kev said:
...
It is also known that objects dropped from the same height simultaneously reach the floor simultaneously regardless of their individual masses and in GR this idea extends to objects with zero mass.
MeJennifer said:
That is simply untrue, both the mass of the planet and the mass of the object in question contribute to the time it takes for them to come together. Of course it is true that the small mass is negligible compared to the large mass but we are talking principles here.
Yes, the principle we are talking about is the equivalence principle.
Imagine a rocket of unladen mass of 1000 kgs with a payload of another 1000 kgs and a test mass of 1 kg near the nose. (Total mass =2001 kgs) Say the payload and test mass are released when the nose is moving at 0.6c. The released masses both continue at 0.6c while the rear of the accelerating rocket catches up with the freefalling masses. The rear (floor) of the rocket arrives at both the large mass and the small mass simultaneously. The discovery by Galileo that objects released together, fall at the same rate irrespective of their mass, over 500 years ago is still true today even in General Relativity.
Please note I was careful to use the word "simultaneously" (twice) in my statement "It is also known that objects dropped from the same height
simultaneously, reach the floor
simultaneously regardless of their individual masses" but the missing comma may have made the meaning unclear. Anyway, the fact the masses are released simultaneously is the key point, as Dalespam noted. I covered the case where objects are released one at a time in the earlier post to try and make the issue clear.
If the 1000 Kg payload of the accelerating rocket is released by itself the engine of the rocket has less mass to accelerate and the rocket accelerates faster towards the released payload than it would if the 1kg terst mass was releaed by itself. The equivalence principle shows that two masses released
together, fall at the same rate, but masses dropped
one at a time may fall at different rates with larger masses falling faster. The same is true in a gravitational field. The planet accelerates up towards the combined mass of the released falling objects when they are released together and the falling objects fall at an equal rate determined only by the mass and distance of the planet.
The full equation for the Newtonian gravitational acceleration is a = \frac{G(M + m)}{R^2}
ref
http://en.wikipedia.org/wiki/Newton's_law_of_universal_gravitation
When the mass (m) of the falling object is significantly smaller than the mass (M) of the gravitational body, the equation a = \frac{GM}{R^2}
is a reasonable aproximation, which is only exactly true in Newtonian physics when the mass of the falling mass (m) is exactly zero.
For the case of two objects (m2 and m3) falling together towards a large massive body (M), the acceleration of the planet towards the falling objects is:
a1 = \frac{G(m2+m3)}{R^2}
The acceleration of object m2 towards the original position of the planet is:
a2 = \frac{GM}{R^2}
The acceleration of object m3 towards the original position of the planet is:
a3 = \frac{GM}{R^2}
The total acceleration of object m2 towards the planet when the acceleration of the planet towards the object is taken into account is:
a2' = a2 +a1 = \frac{GM}{R^2} + \frac{G(m2+m3)}{R^2}
The total acceleration of object m3 towards the planet when the acceleration of the planet towards the object is taken into account is:
a3' = a3 +a1 = \frac{GM}{R^2} + \frac{G(m2+m3)}{R^2}
It can be seen that a2' = a3' and Galileo's claim that objects falling together, fall at the same rate regardless of their individual masses is true. It can further be seen that it is true that the equations for acceleration of a falling body shown above, are equally valid when the mass of the falling body is zero by setting the value of m2 or m3 to zero.
I am sure you will agree with the arguments stated above and that you simply misunderstood what I was getting at, due to a punctuation error on my part.
