I Dropped objects hitting the ground at the same time?

[OldYat47]

Of course not. Collisions are simultaneous events. And there is a basic fault in my original logic. I'm pretty rusty, it seems. I guess I should be consoled that I caught it first.

A true comparison is when the force and distance are the same but the masses of the objects change. That is, the product of mass 1 and mass 2 is the same in both cases.
So let m1 = 2 and m2 =2.
F = (G*4)/d^2, the acceleration of each mass is (G*2)/d^2, and
A closing = (G*4)/d^2

Now let m1 = 1 and m2 = 4
F doesn't change, = (G*4)/d^2
But the accelerations do change:
Am1 = (G*4)/d^2
Am2 = (G)/d^2, and
A closing = (G*5)/d^2
So in cases where the gravitational force is equal, where masses are equal closing acceleration is lower than where masses are unequal. The greater the difference in masses, the greater the closing acceleration.

 

[Tom Kunich]

okay, so I’ve had this random thought. We have all been told that objects fall to the ground at the same speed, even if they have different masses. While it’s true that any two objects, regardless of mass, will accelerate towards Earth at the same speed, that doesn’t mean the Earth is accelerated towards the object at the same speed. Heavier objects have a higher gravitational force, and because of Newton’s Third Law, the Earth expirences the same force towards the object. Because Earth will accelerate towards heavier objects faster, heavier objects dropped in a perfect vacuum will hit the ground before a lighter one.

Gravity is different and can be calculated as GMm/r^2 where G is the gravitational constant, M is one mass, m the other and r the distance between them. Because in the real world the ratio of M to m is so huge you cannot measure differences of acceleration in a gravitational field with the skin friction of an atmosphere.

 

[OldYat47]

Here's some numbers expanding on what Mr. Kunich said above. What happens if you have one really big mass and two much smaller unequal masses at the same height?
Suppose we have two object with masses m1 and m2 = 4*m1. We have a third object with mass m3 = 10^24*m1. The objects are spherical, m1 and m2 are some distance d above the surface of m3, and there is no atmosphere.
We drop m1 onto m3.
Attractive force = [(G * (10^24 * m1) * m1)] / (d^2) or
Attractive force = (G * 10^24 * m1^2) / (d^2)

Acceleration of m1 = (G * 10^24 * m1) / (d^2)
Acceleration of m3 = (G * m1) / (d^2)
Closing acceleration (m1 to m3) = [G * m1 * (10^24 + 1)] / (d^2)

We drop m2 onto m3.
Attractive force = [(G * (10^24 * m1) * (4 * m1)] / (d^2) or
Attractive force = (G * (40^24 * m1^2) / (d^2)

Acceleration of m2 = (G * 10^24 * m1) / (d^2)
Acceleration of m3 = (G * 4 * m1) / (d^2)
Closing acceleration (m2 to m3) = [G * m1 * (10^24 + 4)] / (d^2)

So m2 and m3 would collide in less time than m1 and m3, but the difference in time is tiny (certainly beyond our capacity for measurement) until d gets quite large, and m3 moves a relatively tiny amount. This is different from my previous example because the attractive forces are different in this case.

 

[Sudha Sarita]

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.

 

[OldYat47]

Sudha Sarita, no, they don't hit the ground at the same time, the heavier object hits first. The time difference is immeasurably small, but it's there.
The algebra is not complex, but perhaps I could have included the non-simplified factors.
We have two objects, m1 and m3. m3's mass is 10^24 times m1's mass or (10^24 * m1). The attractive force is [G * m1 * m3 / d^2].

So the attractive force = G * [(10^24 * m1) * m1] / (d^2) = G * (10^24 * m1^2) / (d^2)

The acceleration of m1 is the attractive force divided by m1 = [G * (10^24 * m1^2) / (d^2)] / m1 = G * (10^24 * m1) / (d^2) = (G * m1 / d^2) * (10^24)

The acceleration of m3 is the attractive force divided by m3 = [G * (10^24 * m1^2) / (d^2)] / (10^24 * m1) =
(G * m1 / d^2) * (1)

The closing acceleration is the sum of those two accelerations, (G * m1 /d^2) * (10^24 + 1)]

Similar for m2 and m3.

 

[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.