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) = - xsòx F(x) dx = - xsòx -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.
Solving this for v gives:
v(x) = [(2/m)(E - V(x)]1/2 = dx/dt
toòt dt = xoòx dx / [(2/m)(E - V(x))]1/2 = (m/2)1/2 xoòx dx / [E + GMm/x]1/2
Let's now take a -E1/2 out of the denominator (remember that -E is a positive number) to get (with to=0):
t =
(-m/2E)1/2 xoòx dx / [-1 - GMm/Ex]1/2
Notice that V(x) must be equal to or less than E. Since V(x) is negative, and E is negative, this means that -V(x) >= -E. This means that -GMm/Ex >= 1 and so
-Ex/GMm <= 1.
Let's try this weird substitution: let cos2(q) = 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.)
We need to find dx in terms of dq, so we take the differential of both sides of the above expression:
{d
[-Ex/GMm ] / dx }dx = {d [cos2q] / dq} dq , or
-(E/GMm)
dx = -2 cos(q) sin(q) dq , or
dx =
(2GMm/E ) cos(q) sin(q) dq .
Further, since cos2(q)
= -Ex / GMm , we have for q:
q = cos-1 [-xE/GMm]1/2 .
Our expression, t = (-m/2E)1/2 xoòx dx / [-1 - GMm/Ex]1/2 now becomes:
t
= (-m/2E)1/2 qoòq {(2GMm/E) cos(q) sin(q) dq } / [-1 + (1/cos2(q)]1/2 .
Multiplying numerator and denominator in the integral by cos(q) gives:
t =
(-m/2E)1/2 qoòq {(2GMm/E) cos2(q) sin(q) dq } / [-cos 2(q) + 1]1/2 .
We recognize [-cos 2(q) + 1]1/2 = ± sin(q), so our expression simplifies to:
t = ±
(-m/2E)1/2 qoòq {(2GMm/E) cos2(q) dq } where q = cos-1
[-xE/GMm]1/2 .
To
integrate qoòq cos2(q) dq , we use the trig
identity: cos2q = ½ [1 + cos(2q)] .
qoòq cos2(q) dq = qoòq ½ [1 + cos(2q)] dq =
½ [q + ½ sin(2q)] - ½ [qo + ½ sin(2qo)] .
Putting this altogether gives:
t = ±
(-m/2E)1/2 (GMm/E) { [q + ½ sin(2q)] - [qo + ½ sin(2qo)] }
where q = cos-1
[-xE/GMm]1/2 and qo = 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 qo gives:
qo = 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) { [q + ½ sin(2q)] } where q = cos-1 [x/xo]1/2 .
When
x = xo, t = 0 as it
should. 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).
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) { [q + ½ sin(2q)] } where q = 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.