Orbital Mechanics - Part 3
This derives an expression used in part 2 of the series that started here.
Today, I will be going through the reasoning that lead to Newton’s Law of Gravitation. The previous parts are linked at the bottom of this post.
First, imagine an object falling near the surface of Earth. We are going to neglect air resistance.
The acceleration of an object near the surface is 9.8 m/s², (or 32 ft/s² if you prefer daft units).
The distance fallen by the object is given by s = ut + ½at², where u is the initial velocity, and s is the displacement (distance with direction).
If we drop an object, u=0, so the distance fallen is s = ½at². This is usually written as s = ½gt² as g is the symbol used for the acceleration due to gravity.
After 1 second the object falls 5 m, after 2 seconds, 20 m and so on.
This is true even if the object moves sideways.
The pattern of falling
If an object moves sideways, the pattern of falling is the same as an object dropped directly (if air resistance is unimportant). This means the path becomes curved.
The faster the horizontal speed, the further the object can go before it hits the surface. But the surface is curved… so this gives us an extra distance before impact due to the curvature.
The curved surface of the planet gives extra distance - and the faster you are, the bigger the effect.
If the ‘sideways’ speed is high enough, the curvature of the path is so slight that the object does not hit the surface, and we have an orbit.
Let’s imagine the speed is enough that we have an orbit, and let’s look at a one second ‘slice’ of the motion.
We are imagining the satellite is at an orbit of radius ‘R’. Let’s say that R is enough that we are just above the atmosphere, let’s say 200 km about the surface. As the radius of the planet is about 6400 km, R is 6600 km, or R = 6.6 × 10⁶ m
Thus, we can say that R cos 𝜃 = R - 5
You should verify the calculation for yourself, but this gives 𝜃 = 0.07° in one second (I have done all these calculations without rounding off, but I have rounded off any results).
Given a full orbit is 360°, this means that it will take 5100 seconds for one orbit, or 85 minutes. For comparison, the ISS orbits in 93 minutes (it is a little higher at about 400 km above the surface), Sputnik orbited in 98 minutes, (it went from 230 km to 950 km above the surface).
We can also work out the speed that the object needs to have. The distance labelled s on the diagram is given by R × 𝜃 if 𝜃 is in radians or R × 𝜃 × 𝜋 ÷ 180 if we are still in degrees. This means that ‘s’ is 8120 m, or the speed is about 8 km/s.
As a double check, as 𝜃 is small, x ≅ s, so if we calculate R × sin 𝜃 we should also get 8 km., and we do.
Now, let’s re-use our diagram for the moon. We know the moon is 60 Earth radii away (this is something the ancient greeks established, perhaps a story for another day). So R = 384000 km.
We know the moon orbits once in 27.3 days (due to the motion of the Earth and moon around the sun, it is a few days longer to go from full-moon to full-moon).
We know that 𝜃 = 0.07° (remember that I am using the unrounded value). This then means the time in days taken for the moon to move through the diagram is 27.3 × 0.07 ÷ 360. This is 5.34 × 10⁻³ days or 460 seconds.
We can also calculate how far the moon ‘falls’ under the Earth’s gravity by finding ‘y’.
R - y = R × cos 𝜃
so y = R × ( 1 - cos 𝜃 ) = 0.29km = 290m
So, we can now find the gravitationational field that the Earth exerts on the moon.
Looking up to the top of this post, s = ½gt²
so g=2s ÷ t²
g = 2 × 290 ÷ 460²
g = 2.7 × 10⁻³ m/s²
This is roughly 3600 times lower that the value of g at the Earth’s surface (close to 10 m/s²).
Now, the moon is 60 earth radii from the centre of the Earth, and 60² is 3600.
Thus it looks like g varies with 1/r²
If we take this at face value and assume that the gravitational field of the sun varies in the same fashion, we can derive the already-known Keppler’s laws for planetary motion. this provided confirmation.
What has not been done is find M, the mass of the big object, or G, the gravitational constant. Newton never knew G or M, but neither is needed to do calculations where you compare one object to another (e.g. compare the motion of mars around the sun to the Earth)
Links
Image used as thumbnail (public domain)
More Physics
More Space
.jpg){kind=link}