tag:blogger.com,1999:blog-8807287158334608095.post7190703233813959124..comments2024-03-28T07:13:29.424-04:00Comments on The Virtuosi: Laser LaunchingAlemihttp://www.blogger.com/profile/15394732652049740436noreply@blogger.comBlogger8125tag:blogger.com,1999:blog-8807287158334608095.post-46214694648045536752021-07-19T04:20:12.881-04:002021-07-19T04:20:12.881-04:00เว็บตรง เครดิตฟรี 2021
ufabet เครดิตฟรี ไม่ต้องฝาก...<a href="https://ufabet168.io/" rel="nofollow">เว็บตรง เครดิตฟรี 2021</a><br /><a href="https://ufabet168.io/" rel="nofollow">ufabet เครดิตฟรี ไม่ต้องฝาก 2021</a><br /><a href="https://ufabet168.io/" rel="nofollow">Ufa แจกเครดิตฟรี 2021</a><br /><a href="https://ufabet168.io/" rel="nofollow">ufabet</a><br /><a href="https://ufabet168.io/" rel="nofollow">::ufabet:: ทางเข้า</a>thanonhttps://www.blogger.com/profile/05198289642100180175noreply@blogger.comtag:blogger.com,1999:blog-8807287158334608095.post-81556813972229081302010-04-22T15:34:18.604-04:002010-04-22T15:34:18.604-04:00Ah, yes, I missed the part about only pushing it f...Ah, yes, I missed the part about only pushing it for 15 km. In that case your derivation is correct. I suppose another way to say it is that you can either supply a little energy over a long time or a lot of energy over a short time. I suspect that totaling the energy from both methods will give you the potential difference you need to overcome to break free of the Earth.Yarivhttps://www.blogger.com/profile/15599587931578476824noreply@blogger.comtag:blogger.com,1999:blog-8807287158334608095.post-55122680175269056162010-04-22T14:44:40.033-04:002010-04-22T14:44:40.033-04:00Well, just pushing harder than the earth's gra...Well, just pushing harder than the earth's gravitational force doesn't guarantee that something won't fall back to the ground, unless you keep on pushing. The essence of my claim is that I want to supply enough force to get this thing to escape velocity when it is at 15 km up, as that is when I'm 'letting go'. I figured that would be a good estimate of the amount of energy to get into a high orbit, because we've been neglecting drag forces, so it really won't hit that escape velocity, it'll have an energy somewhat less than this.<br /><br />So, I did play a little fast and loose with the numbers. What I really wanted was to achieve the escape velocity of 15 km above the surface of the earth at 15 km above the surface of the earth. However, if you look at the escape velocity at 15 km up, that's given by<br />\[v_{esc}=\sqrt{\frac{2GM_{earth}}{R_{earth}+15km}} \approx 11.2 km/s\]<br /><br />This is because the radius of the earth is ~6.3 Mm. So a 15 km difference doesn't really change the escape velocity.<br /><br />Hopefully that clears up my calculation a little bit.Jessehttps://www.blogger.com/profile/16335133534234025744noreply@blogger.comtag:blogger.com,1999:blog-8807287158334608095.post-29303742000634843372010-04-22T14:14:03.317-04:002010-04-22T14:14:03.317-04:00also, I have some issues with your derivations; na...also, I have some issues with your derivations; namely, escape velocity is only relevant when you are supplying some initial velocity and then letting go. Here what we're doing is constantly supplying force - so what we really need is just to push harder than g, the earth's pull.<br /><br />We want <br />\[F > mg = m\frac{MG}{R^2}\]<br />which leads to a result similar to yours but slightly different,<br />\[n > \frac{MGm\lambda}{2R^2 h} \]<br />or<br />\[ P > \frac{MGm c}{2R^2}\]<br />which comes to about 1.1 TW. If we want to push it fast we'll probably need double that - so let's say 2 terawatts. A couple of orders of magnitude easier, but still not currently achievable.Yarivhttps://www.blogger.com/profile/15599587931578476824noreply@blogger.comtag:blogger.com,1999:blog-8807287158334608095.post-78341472541538882202010-04-22T13:54:00.568-04:002010-04-22T13:54:00.568-04:00"To me, the most remarkable feature of the es..."To me, the most remarkable feature of the escape velocity is that it is independent of the mass of the object!"<br /><br />Ah, my friend, that's what the equivalence principle is all about!Yarivhttps://www.blogger.com/profile/15599587931578476824noreply@blogger.comtag:blogger.com,1999:blog-8807287158334608095.post-62111973244453196272010-04-22T11:32:21.895-04:002010-04-22T11:32:21.895-04:00I did actually see that brief on wikipedia when I ...I did actually see that brief on wikipedia when I was doing a little background reading on laser propulsion systems. I've just grabbed a few recent articles on ablative laser propulsion, and I'll see if I can come up with a review of the physics and the practicality in the next day or two.Jessehttps://www.blogger.com/profile/16335133534234025744noreply@blogger.comtag:blogger.com,1999:blog-8807287158334608095.post-57672881756030405742010-04-22T10:52:36.693-04:002010-04-22T10:52:36.693-04:00I am very much not a math guy (to my chagrin), but...I am very much not a math guy (to my chagrin), but it seems all of this only takes into account light pressure.<br /><br />However, there is another form of light propulsion called Ablative Laser Propulsion, and I wonder what the numbers on that would reveal.<br /><br />The Wikipedia link: http://en.wikipedia.org/wiki/Laser_propulsion.<br /><br />This method involves using a ground-based laser to convert material on the launch vehicle into plasma which is then vented to create thrust. I've seen footage of scale model tests, and it worked reasonably well. <br /><br />Might the numbers be more favorable using this methodology, and something that we could hope to see as a usable launch method in the near future?<br /><br />I would certainly be interested to see how the math works out...coolmajakahttps://www.blogger.com/profile/06108695850421099652noreply@blogger.comtag:blogger.com,1999:blog-8807287158334608095.post-53612895666421676592010-04-22T10:46:46.677-04:002010-04-22T10:46:46.677-04:00I heard an interesting talk a while back on a vari...I heard an interesting talk a while back on a variant of the laser launching idea, which was to use a high-powered ground-based laser as a kind of heating element. The idea is that you blast your "rocket" with a laser from the ground, heating a store of inert gas with the light, and shooting it out the back end at high speed, providing propulsion.<br /><br />This is a net win because it eliminates the need to carry two components of fuel (both a fuel and an oxidizer), and a lot of the safety problems that go along with that. And it cuts down on the weight needed.<br /><br />Of course, it means that any space launch is accompanied by a giant laser beam tracking the launch vehicle across the sky, which I'm sure would be a huge hit with the FAA...Chad Orzelhttp://scienceblogs.com/principles/noreply@blogger.com