It has been suggested that a baseball, thrown ‘backwards’ out of the shuttle would burn up on re-entry — but (if you consider its mass, density, etc) would it?
Ok, this is based solely on intuition, but it seems like the baseball would reach a ‘terminal velocity’ based on atmospheric drag long before it starts ‘burning up’. Normal shooting stars are travelling much faster than the aforementioned basball, and therefore ‘burn’ before slowing to their ‘terminal velocity’, and the (much slower) de-orbiting satellites are far more dense (metal) and would have a higher terminal velocity.
Well, we’ve had small objects in low orbit for forty years now, and only some of the bigger or denser bits have hit the ground. And you may have noticed that re-entry vehicles (Apollo for example) have heat shields on their undersides. So I’d say yes a small light object probably will burn up.
What Tim and Bert ought to do, though, is go partway up the Space Elevator when it’s built, and drop from there. I’m too lazy to compute it again but an object released from the lower two-thirds or so will hit air, no MLB-style hurling required. And a ride on the Space Elevator will be cheaper than any rocket ride.
No argument that we’ve had small items falling for forty years - but your observation that ‘only the larger and denser bits have hit the ground’ only furthers my original intuitive feeling.
I would argue that there has never been been a ‘baseball-like’ item that has fallen from space over the last forty years. (Baseball-SIZED yes, but not baseball-LIKE).
Yes, re-entry vehicles have heat shields, but that’s because the vehicles are designed to ‘slice (or skip) through’ the atmosphere and maintain a high terminal velocity (and therefore need to withstand a high temperature).
One could easily argue that if a re-entry vehicle had enough drag, it wouldn’t ‘burn up’ - it would just ‘fall down’ (well below the speed of ‘burn-up’.) Just look at the Mars program ‘parachute-type’ re-entry vehicles, for example.
So - having shown that some ‘normal’ objects would actually make it to the ground (because their terminal velocity is below the ‘burn-up’ velocity) — the question remains: Which type of re-entry vehicle is a baseball? Is it closer to a solid metal ball (fast) or closer to a parachute (slow)?
Welcome to the Straight Dope Message Board, GNL, glad to have you with us.
When you start a thread, it’s helpful to others if you provide a link to the Staff Report upon which you are commenting. In this case, it’s the report that will not appear on the website officially until Tuesday June 24, and I have edited your post accordingly.
I am not convinced. Meteors are hitting the atmosphere at much higher velocities than would anything ‘thrown’ from orbit. Indeed, consider how much stuff was recovered essentially intact from the recent shuttle disaster - including live worms. I imagine that a 7000mph baseball would be severely singed; but, unlike the grain of sand at meteoric velocities, the process might not dissipate enough energy to be visible from the ground.
In the extreme case, suppose that the ball could be released from shuttle-orbit altitude with a horizontal velocity of zero relative to the earth. It would not achieve any great velocity at all before it started getting slowed by friction with an initially very tenuous atmosphere. I would not be surprised if this sort of “re-entry” would not even damage the ball, assuming you could find where it landed.
First a small correction to the column, then on to answer the “Would a baseball burn up?” question.
From the column:
Actually, 6,378km is the radius of the earth itself, so the space station is orbiting at 6378+390=6768km. With this correction, the rest of the numbers in the column make sense.
Now, regarding the baseball. My degree is in Aerospace Engineering, although I’m not an expert on re-entry. My semi-professional opinion is that I’m quite certain that the baseball would fry.
At some point, the ball will reach a state called “Max Q”, which is the maximum combination of velocity and atmospheric density. The drag is proportional to the “dymamic pressure” (“Q”), which is
[ul]1/2 x Rho x V-squared[/ul]
where Rho is the atmospheric density and V is the velocity with respect to the atmosphere. Max Q gets you max drag gets you max heating, since the drag is dissipating kinetic energy as heat.
Think of it this way. Out in space, “V” is high (7700m/s, or 17,000mph), but drag is nearly zero, because “Rho” is nearly zero. If it makes it to the ground, “Rho” is high, but drag is again zero, because “V” is zero. In between, there will be some point where the combination of “Rho” and “V” is highest, hence “Max Q”.
And for an object re-entering the atmosphere from near-earth orbit, Max Q is pretty max. Columbia was going about 12,500mph at an altitude of about 40 miles when it broke up. At that altitude, with that velocity, the drag is enough to melt steel, as we saw.
It would be about the same with the baseball. I’m guessing that the Space Shuttle isn’t particularly more/less dense than a baseball, on average. In the Shuttle there’s alot of steel balanced by alot of empty space, versus a baseball of more-or-less uniform but moderate density. The Shuttle’s also not particularly more/less aerodynamic than a baseball. Someone once told me that it’s basically as aerodynamic as a half of an apple, with the cut side down.
