Free Fall from a great height without air resistance
We can make this falling problem more doable by removing air resistance. But by ignoring air resistance,
this problem is not very realistic for the earth. It should be a lot closer to realistic for the moon where there
is essentially no air.
To make the problem more realistic for the earth, we would have to include an air resistance term into
Newton’s 2nd law. To make the air resistance term more realistic, we would have to take away the
assumption that the density of air is constant (and thus the coefficient of air resistance is not constant) for
great heights where g deviates from its 9.8 m/s2 value at the earth’s surface. As a first approximation,
assuming the mix of gases in the atmosphere remains the same and that the temperature of the air remains
the same, then the air density (and hence the air resistance coefficient) would decrease exponentially with
height. (We’ll look more closely at this near the end of the course.) Newton’s 2 nd law would then look
like: -GMm/(r+y)2 – (1/2 C ro e-ay) (dy/dt)2 = m (d2y/dt2) . This doesn’t look like it would be easy to
solve.
Free Fall from a great height without air resistance
Fgravity = -GMm/x2 = F(x)
To follow the procedure outlined for F(x), we define a potential energy:
V(x) = - xsx F(x) dx = - xsx -GMm/x2 dx = -GMm/x + GMm/xs .
Thus the first integral was relatively easy to do!
If we choose xs to be infinity, then GMm/xs = 0 so V(x) = -GMm/x .
Next, we use Conservation of Energy to relate v to x:
½mv2 + V(x) = ½mvo2 + V(xo) = E = constant.
Note that if the object does not have enough energy to reach infinity, then E < 0. Keep
this in mind that E is a negative number in what follows. [Here is an analogy to help with
negative energies: think of money. Can you have negative money? Not with cash, but
you can with debt. Having negative gravitational energy is like having an energy debt to
the earth. In order to escape from the earth, you need to add energy to reach a zero debt
which would be a zero energy. If you have less debt, you are richer; that is, if you have
less negative gravitational potential energy, you have more total energy (but the total
energy may still be negative).]
Solving this Conservation of Energy equation for v gives:
v(x) = [(2/m)(E - V(x)]1/2 = dx/dt
to
t
dt =
xo
x
dx / [(2/m)(E - V(x))]1/2 = (m/2)1/2 xox dx / [E + GMm/x]1/2
Let's now take a (-E)1/2 out of the denominator (remember that E < 0 so -E is a positive
number) to get (with to=0):
t = (-m/2E)1/2 xox dx / [-1 - GMm/Ex]1/2
Notice that V(x) must be equal to or less than E (since KE>0, E > V(x) ). Since V(x) is
negative, and E is negative, this means that -V(x) >= -E. This means that -GMm/Ex >=
1 and so 1/(-GMm/Ex) <= 1.
Let's try this weird substitution: let cos2() = 1 / (-GMm/Ex) = -Ex / GMm .
By doing this, we can make the denominator in the integral simpler, and hopefully make
the integral itself simpler to integrate: [-1 - GMm/Ex]1/2 = [-1 + (1/cos2()]1/2
= [{-cos2() + 1}/cos2()]1/2 = [sin2()/cos2()]1/2 = sin()/cos() .
We need to find dx in terms of d, so we take the differential of both sides of the above
expression: df((x))/dx = df()/d)*(d/dx), or {df(x))/dx}*dx = {df()/d}*d , so
{d [-Ex/GMm ] / dx }dx = {d [cos2] / d} d , or
-(E/GMm) dx =
-2 cos() sin() d or
dx = (2GMm/E ) cos() sin() d
Our expression, t = (-m/2E)1/2 xox dx / [-1 - GMm/Ex]1/2 now becomes:
t = (-m/2E)1/2 o {(2GMm/E) cos() sin() d / [sin()/cos() ] ,
and our expression then simplifies to:
t = (-m/2E)1/2 o {(2GMm/E) cos2() d
where  = cos-1 [-xE/GMm]1/2 .
To integrate o cos2() dwe use the trig identity: cos2 = ½ [1 + cos(2)] .

o
cos2() do ½ [1 + cos(2)] d½ [ + ½ sin(2)] -½ [ + ½ sin(2)] .
