How complicated is Rocket Science?

Discussion in 'General Science & Technology' started by Captain Kremmen, Dec 10, 2009.

  1. D H Some other guy Valued Senior Member

    Messages:
    2,257
    All you are doing with your snarky comments is making a fool of yourself.

    A cartoon description of the equivalence principle:
    http://www.pbs.org/wgbh/nova/einstein/rela-ele-q-300.html

    The normal force is a manifestation of the Coulomb force. Suppose you are standing on the floor. The Coulomb force between the electrons at the surface of the floor exert an upward force on the electrons at the bottom of your feet. Now suppose you grab a pull-up bar and slowly start lifting yourself off the floor. The ground will continue to exert an upward force (but reduced in magnitude) up to the moment that the upward force from the pull-up equals the downward gravitational force. At this point you will be lifting your feet off the ground.

    While the Coulomb force is an inverse force law, the total Coulomb force between you and the ground falls off much, much faster than 1/r[sup]2[/sup].
    This is because you and the ground are more or less electrically neutral. From a quasi-classical point of view, the normal force is essentially a high order multipole interaction.

    The normal force is a constraint force, as is static friction. Re static friction: A box is sitting on the floor. Exert a small horizontal force on the box and it will not move, seemingly in violation of Newton's second law. Newton's second law pertains to the net force acting on an object rather than the individual components of the net force. Via static friction, the floor exerts a horizontal force on the box that exactly opposes your tiny horizontal force on the box. (Note well: Even though this is an equal-but-opposite force, it is not a third law interaction.) Up to a point, the opposing force by the ground increases as you increase your horizontal force on the box. This point where the ground can no longer match the force you are exerting is described in terms of the static coefficient of friction.

    Back to the normal force: It too has a breaking point. The normal force decreases in magnitude when start doing the pull-up. It reaches zero when you reach the point where your weight is fully suspended from the pull-up bar. Add a tiny bit of force to the pull-up bar and you will lift yourself off the ground. The normal force does not suck you back to the ground. The breaking point for the normal force is the point at which the normal force would have to be directed inward rather than outward.

    Good question! Weight is a tricky concept. Legally and colloquially, weight is a synonym for mass. A one pound can of peas "weighs" one pound at the North Pole, at the top of Mt. Everest, and on the surface of the Moon. Ignoring that distraction, weight in physics has units of force. Even then, there are two distinct meanings for weight in physics. One meaning is gravitational force, aka "weight", aka "actual weight". This is simply the mass of the object times the gravitational component of acceleration: W=mg. The other meaning is the net force acting on an object less the gravitational force. This is the "apparent weight" or "scale weight" of the object. To illustrate the difference, consider some astronauts floating around inside the International Space Station. Their actual weight is about 90% of their actual weight on the surface of the Earth. Their apparent weight is essentially zero.

    Suppose you are standing still on a scale on the surface of the Earth. That scale does not measure your actual weight. It instead measures your apparent weight. This is why another name for apparent weight is scale weight. Your actual weight is directed down while your scale weight is directed upward. If the Earth wasn't rotating your actual and scale weight would be equal but opposite to one another. However, the Earth is rotating. There is a net force acting on you, just enough to keep you in uniform circular motion about the Earth's rotation axis. Your actual and scale weight are not quite equal but opposite to one another.

    So what is it that you feel? You feel scale weight, not actual weight. When you are standing on the scale, the scale is exerting an upward force on your feet. It is this upward force that you feel in your feet. This force propagates throughout your body. You feel it in your guts and in your head. The vestibular system is a part of inner ear. The otolithic organs and semicircular canals in your inner ear are very much akin to the accelerometers and gyroscopes in an airplane's or spacecraft's inertial guidance system.

    Place an accelerometer at rest on the surface of the Earth and the accelerometer will register a one g acceleration directed upward. The accelerometer, like your otolith organs, is insensitive to the gravitational force. They are instead sensitive to the net sum of every force but gravity: scale weight. What you feel as weight is scale weight.
     