Even if you could reverse the ball’s velocity completely, so that it was dropping straight down with respect to a target below, it would still build up an extreme amount of velocity before it started to “bite” into the atmosphere. “Terminal velocity,” where drag balances gravity, is about 100mph for average-sized objects at sea level, but if there’s no atmosphere, then terminal velocity is essentially infinite. So the baseball would build up alot of speed, which would have to be dissipated before the ball reached its new terminal velocity in the atmosphere.
But you’re right that, at some critical initial altitude (less than 390km, I’m guessing), you could drop the ball and it would never get going fast enough to burn up. Not sure what that altitude would be. Maybe I’ll run a little simulation and see. If so, I’ll report back to you my results.
Hope this explanation helps. If things are still murky, let me know; I’m happy to try again.
OK, so I ran the simulation, and here’s what I found:
If dropped straight down from 390 km (240 miles), the baseball reaches a maximum speed of about 5500 mph at about 45 miles altitude. It accomplishes this in about four minutes. (This speed is about 1/3 the orbital speed mentioned in the column.)
Then it hits the wall. It decelerates to about 400 mph in about 40 seconds. Its maximum deceleration is over 400 mph per second. During the peak of the “Max Q” event (around 270 seconds into its fall), the ball and the surrounding air are absorbing 53,000 watts of power, or 70 HP, or 50 BTU/sec.
During the middle 20 seconds or so of the Max Q event, it’s averaging about 25,000 watts. So think of removing the heating elements from 20 1200W hairdryers, and wrapping them around a baseball, and running them at high heat for 20 seconds. I think the baseball is toast.
The cinder lands about 15 minutes after you first dropped it, hitting the ground at a leisurely 50-100mph.
Are there any thermodynamics-oriented folks out there? Do you have a better gut feel for the power/heat numbers?
I’m not really thermo-oriented, but I can add a little info:
If I’ve Googled correctly, an official baseball weight 5 oz (0.14 kg) and is made out of cork, wool windings, and a leather cover. The specific heat of cork is 2.03 kJ/(kgK), wool is ~1.50 kJ/(kgK), and leather is 1.38 kJ/(kgK). I imagine that the wool makes up most of the weight, so the total heat capacity is, say 1.6 kJ/(kgK). For the 5 oz ball, then, it would take 1.6*0.14 = 0.224 kJ to raise its temperature one degree K.
Now, what temperature rise is required to ignite the ball? The ball starts at what? -200C? During flight, the leather covering would ignite at 212 degrees C (from here). So a temperature increase of 400-450 degreesC (or K) would ignite the ball.
Now, you’ve calculated an average of 25,000W over 20 sec, which totals 500kJ during that time. If all that heat went into the ball, the ball temp would rise by ~2200 degrees. I don’t have a good intuituve feel for what proportion of the heat goes into the ball, but maybe 25% would be reasonable, which would give over the required ~400 degree temperature rise.
I imagine that there’s other complicating factors, but this ought to do for a start.
Good catch on the misplaced radius. I used all the right figures for my calculations, but I trimmed it down a good bit when I wrote the Report. Apparantly I trimmed the wrong bit there. Admins, could you change the 6378 in the report to 6768?
And I hadn’t actually considered that the baseball would be going much slower than a typical meteor, but I’m glad to see that other folks did. I knew I could count on you’uns Dopers :).
I found an online source stating that 10-40 km/sec is a typical speed for meteors entering the atmosphere. The lower value is just slightly higher than the space station’s speed, while the higher value is about 5x the space station’s speed.
So faster, but not remarkably faster. Anything moving much faster than that in the neighborhood of Earth would be on a one-way trip out of the Solar System, I think. The only way to get that kind of speed is to come from outside the Solar System, or to be sling-shot around another planet, both unusual events compared to the number of run-of-the-mill meteors.
Actually you could get up to about 70 km/s, if an object came in at escape speed (or just under), and in the opposite direction to the Earth’s orbit. Remember, we’re tooling along at 30 km/s, ourselves. But then, retrograde orbits are also pretty rare in the Solar System.
Close, but remember that the ball is not starting out at 0 MPH. It’s starting out at an orbital velocity of about 17,500 MPH. If you throw a 75 MPH pitch backwards, it’s now going about 17,425 MPH in roughly the same direction you were originally going. (If you simply drop a baseball from the space shuttle, it will be travelling exactly the same speed and direction as you – in other words, from your perspective it will appear to float along with you.)
Bits of cloth have reentered before, mostly lost bits of insulation. They do indeed burn up; they are travelling awfully fast when they hit the atmosphere, and generally are consumed long before they can reach terminal velocity. (Larger, denser objects such as Volkswagen-sized fuel tanks, are known to survive long enough to reach terminal velocity. Because of this, they are generally quite cool when they hit the ground, which many find surprising. Meteorites large enough to survive that initial reentry also tend to be quite cool on impact and do not start fires, contrary to what common sense tells us.)