Putting this altogether gives:
t = (-m/2E)1/2 (GMm/E) { [ + ½ sin(2)] - [ + ½ sin(2)] }
where  = cos-1 [-xE/GMm]1/2 and  = cos-1 [-xoE/GMm]1/2 .
We have t(x), but you can see that it does not look promising to find x(t).
Special Case: vo = 0 (dropped)
From Conservation of Energy, we have:
½mv2 + V(x) = ½mvo2 + V(xo) = E = constant
If vo = 0 (object is dropped), E simplifies to: E = V(xo) = -GMm/xo . Putting this into
our expression for o gives:
 = cos-1 [-xoE/GMm]1/2 = cos-1 [-xo{-GMm/xo}/GMm]1/2 = cos-1 [1]1/2 = 0,
Therefore, our expression for t simplifies to:
t = ± (xo/2GM)1/2 (-xo) { [ + ½ sin(2)] } where  = cos-1 [x/xo]1/2 .
Note that the ± sign indicates that the time going up is symmetrical to the time going
down (remember solution is for the case of no air resistance).
Limiting cases:
When x = xo, t = 0 as it should.
For x just below xo,  is small, and sin(2) is approximately 2so our solution becomes
t ≈ ± (xo/2GM)1/2 (-xo) { [ + ½ (2)] }
= ± (xo/2GM)1/2 (-xo) (2) =
± (2xo/go)1/2 cos-1 [x/xo]1/2
where go = GM/xo2 = the acceleration due to gravity at the initial height. Proceeding
using only the + sign, and expanding cosine in a power series:
x/xo = cos2[t(go/2xo)1/2] ≈ {1 – ½ [t(go/2xo)1/2]2}2
= {1 – ¼ t2go/xo}2 = 1 - ½t2go/xo + (1/16)t4go2/xo2), or
x ≈ xo – ½ got2 ,
which is the case for constant acceleration using the value of the acceleration due to
gravity at the initial height.
It looks like the next order correction is positive which would reduce the amount of fall,
but keep in mind that we only kept terms to first order when we expanded sin(2) earlier,
so if we really want to see the next order correction, we would have to include the next
order term in that expansion also.
Numerical check:
This answer looks strange. Is it correct? We can put in a trial distance, and check the
time. The correct time should be larger than the time to fall with constant gravity at the
earth’s surface, g = -GMm/Re2 = -9.8 m/s2 , and it should be smaller than the time to fall
with constant gravity at the initial height, go = -GMm/(Re+xo)2 .
Let’s try h = xo – Re = 64 km = 6,400 m.
Using gravity at the earth’s surface (with Re = 6,400 km) gives ge = (6.67 x 1011
Nt*m2/kg2) * (6.0 x 1024 kg) / (6.4 x 106 m)2 = 9.77 m/s2 . Now using the equations for
constant acceleration, we have x = xo + vo*t + ½ a*t2 or h = ½ gt2 , or t = [2h/g]1/2 =
[2*6.4x104 m / 9.77 m/s2]1/2 = 114.46 seconds.
Using gravity at the release point (with r = Re + h = 6,464 km) gives go = (6.67 x 1011
Nt*m2/kg2) * (6.0 x 1024 kg) / (6.464 x 106 m)2 = 9.578 m/s2 . Now using the equations
for constant acceleration, we have x = xo + vo*t + ½ a*t2 or h = ½ gt2 , or t = [2h/g]1/2
= [2*6.4x104 m / 9.578 m/s2]1/2 = 115.60 seconds.
The derived formula should give a time between 114.46 seconds and 115.60 seconds.
Let’s see if it does:
t = ± (xo/2GM)1/2 (-xo) { [ + ½ sin(2)] } where  = cos-1 [x/xo]1/2
t = ± (6.464 x 106 m / [2 * 6.67 x 10-11 Nt*m2/kg2 * 6.0 x 1024 kg] )1/2 * (- 6.464 x 106 m)
*{ cos-1 [6.400 x 106 m / 6.464 x 106 m] + ½ sin { cos-1 [6.400 x 106 m / 6.464 x 106 m]})
= 580.896 sec * { .09967 + .09901} = 115.41 seconds.
Because the mass is moving slower at the higher locations, it spends more time at the
higher locations, and we might expect the actual time to be closer to the 115.41 seconds
than the 114.46 seconds – which the 115.41 seconds is.