The weight at the height of the Earth's surface is Mg and the aerodynamic drag of an object is proportional to its speed squared. Setting up Newton's equations
$$ mg-kv ^ 2 = ma $$
When the dimensions of two test objects are equal ## k ## is a constant.
When the equilibrium of forces is reached, the acceleration ## to ## becomes zero or it descends with a constant velocity, whose value is obtained by solving
$$ mg-kv ^ 2 = 0 $$
$$ mg = kv ^ 2 $$
$$ v = \sqrt {\dfrac {mg}{k}} $$
According to this, the more massive, the higher terminal velocity, then it falls faster.
But as I said this is only true if k is constant (see drag coefficient),
Massive objects do not always fall first,
• A 747 airplane has more mass than a car, however both launched from a certain height and with horizontal speed, the car will fall first.
• An empty balloon falls before the same inflated balloon of the same mass ...
• A satellite with a certain tangential velocity does not fall, no matter how much its mass is ...
alan123hk said:
Since seeing this thread yesterday, I have been trying to derive the time equation for the collision of two masses due to Newtonian gravity. Unfortunately, this seems to be much more difficult than I thought before, so I haven't made much progress yet.
To know the time of fall of an object, taking into account that gravity varies with height and in the absence of friction (this almost leaves the thread) is
We already know the modulus of acceleration of gravity falls with the square of the distance to the center.
$$ a_g = \dfrac {GM_T}{r ^ 2} $$
It is logical to think that the relative acceleration with respect to the surface for an object at a higher altitude is less than one at a low altitude. these types of corrections are made by means of differential equations.
Let's see the simple calculation of frictionless free fall, An object of mass ## m ## is dropped without initial velocity from a height ## r_o ## with respect to a massive body of mass ## M ##
Using the conservation of energy you have a first order differential equation that is relatively easy to integrate.
$$ \displaystyle \frac {1}{2} v ^ 2-G \frac {M}{r} = - G \frac {M}{r_0} $$
from where
$$ v = - \displaystyle \frac {dr}{dt} = \sqrt {2 GM} \sqrt {\frac {1}{r} - \frac {1}{r_0}} = \sqrt {\frac { 2 GM}{r_0}} \sqrt {\frac {r_0-r}{r}} $$
if we try to integrate
$$ \displaystyle- \int_{r_0} ^ r {\sqrt {\frac {r}{r_0-r}} dr} = \sqrt {\frac {2 GM}{r_0}} \int_0 ^ t {dt} $$
The integral on the left is a bit heavy to perform. My source, friend from another forum has made, the change of variable $$ u ^ 2 = r / (r_0-r) $$.
Finally
$$ \displaystyle- \int_{r_0} ^ r {\sqrt {\frac {r}{r_0-r}} dr} = r_0 \left (\frac {\pi}{2} - \arctan \sqrt {\frac {r}{r_0-r}} + \frac {\sqrt {r} \sqrt {r_0-r}}{r_0} \right) $$
then
$$ t (r) = \displaystyle \sqrt {\frac {r_0 ^ 3}{2GM}} \left (\frac {\pi}{2} - \arctan \sqrt {\frac {r}{r_0-r }} + \frac {\sqrt {r} \sqrt {r_0-r}}{r_0} \right) $$ (this makes sense as long as ## r ## is greater to the radius of the earth).