## Monday, November 28, 2011

### Problem

Compute the limiting velocity vl of impact for a body in free fall toward the earth, neglecting any drag due to atmospheric friction.

### Solution

#### A hint from dimensional analysis

Assuming that such a limiting velocity (vl) exists and it depends only on the earth radius (r) and the value of gravitational acceleration on the earth surface (g), a bit of dimensional analysis will suggest how these quantities relate. I chose to operate in the mks system, which is the one preferred by all physicist. [ ] is an operator that is spelt "the dimension of":

[v] = [l] / [t] = m/s

[g] = [a] = [v] / [t] = m/s^2

It seems plausible that:

[vl] = [g]^a [r]^b

that is vl depends both on the surface value of g and the radius of the planet r. It depends on r because the larger the value of r is, the more slowly does the value of g fade off with distance relatively to the g value on the earth surface and being everything else - like the planet mass - kept constant. On the other hand vl cannot depend on the mass of the falling body, since there is no mass unit on the left-hand side. That is there is no mass unit in a speed unit: the dimensions of speed is the dimension of length divided by the dimension of time.

m/s = (m/s^2)^a (m)^b

Balancing dimensions we get a very simple linear system:

{ a + b = 1
{ 2a = 1

{ a = 1/2
{ b = 1/2

Therefore we just need to confirm the following result and determine the unknown constant k:

vl = k sqrt(gr)

#### Demonstration using Newton's laws

Newton's second law, which is a general law valid for any kind of force, states that the gravitational force exerted by the earth upon a body is:

F = m a = m g

being the body's mass m a constant (at least for speeds far less than the speed of light). On the contrary, both F (the body's weight) and g (the gravitational acceleration) are variable when large distances are involved. Indeed, for determining the limiting velocity vl we have to drop a body of mass m from the outer space, ideally (or mathematically) from an infinite distance, assuming the earth gravitational field extends all over the space. As it does, in principle, although it gets weaker and weaker with distance. We can easily see that from another Newton's law, the law of universal gravitation:

F = G me m / d^2

In the mks system, G is a constant approximately equal to 6.67 10^-11 m^3/kg-sec^2 and me is the earth mass, about 5.9742 10^24 kg, both first determined by Henry Cavendish in his famous experiment using a torsion balance. d is the distance from the centre of the earth and is equal to the earth radius if we are on the earth surface. The earth radius is known with remarkable accuracy since about the 2nd century BC. It was Eratosthenes of Cyrene who measured it using purely geometric methods. Combining these two laws we get an expression for g in function of the variable d, the distance of the body from the centre of the earth:

g(d) = G me / d^2

Here we have divided through by m and we can since m is a non-null constant. We will see that the velocities involved aren't so big to cause masses to change appreciably according to Einstein's equations.

Therefore we have an acceleration g that varies inversely as the square of the distance (d) from the earth's center. We can say the same for the weight of the body, viewing F in the Newton's law of gravitation as a function of d:

F(d) = m g(d) = G me m / d^2

Now recall that in mechanics the general definitions of velocity and acceleration as function of time are expressed using derivatives:

v(t) = ds/dt

a(t) = dv/dt

For our problem, it seems quite convenient to establish a one-dimensional coordinate system with origin at the centre of the earth and running through the point D where we let the body drop. In this system d=s (that is the spatial coordinate is exactly the distance from the centre of the earth) and the previous formula for the acceleration becomes:

g(s) = dv/dt

Note g(s) in this reference system is always negative for any value of s that makes sense:

- G me / s^2 dt = dv

Now, let's plug in v(t) = ds/dt to eliminate dt so we don't have any time dependency:

- G me / s^2 ds = v dv

This is a very easy differential equation we have to solve. The solution is a function v of s, or v(s) which tells us how velocity changes during the free fall. We have to integrate both sides and apply initial and final conditions. Namely on the right-hand side the free variable is v and the integration interval will be from 0 (since the body starts to drop from rest, the initial velocity is 0) to vl, which is the limit velocity we need to find. We will find its value is negative because of our x-axis orientation, which is always opposite to the body motion (so that ds's are negative). On the left-hand side the integration variable is the space or distance (s) and the lower limit is d since the body starts from a very far away point D of abscissa d, whilst the upper bound is the radius of the earth r. Ideally, the body cannot go lower than that because we are supposing the earth as perfectly rigid.

We also need a couple of simple indefinite integration formulae from calculus:

Int 1/x^2 dx = - 1/x + k
Int x dx = x^2/2 + k

These formulae are just the inverse of another couple of differentiation rules:

d(-1/x + k)/dx = 1/x^2
d(x^2/2 + k)/dx = x

which in turn are just variations on these two derivative properties:

dk/dx = 0 if k does not depend on x
d(x^n)/dx = n x^(n-1) valid for any natural number; when n=0, it becomes the previous rule

You see that, ultimately, we are applying only one property here, the derivative of the power function.

Solving the integrals we get:

(vl)^2/2 - 0^2/2 = (G me / r) - (G me / d)

To find an exact value for vl we have to take the limit as d approaches infinity. We simply have that G me / d becomes zero and we are left with:

vl^2 = 2 G me / r

if we now multiply both the numerator and the denominator of the right-hand side by r we recognize that G me / r^2 equals the gravitational acceleration on the earth surface. We call this simply g, practically a constant.

vl^2 = 2 g r

Solving for vl we have two values:

vl = +- sqrt(2 g r)

Because of how we chose our reference system, we should discard the positive value. Anyway we can always invert the orientation of our coordinate system at the last minute and take the positive value as the only good one instead:

vl = sqrt(2 g r)

This tells us that the k constant from the dimensional analysis above is equal to sqrt(2).

Using the values 9.8 m/s^2 for g and 6378000 m for r we get a speed limit of about 11181 m/s or 11 km/s or (just multiplying by 3600) 40250 km/h. Although it seems to be a very big speed, it is still about only the 0.0037% of the speed of light, confirming that this problem can correctly be treated using classical mechanics.

As a comparison, the average speed a passenger airplane flies at is less than 1000 km/h, but the new US unmanned military plane can reach about 21000 km/h!