Gravitational force on objects

CAF123

If I take a feather and a rock and drop them at the same time, because of the effect of the gravitational force, we know that the rock will hit the ground first. (The rock experiences a larger gravitational force).

My question is if these objects are dropped at the same time and vertically, then the velocity can be described as [itex] v = gt [/itex], for both objects. Therefore, at [itex] t = 1, v = 9.81 m s^{-1} [/itex] and at [itex] t = 2, v = 19.6 m s^{-1} [/itex]. However, i find it difficult to believe the feather will have this velocity. So how does the differences in gravitational force be accounted for in this equation?

Many thanks

Doc Al

The reason that the rock hits first is not because of its greater gravitational force. If only gravity acted, then both would have the same acceleration = g. But other forces are involved, in particular air resistance. Take away the air and the feather and rock will fall together.

CAF123

I see, many thanks.
So essentially the formula only works when considering zero air resistance

CWatters

yes and no...

Just before an object is dropped it has some energy lets call it E_start where

E_start = PE_start + KE_start

Since it's stationary KE_start = 0 so

E_start = PE_start = mgh

When it hits the ground it has some energy, lets call it E_end then

E_end = PE_end + KE_end

but PE_end = 0 so

E_end = KE_end = 0.5mv²

If we assume there is no air resistance to be overcome (or any other way for energy to leave the system) then the law of Conservation of Energy says

E_start = E_end

or

mgh = 0.5mv²

Notice how the mass cancels and the velocity of impact is independant of mass

v = SQRT(2gh)

CAF123

And if it takes [itex] t [/itex] seconds to cover this distance [itex] h [/itex], either [itex] v = gt [/itex] or [itex] v = \sqrt{2gh} [/itex] can be used to determine this impact velocity, I presume?

CWatters

That's what i said.

Morgoth

you have two objects of mass M and m, with M>m .
if you let them fall, you get from Newton's 2nd law:
M a1= - M g
m a2= - m g
from that you see immediately that a1=a2
which means the speed of the mass M will change by the same ammount in the same time interval as will the speed of mass m.

The mass of the falling object, does not appear in your motion equations. It doesn't affect it.
Even if I didn't use the "static model", that the grav force is mg, but used the force:
F= G M' m /r (M':Earth's Mass)
the masses of the falling objects would again drop out.
Of course that is not obvious, it comes thanks to the equivalence principle, which says that the mass that appears in ma, and the mass that appears in m* (GM'/r) is the same
m=m*