Orbital motion "explained".

One of the first things Newton's equation of motion F=ma leads to is how far (h) a body falls in time t when a is constant, namely one-half gee tee-squared (h=½gt2) where gee (g) is the local acceleration of gravity, about 9.8 metres per second per second (g=9.8ms-2). So in the first second a freely falling body falls about 4.9 metres and is moving 9.8 ms-1.

Now let us find a critical horizontal velocity vc sucn that in one second a body travelling with speed vc will move a distance x such that the earth's surface will fall 4.9 meters, the same distance the body will fall:

On our spherical world if your eyes are at a height h above the surface (h is -say- 4.9 metre) you can see the surface out to a distance x, the horizon distance or "dip". Pythagoras could work this out, just solve the right triangle whose sides are Earth's radius R=6378000m and horizon distance x and whose hypotenuse is (R+h) vis

Pythagorean theorem R2+x2=(R+h)2
algebraic expansion of (R+h)2 R2+x2=R2+2Rh+h2
subtract R2 from both sides x2=h2+2Rh

Now for the trick. Ignore h2 since 2Rh is much greater than h2 (2Rh>>h2) leaving

x2=2Rh.

If a body has a horizontal speed of x metres per second then both it and the surface will fall a distance h, the body is in circular orbit! Substituting 4.9 for h and 6378000 for R we find x=7900 so if the horizontal velocity is 7900 metres per second it is in orbit. Remember: near earth orbital velocity is 7.9 km/s.

Now we can get a general formula by substituting ½gt2 for h, x2=2R(½gt2), or, simply, circular velocity squared vc2=(x/t)2=gR.

The acceleration of gravity sometimes called little g (g) follows immediately from Newton's law of gravitation, g=F/m=GM/R2 so we can write down the general formula for circular orbital velocity vc2=(x/t)2=gR=GM/R. That is,

vc=(GM/R)½.

Big G=6.672 times 10 to the negative eleventh power, G=6.672x10-11. The Earth's mass is 5.977x1024kg so at the surface we get vc=(6.672x10-11x5.977x1024/6378000)½=7907ms-1. Just like we just showed, but now from "first principles".

We see the circular orbital speed falls as 1/distance½. For near Earth orbits, if we measure distances A in Earth radii, the circular orbital speed is just vc=7.907A-½km/s.

How long does it take a satellite to orbit the earth? time=distance/rate=40000/7.9=5063s or about 84 minutes. As an exercise, do an accurate calculation for a satellite 200 km above the Earth's surface (R=6578000m).

The Moon is Rm=3.844x108m from the earth and has an orbital velocity of 2p Rm (the circumference of the Moon's orbit) divided by 27.32x86400 (the time in seconds for one orbit) or v=1023ms-1. The value predicted by our (Newton's) formula is...the same! The square of the orbital velocity goes as the reciprocal of the orbital distance, which observation verifies Newton's inverse square law. Copernicus and Kepler and Galileo would have recognized this fact.

Say, what if we set vc=c, the speed of light? Then we could solve for Rc=GM/c2. This is the "photon circle", a distance at which light itself is in orbit! Note that all of the mass M must be inside the photon circle for this to work, a situation never found in the solar system.

Now a little more physics phor phun.
How much work W does it take to move from a distance R from the centre of the Earth to infinity? We call this the gravitational potential W. It is the integtal (since F varies with r) of force times distance (element - dr)... W=integral(F dr) where F=GMm/r2. The work is thus (integral of 1/r2 is 1/r) GMm/r:
W=GMm/r

If a body is given enough kinetic energy ½mv2 it will just escape. Setting that critical escape velocity equal to the gravitational potential gives
v2=2GM/r.

Compare that with the circular velocity, and we see that the escape velocity is just the square root of two times the circular velocity, about 11.18km/s for the Earth.

Finally, why not take v=speed of light... nothing beats that! Set v=c then solve for r, call it RS....

RS=2GM/c2.
Of course for this to work all the mass M must be within RS and for the Sun this works out to be about 5km, pretty dense! Then even light could not escape. This Radius is called the Schwarzschild radius or "black hole" radius. We are pretty sure these things actually exist!