Galileo's experiments (either thought or real) led to the following prediction:- Suppose two objects (a 1kg and 2kg mass) are dropped simultaneously from the same height above the earth in a vacuum, then the time for each to reach the earth would be the same.
1. I say that this prediction is only partially true.
2. Suppose the same experiment was repeated with each mass being released separately.
I say that the 2kg mass would collide with the earth in less time than the 1kg mass.
3. Suppose the same experiment was repeated with the 2kg and 1kg masses being released simultaneously but on opposite sides of the earth. "Obviously" the 2kg mass will collide with the earth before the 1kg mass.
The explanations are as follows:-
Let the mass of the earth be M and assume that the earth generates a gravitational field of strength g.
Suppose we drop an object of mass m from a height of 1m above the earth in a sealed vacuum. Let the gravitational field strength of the object be g'.
The force of the earth on the object will be F(e) = mg.
The force of the object on the earth will be F(o) = Mg'.
According to Newton's third law of motion F(e) = F(o) and hence mg = Mg'
The displacement of both the earth and the object may be calculated using
s = ut + 1/2 x a x t x t
Let the earth move X metres toward the object. Hence, in time t (time for collision) the object will move 1 - X metres.
For the earth:- X = 0 x t + 1/2 x g' x t x t
For the object:- 1 - X = 0 x t + 1/2 x g x t x t
Eliminating X between the two equations (just add them together) we get :-
t x t = 2 / (g + g')
Hence, by taking the square root of both sides we can find t (the time taken for the object and earth to collide). Notice that the time to collide is a function of g and g'.
PREDICTION NUMBER 1
If we increase the mass of the object to 2m its field strength will increase to 2g'. Gravity is a vector quantity. Hence, when we drop masses m and 2m one metre above the earth the total field strength, generated by the two masses, is 3g' (both fields are acting in the same direction). The net field acting on the earth is 3g'. Hence, the earth falls towards both masses at the same rate. The earth's field acting on both objects is g. Hence, both masses fall towards the earth at the same rate. The time to collide is the same for each mass and is given by
t x t = 2 / (g + 3g')
PREDICTION NUMBER 2
Dropping mass m by itself will result in a time of collision equation of:- t x t = 2 / (g + g').
Dropping mass 2m by itself will result in a time of collision equation of:- t x t = 2 / (g +2g').
Hence, the 2kg mass hits the ground in less time than the 1kg mass. The earth falls at a faster rate toward the 2kg mass compared to the 1kg mass because of the greater field strength generated by the 2kg mass.
It might be argued that for small masses the time difference is not measurable. This, I say, is irrelevant.
PREDICTION NUMBER 3
Suppose masses m and 2m are dropped simultaneously, in a vacuum, from a height of 1 metre on opposite sides of the earth. Mass m generates a field of strength g' and mass 2m generates a field of 2g'. Hence, the earth will fall faster toward the 2kg mass resulting in a lesser collision time.
SUMMARY
It should now be quite evident that the only instance when two objects of differing masses, released simultaneously 1 metre above the ground in a vacuum, will hit the ground at the same time is if they are in close proximity to each other.
Anyone who views force as a push or pull might have problems following the above. I suggest viewing my Blog regarding the concept of Force.
