Light at Light Speed

Discussion in 'Physics & Math' started by Bowser, May 1, 2011.

  1. arfa brane call me arf Valued Senior Member

    This question: How do you know you aren't in free fall towards the surface of the earth?

    IS a trick question. When you are free falling, say after jumping from a plane, you accelerate towards the ground at 9.81 m/s. If there was no atmosphere you would accelerate until you reached the ground.
    Your weight appears to vanish while you accelerate, because the forces acting on you are in equilibrium. In free space, far from any gravitational forces, you experience the same thing. Equilibrium is just the condition where all forces cancel, there is no residual force accelerating your mass, so in effect an equilibrium condition is equal to no forces at all, which is equivalent to all forces cancelling (at a common centre).

    On the surface you experience weight. This is because you aren't "at rest" even when you stand or sit still. You aren't at rest because parts of you are still in motion--your heart, your lungs, your muscles. Parts of your body are in free fall, and you can only keep some of your body motionless--this actually requires that your muscles react constantly, so parts of you accelerate, and this is the 'sensation' of weight.

    What about passive objects, which are 'inert'? These can be motionless with respect to the surface, and Newton's laws of motion don't discriminate--all objects with mass accelerate independently of their material structure.

    We can 'sense' the weight of inert masses by 'weighing' them in our hands. Using a spring balance means having to 'modify' the equations of motion for a mass in free fall, which changes its position in time, for weights which are not changing their position in time--the time dimension needs to be eliminated from the equations.

    However, the mass, Newton's constant and the gravitational strength, g, are not eliminated when forces are in equilibrium. What you need, obviously, is a static solution that corresponds to the kinematic one.

    Terminal velocity is the situation in which the force of gravity is 'reduced', not eliminated, by a force of friction that opposes the acceleration, the first derivative of velocity, so dv/dt is zero and you don't accelerate. But you experience some of your 'weight' in that case.

    When at the surface your downward velocity is zero, you experience your full weight. This suggests that the surface is just a strongly viscous medium that gives you a terminal velocity 'very close' to zero.
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