Dropped objects hitting the ground at the same time?

[ZapperZ]

Dropped objects hit the ground same time. Objects which are of point size with respect to Earth are attracted towards ground due to force of gravitation. Force of gravitation is given by formula F = G\frac{m_1m_E}{R^2}. When an object of mass m_1 is dropped means it has zero initial velocity. It experiences an acceleration towards Earth which is given by a = \frac{F}{m_1}. Depending on the mass of every object there can be variation in acceleration towards Earth over a wide range. Assume that all objects are dropped from the same height then on the basis of equation h = ut + 1/2 (at^2), time can vary over a range. I have not considered force due to air resistance.

This is very confusing and contradictory.

First of all, for m<<M, where M is mass of the Earth, then "g" is a constant. That is what we use in intro General Physics.

However, if you want to deal with this in general by stating that "... there can be variation in acceleration towards Earth over a wide range ...", then you can no longer use "h = ut + 1/2 (at^2)", because that kinematical equation was derived under the condition that acceleration is a constant!

Zz.

 

[OldYat47]

There can be variations in acceleration "towards Earth". Suppose we have two Earths. The mutual attraction is greater than g.
G * (Earth's mass * Earth's mass) / d^2 = g * (Earth's mass^2) / d^2.
The acceleration of each Earth = G * (Earth's mass) / d^2
The closing acceleration is the sum of both accelerations, in this case the closing acceleration = 2 * G * (Earth's mass) / d^2
If one of the objects is (Earth's mass / 2) the closing acceleration is greater. If you are sitting on "our" Earth the acceleration you observe will be different in the two cases.

 

[Ty Wendland]

There can be variations in acceleration "towards Earth". Suppose we have two Earths. The mutual attraction is greater than g.
G * (Earth's mass * Earth's mass) / d^2 = g * (Earth's mass^2) / d^2.
The acceleration of each Earth = G * (Earth's mass) / d^2
The closing acceleration is the sum of both accelerations, in this case the closing acceleration = 2 * G * (Earth's mass) / d^2
If one of the objects is (Earth's mass / 2) the closing acceleration is greater. If you are sitting on "our" Earth the acceleration you observe will be different in the two cases.

So I guess you agree with my initial hypothesis?

 

[RandyD123]

Not sure if this applies but I saw this on YouTube:

 

[OldYat47]

Yes and no. The heavier object will hit first, but Earth's movement will "contribute" only proportionally to the relative magnitudes of the masses. Imagine a planet identical to Earth, but with zero atmosphere. Take two objects identical except for mass. Take them to the same height and drop them. The more massive object will hit the "Earth2" first, but the difference in time between the two impacts will be in proportion to the mass of the objects relative to the mass of Earth2 [(In the case of my last example, The difference between (10^24 + 1) and (10^24 + 4), a really small difference]. The total travel distance of Earth2 before impact will be similarly proportional by comparative masses to the movement of the other mass.

 

[Janus]

Yes and no. The heavier object will hit first, but Earth's movement will "contribute" only proportionally to the relative magnitudes of the masses. Imagine a planet identical to Earth, but with zero atmosphere. Take two objects identical except for mass. Take them to the same height and drop them. The more massive object will hit the "Earth2" first, but the difference in time between the two impacts will be in proportion to the mass of the objects relative to the mass of Earth2 [(In the case of my last example, The difference between (10^24 + 1) and (10^24 + 4), a really small difference]. The total travel distance of Earth2 before impact will be similarly proportional by comparative masses to the movement of the other mass.

If you drop the objects individually and at different times, then the more massive object will take slightly less time to impact. However, if you drop them simultaneously, and side by side, then they will impact at the same time. The Earth cannot have two different values of acceleration over the same interval. Instead, it will accelerate towards the falling objects at a rate that is due to the sum the masses of the objects.

 

[berkeman]

If you drop the objects individually and at different times, then the more massive object will take slightly less time to impact. However, if you drop them simultaneously, and side by side, then they will impact at the same time. The Earth cannot have two different values of acceleration over the same interval. Instead, it will accelerate towards the falling objects at a rate that is due to the sum the masses of the objects.

And with this clarification restated, I think the question has been answered well. Thanks everybody for a surprisingly interesting discussion of a topic that seems obvious on the surface of it, but requires a careful problem definition to give a correct answer.