For the primary time, people have detected an interstellar asteroid—an area rock they’re calling ‘Oumuamua, which is a Hawaiian phrase which means “scout.” It is the one object we have ever seen that entered the photo voltaic system from past our little assortment of planets. That is a fairly large deal by itself. However on prime of that, this asteroid has a extremely attention-grabbing form: It’s extremely lengthy and thin, with a width to size ratio of about 1 to 10.
Principally, it seems like a cigar—or a minimum of that is what everyone seems to be saying. The one pictures we have now that present its form intimately are artistic renderings. As a result of the asteroid is so comparatively small and much away, you possibly can’t simply see it with a visible-light telescope.
However if you cannot see it, how will you describe it? The reply to this (as in lots of conditions in science) is to make use of oblique observations. The one factor that can be measured is the brightness of the thing. As a result of this rock can also be spinning, the sunshine it displays from the solar adjustments over time. By wanting on the ratio of the brightest to weakest observations, you may get an estimate of largest to smallest dimension. When you estimate the albedo (a measure of reflectance), it’s also possible to estimate the whole dimension. Increase. There you will have it—a cigar-shaped asteroid.
If you wish to be taught the solutions to extra “how have you learnt”-type questions on ‘Oumuamua, take a look at this awesome NASA FAQ. However if you wish to calculate some solutions for your self—effectively, simply hold studying.
May this be a spacecraft?
OK, everybody settle down. This is not a web page from Rendezvous with Rama—an Arthur C. Clarke novel that depicts an interstellar object that occurs to be an alien starship. However what if it was a spaceship? May its rotation make a sort of synthetic gravity?
From the info, we all know the next two necessary issues about ‘Oumuamua: the size and the rotation fee. OK, we do not really know the size, but it surely’s someplace between 200 and 400 meters. I’ll use the larger dimension. The rotation fee is thought with extra precision because it rotates once every 8.1 hours.
How would this make synthetic gravity? Let me begin with a diagram.
When this “spacecraft” is much away from a star or different huge objects, there’s primarily no gravitational drive appearing on folks inside. You may make faux gravity by spinning the spacecraft (as proven above). This spinning movement creates a drive (labeled F above) that reproduces the impact of the ground pushing on you to counteract gravity. This drive from the ground is what we really feel on Earth. If you need an extended clarification, take a look at my older post about weight and weightlessness.
However why is there a drive between the human (or perhaps an alien) and the top of the rotating asteroid? It is as a result of the human is shifting in a circle. As a way to transfer in a round movement, you want a drive pointing in direction of the middle of the circle. We frequently name this a centripetal drive which suggests “middle pointing drive.” The magnitude of this drive is determined by the mass of the thing, the scale of the circle and the rotation fee (ω).
In fact, totally different people of various plenty can have totally different forces—identical to on Earth. So actually it is simply the acceleration that issues for synthetic gravity; that is simply the product of ω squared and the radius. When you get an acceleration of 9.eight m/s2, then you will have efficiently reproduced Earth-like gravity (however probably not gravity—simply obvious weight).
Now for the acceleration of the top of ‘Oumuamua (it is kind of enjoyable to say that out loud after you get used to it). This rock has a round radius of 200 meters (I am utilizing the largest worth), however what in regards to the angular velocity? Because it takes eight.1 hours for one rotation, ω could be 2 occasions π divided by eight.1 hours. In fact I actually need this time in seconds, so that provides me:
Sure, that is a reasonably gradual rotation fee. Only for comparability, it takes the Earth about 24 hours to rotate. Now I can put this together with the radius to calculate the acceleration. This offers a centripetal acceleration of 9.three x 10-6 m/s2 or an obvious weight that’s simply 9.48 x 10-7 of that on Earth—in different phrases, it is tremendous tiny.
There, I simply did the primary homework query for you. Now for some extra. Word, a few of these would possibly require some complicated calculations and estimations—that is what makes them enjoyable.
- How briskly would this asteroid must rotate such that folks inside would expertise an obvious weight half that of on Earth? Word: Listed here are some rotating space craft from science fiction—only for enjoyable.
- If a human was standing inside this asteroid (that is really a spacecraft), may that human leap from one aspect of the ship to the opposite?
- Assume ‘Oumuamua has a uniform density like a rock (you decide your favourite rock). How a lot power would it not take to get it rotating on the present angular velocity?
- If you’re standing at one finish of the rotating spaceship, how briskly would you must throw a baseball such that it makes it to the opposite aspect with out hitting the wall? Trace: Think about the Coriolis force. You would possibly must make a numerical mannequin, however you may most likely do that with a tough estimation.
- Create a Python mannequin displaying the rotating ‘Oumuamua because it strikes via the photo voltaic system. You possibly can mannequin the photo voltaic system as simply the solar and Jupiter.
- Suppose we need to rendezvous with ‘Oumuamua—which might be a cool title for a ebook. How briskly would the spacecraft from Earth need to journey as a way to meet up with ‘Oumuamua earlier than it leaves the photo voltaic system? How lengthy would this journey take? This one is troublesome.
- Super large asteroids are spherical, however this one just isn’t. Calculate the web gravitational drive on part of the asteroid to indicate that it’s a lot smaller than the bonding forces of a typical rock.
- As ‘Oumuamua strikes previous the solar, it has two varieties of angular momentum. There may be the orbital angular momentum and the spin angular momentum. I consider (however I am not 100 p.c positive) that the whole angular momentum of spin plus orbit ought to be fixed. Estimate the change in spin angular momentum because it passes the solar (on account of tidal gravitational forces). Ought to this modification the orbital angular momentum?
- Why is that this asteroid so skinny? Make up a believable story to clarify its form.