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.  Also keep in mind that V(x) is also a negative number.

Solving this Conservation of Energy equation 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 (-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 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 ; so

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   .

Limiting cases:

When x = xo,  q =0 and so 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).

For x just below xo, q is small, and sin(2q) is approximately 2q, so our solution becomes

t     ± (xo/2GM)1/2 (-xo) { [q + ½ (2q)]  }   =   ± (xo/2GM)1/2 (-xo) (2q)  =     ± (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(2q) 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) { [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]1/2 + ½ sin {2*cos-1 [6.400 x 106 m / 6.464 x 106 m]1/2 }1/2 )

=  580.896 sec * {.099669  + .099010}  =  115.41 seconds.

The above result from the function is between the two times, and it is closer to the longer time which corresponds to the acceleration at the higher location.  This makes sense since the object is falling slower at the higher height and hence spends more time at the higher height.