  2. Google AdSense Guest Advertisement



    to hide all adverts.
  3. noodler Banned Banned

    Messages:
    751
    Right. That's something many people naively overlook, when developing their "algorithm" to explain weight. If you suspend yourself from a tree branch, or suspend a weight you are accounting for the downward accn. of the weight. But not for the total energy because of course, although the tree and you are "motionless" both are moving through space (or, you could get all technical, build a frame and use springs to suspend yourself, or just use a standard set of scales to satisfy the technical "requirements").

    But, you can say that "weight" is a first order consequence of gravity. Ok so weights and springs: when you weigh yourself on scales, the tension and or torsion in a spring or some other elastic medium, balances the force you apply, so F = mg = -kx.
    The "x" in this case is your position, or the position of the "scale display" is equivalent to a displacement ds.

    Since "x" is now constant, and you know the spring constant k you know how to calculate your weight. If you already know the weight (of anything) you can use this relation to calculate G, since g is also a known quantity, or at least is a determinable quantity (with observations of motion).

    As you note, the apparent "motionlessness" is because of net forces. But a weight hanging from a spring is at an equilibrium (at least locally), and we know that this means mg = - kx. So now we can design a machine that has a large enough k so a mass m will have greater "energy" than -kx and "escape from" g. This is physically equivalent to displacing a mass which is at equilibrium at the end of a spring, so the mass travels upward and eventually the mass is (temporarily) weightless, or experiences no forces.

    Of course, because the mass doesn't travel straight "up" there are corrections needed (but not for small distances ds). So to first order we can say that small displacements of mass (at the end of springs) means that mg = -kx and "force" is not required, and is seen as a formal variable or label we attach to the quantities either side of the equation.
     
  4. Google AdSense Guest Advertisement



    to hide all adverts.
  5. D H Some other guy Valued Senior Member

    Messages:
    2,257
    More nonsense. Stop trolling.
     
  6. Google AdSense Guest Advertisement



    to hide all adverts.
  7. noodler Banned Banned

    Messages:
    751
    I'm sorry, I don't understand that.
    Are you saying that building a machine that has a large enough k so it can give a mass energy, means it can't have a kx? Or do you mean that mg = -kx is nonsense?

    What happens to a weight suspended from a spring if it moves up and down, periodically?
    At the highest point, does the mass weigh anything?

    Or is it that you simply don't understand that launching a rocket is equivalent to displacing a mass so that mg = -kx is "canceled" and the mass experiences no net forces? Is a body in orbit experiencing any net forces, or is that too naive?
     
  8. mananmater Registered Member

    Messages:
    46
    About as complicated as trying to walk on 1 toe, but the easy ness of folding clothes is inside it, you just have to learn to love it.
     
  9. Dywyddyr Penguinaciously duckalicious. Valued Senior Member

    Messages:
    18,645
    Was that supposed to make sense?
     
  10. cellphoneman Registered Member

    Messages:
    7
    It is very complicated.
     
  11. kmguru Staff Member

    Messages:
    11,757
    It is very simple if you know what you are doing. But very complicated even to super powers if you do not have that knowledge and it is not in the books!

    Please Register or Log in to view the hidden image!

     
  12. Dywyddyr Penguinaciously duckalicious. Valued Senior Member

    Messages:
    18,645
    I dunno. NASA has a fairly extensive output these days of PDFs you can download covering most of it...

    Please Register or Log in to view the hidden image!

     
  13. kmguru Staff Member

    Messages:
    11,757
  14. mugaliens Registered Member

    Messages:
    110
    It depends on what you're doing. Computing a simple orbit transfer only requires knowledge of partial derivatives and advanced calculus to account for the changing mass during burn.

    Designing the systems requires the collective efforts of engineers with decades of experience working with PhDs.

    Put simply, college doesn't teach you "rocket science." It teaches you aerospace engineering, which is the building block upon which OJT and advanced degrees teach you the rest.
     
  15. krazedkat IQ of "Highly Gifted"-"Genius" Registered Senior Member

    Messages:
    262
    I have designed a rocket, it's a concept though. The scientists don't build it, others do -.- and the scientists don't fly it.
     

Share This Page