up during trans-atmospheric travel and gravitational effects from the Earth, shape of the meteor (atmospheric resistance) and so on.

terminal velocity (object dependent)

NOOOOOO!!!!

Terminal velocity is the velocity at which the weight of (Earth's pull on) the object is just counterbalanced by the sum of buoyant and friction forces of the atmosphere, so it falls at a constant (approximately) velocity. These velocities range from essentially zero for very small dust particles to something on the order of 300 mph for large hail stones. A dense aerodynamic object like a bullet shot straight up will come back down with a velocity of that order of magnitude.

Any space rock the size of a bus approaching the Earth at over 25,000 mph will have significant friction forces, but nowhere near the weight of the object. Even in a very low angle of approach, it will not spend enough time in the atmosphere to loose all that kinetic energy. Virtually all the atmosphere is below 50 miles altitude, so in a vertical approach, the object has roughly 50/25000=2/1000 of an hour, or 0.12 minutes, or 7.2 seconds between entering the atmosphere and striking the Earth. It would only be slowed down very slightly!

what are the angular circumstances necessary for and asteroid to make a "soft" landing?

I imagine an object coming in a following trajectory, using earth gravity to slow it down just enough, then being caught up by the earth. it could land on the side of the earth receding from the orbital direction, slowing the landing even more.

What would the proverbial farmer in the field see when it came in at thirty or so miles an hour? I bet he would "run all the way to town screaming 'It came out of the sky!'"

-A-

I would think it unlikely that you could get a velocity lower than the free falling terminal velocity...which is probably about 90mph ish?

Unless maybe it's a big soft fluffy asbestos asteroid... ?

Mmmmm flufftastic!

Or conversely, if it lands in a big, soft, fluffy asbestos field.

Let's not joke or sugar coat the situation. A bus sized asteroid would make a mess.

Not if it landed in Iraq ... George and Tony have already made a BIG mess so no one would notice...

If it really is the size of a school bus, it's only allowed to travel at a maximum speed of 50 mph. So you can figure from there.

SCHOOL BUS SIZED....????!!!!

Oh my God! The children! what about the children?

From now on I'm driving my son to school!

cr3

Ya know, he said "have no other gods before me." So, like, does that mean there are other gods? And if so, what kind? How many others are we not to say don't exist. I know this is implicit, but I'm just thinking here.... Hmm?

genesis ch 3

You mean where God proclaims that the one thing we can understand is how he could just kill a god?

But it doesn't say he got them all. He only got the ones that were messin' with his stuff.

perhaps you should read it.

Nah! All the best pictures are already colored in.

I think you're up a creek with the question about velocity, unless you measured many many parameters from the impact site and worked backward to determine it's velocity.

A better question would be: if a bus sized asteroid hit the earth, what would be the damage if it was going at this velocity, then this velocity, then this velocity... There's no telling how fast the asteroid is traveling initially. It could be going fast or slow.

Laden or unladen?

Bus size is minor, we have those for tea. I am just joking it should do a lot of damage.

For a 10km one

www.hartrao.ac.za/other/vredefort/vredefort.html

Or see post #4 for a smaller one.

Did you know that about five or six times a year there is a multi-megaton explosion in our upper atmosphere?

The Federal Government was so convinced that these were caused by the Soviets breaking the nuclear test ban treaty that we launched a satellite to check. Turned out on an average of six times a year a large enough asteroid blows up in our atmosphere to create these multi-megaton explosions.

Kind of makes you fell vulnerable, doesn't it.

No, I hadn't heard that! Sounds plausible. Any reference available?

You're going to have to surf the web. But It was a DOD satellite, probably launched by NASA. I'd limit the time line to 60s - 70s.

you first would have to find out what is closest to its line of locomotion and then you figure out how far this object is from your telescope. then by using the angle that your telescope is positioned determine the angle of sight towards the asteroid with this said you can now use soh cah toa (primary trigonometrical ratios) to find the distance between these objects. with that said you may now wait exactly 2-9 days (makes sure distance traveled is not more then half way to object)to see how far it moved and by accurately doing the same steps as before using the same object as reference you may now subtract original distance from new distance (number should be negative so turn it positive) using the two numbers you now have (distance traveled and distance to be traveled) you may use the time needed to travel to givin point and the the distance traveled to figure out velocity now wait a little longer and figure out velocity again if it is more or less then you are going to be monitoring it untill it is very close and very fast (or clear of gravitational forces) note your most likely not going to be able to use soh cah toa because in that case it will be flying parallel to earth use cosine law or sine law instead for your case. god im still in highschool and i had to point this out to you guys. thats sad( i left out alot of equations for lack of time (and possibly space) so you will have to come up with equation for distance to moderately stationary object) plus it is not science it is math and ratios and use atomic weight of atmosphere layers and atomic weight of asteroid to determine rate of decceleration. use the atom and rate of an object sinking in water as a reference. (compare rates of two symmetrical balls of different matter and compare to figure out atom,electron,proton role in rate).