Here’s something not mentioned in the column…
At the altitudes reachable by the space shuttle, orbits are fundamentally unstable. These are below the Earth’s “Roche limit”, an average altitude which exists for every planet below which orbits are not stable. Above this altitude, orbits are stable. Tidal forces will gradually cause the orbit to decay and eventually the satellite will come down, even if there is no atmosphere to drag it down. This is the ultimate fate of Mars’ moon Phobos, incidentally; it orbits within Mars’ Roche limit, and thus it is doomed to ultimately impact on the surface of Mars, even though today it is far too high to catch much drag from the extremely thin Martian atmosphere.
Tidal interactions can have very interesting effects on orbital mechanics. These sorts of interactions are presently causing our Moon to recede (it is far beyond the Roche limit). They also can be harnessed to accelerate spacecraft in a technique popularily called a “slingshot” but properly called a “gravity assist”. This is because not only does the planet’s gravity tug on the satellite, but the satellite’s gravity also tugs on the planet – and not perfectly evenly, either, since obviously neither is a perfect mathematical point. Momentum (either orbital or rotational) can be transferred through a tidal interaction, speeding up one body and slowing down the other.
If there were no atmosphere on Earth, tidal interactions would eventually bring the baseball down – probably in a few million years.
For a meteor to reach terminal velocity, it doesn’t speed up like a ball dropped from a building would. It has to slow down instead, and it has to slow down from hypersonic velocities. Keeping that in mind can help it make a little more sense.
zut, thanks for the thermo-assist, very interesting.
calliarcale, I was performing the mental experiment suggested by DrHow of somehow stopping the ball dead so that it wasn’t moving with respect to the earth rotating below it. You’d have to shoot it out the back of the Shuttle/Station with some kind of super-gun at about 16,000mph (since the earth is rotating at about 1000mph below you). Once you’ve accomplished that, it should drop more-or-less like a rock.
But that’s not exactly true. At that altitude, the earth’s gravity is reduced by about 12%, just due to your increased distance from the center of the earth. And the 1000mph remaining speed would get you another 4% reduction due to centripetal force.
And you wouldn’t actually shoot it straight out the back of the Shuttle/Station, since they aren’t in an equatorial orbit. You’d have to shoot it out at an angle.
To elaborate on the Roche limit: It has nothing to do with the stability of an orbit. It has to do with the stability of an object in the orbit. Most things in space are held together almost entirely by gravity. For most moons and the like, it’s a decent approximation to treat them as a big pile of gravel, with nothing holding them together. But if you get too close to your primary (closer than the Roche limit), then the tidal forces pulling it apart are stronger than the object’s own gravity holding it together. So it’ll pull apart into chunks small enough that they can hold together. But even then, those chunks will still be in orbit. And even though the Shuttle is below the Roche limit, it’s strong enough that it won’t pull apart.
When you’ve got nothing but Newtonian gravity in a system, with one body orbiting another, there is no such thing as an unstable orbit. You can get an unstable orbit if you have some sort of drag force, or if there are other gravitating objects around, but we don’t have either of those in the case we’re considering. In General Relativity, every bound (i.e., not escape) orbit is unstable, due to energy loss by gravitational waves, but for an object orbiting the Earth, it would take aeons for that to be significant.
And there is such a maneuver as a gravity assist, used in getting spacecraft places, but it has nothing to do with tidal effects. A gravitational slingshot would work perfectly well with a point mass.
One thing that might be relevant;
although they approach the Earth with a very wide range of velocities, meterorites do regularly reach the ground, often with a fused outer surface from the heat;
but the centre of the rock is often unchanged; in fact I am told that the centre can still be very cold when it lands.
Space is very cold, and the heat of entry into the atmosphere largely goes into heating the atmosphere around the bolide,
in a kind of bow wave which reaches very high temperatures and largely heats the meteor by radiation.
It is this bow shock which we see in the sky, and most of this heat flows around the object and dissipates behind it.
Especially if the object is smooth,
like a baseball.
It would help if you allowed the baseball to cool to space temperature for a few days, in the shade somewhere at the back of the ISS perhaps, before you chuck it down to earth.
Okay, I remember reading something about an idea a while back (called Thor, I think) that would use metal rods falling from low orbit to penetrate the top armor of tanks and other things. I would assume that the rods would be designed to have a maximal terminal velocity.
A) I’m pretty sure this would work, but would the aforementioned melting due to atmospheric friction cause problems without having to resort to odd materials? (ie: could it be done with iron or steel)
B) Assuming a shuttle height orbit, and only a minor “kick” charge to move them out the back or down, how long would they take to hit the ground?
Even if the rods did melt, it wouldn’t matter much. A whole bunch of metal, falling to Earth at great speed, is going to do a lot of damage, period, regardless of whether they melt. The only drawback would be that they wouldn’t retain their shape, and therefore might have a slower terminal speed.
And if they’re launched from low Earth orbit, they would take no more than 45 minutes to hit the Earth (assuming we’re not relying on air resistance to bring them down). As mentioned in the Report, an object “thrown to Earth” would take half an orbit or less to hit, and the orbital period for low Earth orbit is about 90 minutes. Probably the rods would be designed to hit quicker, but I couldn’t say how much quicker without knowing just how fast they were launched.
