Question

If I drop 2 objects on earth, with different masses which one hits the ground first? (let one of the object have mass twice as that of the earth)

Answer 1

if you exclude any friction between the air and the objects, both of them will hit the ground at the same time as time in free drop does not depend on mass of the objects.

Answer 2

If one of the object have mass twice as that of the earth, according to F=G(M1M2/r^2)
the attraction force that they act on each other will be very very large, and since the object have a mass twice of earth it have a inertia twice of earth thus earth will move twice as fast to the object than the object is moving to earth.

Answer 3

the force acting on the body would be twice but the acceleration of the body in the free drop would be the force acting on the body divided by the mass of the body, hence the mass terms cancel out and so a body would accelerate irrespective of its mass.

Answer 4

That answer to the question is not as simple as it first appears. there are many factors to consider. The height, the density of the objects, the altitude and the aerodynamics.

If the items are dropped from relatively low heights the aerodynamics are not likely to be of much effect. That said, it is significantly dependant on the objects. For example a single feather has a very high coefficient of drag and also very low density so that would reach its "terminal velocity" far sooner due to the drag of the air around it.

Any object will accelerate on (above) earth at a rate of 9.81metres per second for every second of drop. This assumes a vacuum and no air resistance. As everyone who has peddled a push-bike will be aware, the faster you go the stronger the wind....This is also true for an item in free-fall. The faster it travels through the air, the greater the wind resistance it experiences. eventually, the increase in acceleration will be cancelled out by the air-resistance that is produced as a result. Terminal velocity is achieved when these forces ballance. (every action has an equal and oposite reaction)

If the item's fall from height is sufficinet to allow the air resistance to become a factor, then the denser item will hit first (assuming the cofficient of drag is comparable) as its potential energy (mgh) will be greater than the less dense item.

Again that said, you could have a low drag item that is less dense hitting sooner than a high density item with a high coefficient of drag.

Answer 5

Ooops, just read the mass item....that really does complicate things! Now we also have to consider whether the item is also in solar orbit similar to that of the earth, and aproaching from this solar orbit. The attraction of the Earth to the other object as - with this item having TWICE the mass of the Earth, its inertia would be greater and then the Earth would be pulled out of its orbit towards this "alien" mass...not the other way round...and then you have to ask the question on "what constitutes dropping" as - in this particular instance, the heavier "dropped" mass would have a gravitational pull significantly greater than that of the Earth and actually draw the Earth towards it!!!

The trajectories of these items, rate of spin, amterial constitution at point of impact would have to be taken into account as would the effective Solar gravity and that of the surrounding planets in order to determine the actual outcome following the collision. this effectively starts to sound a bit like a game of cosmic billiards!

Answer 6

this is a very simple concept...provided no friction...bodies take equal time to fall to the ground...no matter what their masses are...

Answer 7

If you neglect the viscous friction with the air. And you drop them from low altitudes .
It's obvious that they will hit the ground in the same time.
As, a=-g => v=-gt (supposing you drop them with no initial velocity) => y=-1/2 gt²
Therefore,\[tf=\sqrt{2h/g}\] and that's independant to the the mass.

Answer 8

Freefall is mass independent.

Say we have M1=x kg and M2 have a mass 2x kg at a height of 100 m
Both of the objects will be under the same acceleration 'g'

Keeping it simple, let's use the equations of Motion.
S=ut+1/2 at^2 [The second Equation]
Deriving the Time Factor
(ut=0 because object under free fall has 0 initial velocity)
So, We get,




Since you see there is no mass concerned with the equation, both the objects will cover the same distance in the same time (Given the Air Resistance is Negligable))