astronomyearthgeometrymoonsun
I know this seems like a simple question, but I'm trying to debate with a flat earth theorist. I asked him to explain why can the ISS visibly be seen orbiting the Earth with the naked eye, and he put this question to me instead.
He asked:
"How can a full moon be observed south of an observer's location, despite the fact that if the moon is illuminated by the sun, an observer has to be almost directly between the sun and the moon to observe a full moon?"
The Kepler orbit of the Earth around the Sun is determined by two constants: the specific orbital energy $E$ and the specific relative angular momentum $h$: $$ \begin{align} E &= \frac{1}{2}v_{r,\oplus}^2 + \frac{1}{2}v_{T,\oplus}^2 - \frac{\mu}{r}= -\frac{\mu}{2a},\\ h^2 &= r^2\,v^2_{T,\oplus} = \mu a(1-e^2), \end{align} $$ where $\mu = G(M_\odot + M_\oplus)$, $r$ is the distance Earth-Sun (at the moment of impact), $a$ is the semi-major axis, $e$ is the orbital eccentricity, $v_{r,\oplus}$ is the radial orbital velocity of the Earth, and $v_{T,\oplus}$ the tangential velocity. Now, suppose that a large asteroid collides with the Earth, with orbital velocity $(v_{T,A},v_{r,A})$ and mass $M_A$. Its relative velocity is then $$ \begin{align} v_{T,A}' &= v_{T,A} - v_{T,\oplus},\\ v_{r,A}' &= v_{r,A} - v_{r,\oplus}. \end{align} $$ We can express these relative velocities in terms of the total impact velocity $v_\text{i}$ and the impact angle $\theta$: $$ \begin{align} v_{T,A}' &= v_\text{i}\cos\theta,\\ v_{r,A}' &= -v_\text{i}\sin\theta, \end{align} $$ where I defined $\theta$ as in Fig. 1 of this article. So we obtain $$ \begin{align} v_{T,A} &= v_{T,\oplus} + v_\text{i}\cos\theta,\\ v_{r,A} &= v_{r,\oplus} - v_\text{i}\sin\theta. \end{align} $$ If we assume that the collision is central, that heat loss is negligible and that the debris remains gravitationally bound to the Earth, then conservation of momentum implies $$ \begin{align} M_\oplus\,v_{T,\oplus} + M_A\,v_{T,A} &= (M_\oplus+M_A)u_{T,\oplus}\\ M_\oplus\,v_{r,\oplus} + M_A\,v_{r,A} &= (M_\oplus+M_A)u_{r,\oplus}, \end{align} $$ with $(u_{T,\oplus},u_{r,\oplus})$ the new orbital velocity of the Earth (and the gravitationally bound debris) after the impact. We get $$ \begin{align} u_{T,\oplus} &= v_{T,\oplus} + \frac{M_A}{M_\oplus+M_A}v_\text{i}\cos\theta,\\ u_{r,\oplus} &= v_{r,\oplus} - \frac{M_A}{M_\oplus+M_A}v_\text{i}\sin\theta. \end{align} $$ So the orbital energy and angular momentum will have changed into $$ \begin{align} E' &= \frac{1}{2}u_{r,\oplus}^2 + \frac{1}{2}u_{T,\oplus}^2 - \frac{\mu}{r}= -\frac{\mu}{2a'},\\ h'^2 &= r^2\,u^2_{T,\oplus} = \mu a'(1-e'^2). \end{align} $$ (the change in $\mu$ is negligible). Right, let's plug in some numbers. Suppose we start with a circular orbit, with a radius equal to the present-day semi-major axis: $$ \begin{align} \mu &= 1.32712838\times 10^{11}\;\text{km}^3\,\text{s}^{-2},\\ r &= a = 1.49598261\times 10^{8}\;\text{km},\\ e &= 0. \end{align} $$ For a circular orbit, it follows that $$ \begin{align} v_{T,\oplus} &= \sqrt{\frac{\mu}{r}}= 29.785\;\text{km}\,\text{s}^{-1},\\ v_{r,\oplus} &=0\;\text{km}\,\text{s}^{-1}. \end{align} $$ The impact velocity of an asteroid will always be at least equal to the Earth's escape velocity $11.2\,\text{km/s}$, which is the speed it takes up as it falls into the Earth's gravitational potential well. The article that I already linked to states that typical asteroid impact velocities are in the range of $12-20\,\text{km/s}$. In theory, the impact velocity can be as large as $72\,\text{km/s}$ in the case of a head-on collision, when the Earth and the asteroid have opposite orbital velocities, thus a relative velocity of ~$60\,\text{km/s}$, augmented with the escape velocity as the asteroid falls into our gravitational potential well. This is very unlikely for asteroids, but it is possible for comets.
So, let us assume a typical impact velocity $v_\text{i}=16\,\text{km/s}$, a mass $M_A = 0.1M_\oplus$ and an impact angle $\theta=45^\circ$. We find $$ \begin{align} u_{T,\oplus} &= 30.813\;\text{km}\,\text{s}^{-1},\\ u_{r,\oplus} &= -1.0285\;\text{km}\,\text{s}^{-1},\\ E' &= -411.87\;\text{km}^2\,\text{s}^{-2},\\ h'^2 &= 2.1248\times 10^{19}\;\text{km}^4\,\text{s}^{-2},\\ a' = -\frac{\mu}{2E'} &= 1.61109\times 10^8\;\text{km},\\ e' = \big[1- h'^2/(\mu a')\big]^{1/2} &= 0.0788,\\ r_\text{p} = a'(1-e') &= 1.48411\times 10^8\;\text{km},\\ r_\text{a} = a'(1+e') &= 1.73807\times 10^8\;\text{km}, \end{align} $$ with $r_\text{p}$ and $r_\text{a}$ perihelion and aphelion. Evidently, the influence on the Earth's orbit is substantial.
In the case of a direct-from-behind collision, we get $\theta=0^\circ$, $v_\text{i}=11.2\,\text{km/s}$, so that $$ \begin{align} u_{T,\oplus} &= 30.803\;\text{km}\,\text{s}^{-1},\\ u_{r,\oplus} &= 0\;\text{km}\,\text{s}^{-1},\\ E' &= -412.72\;\text{km}^2\,\text{s}^{-2},\\ h'^2 &= 2.1234\times 10^{19}\;\text{km}^4\,\text{s}^{-2},\\ a' = -\frac{\mu}{2E'} &= 1.60778\times 10^8\;\text{km},\\ e' = \big[1- h'^2/(\mu a')\big]^{1/2} &= 0.0695,\\ r_\text{p} = a'(1-e') &= 1.49598\times 10^8\;\text{km},\\ r_\text{a} = a'(1+e') &= 1.71958\times 10^8\;\text{km}. \end{align} $$ As expected, the radius at impact has become the perihelion, and the change in eccentricity is lowest.
And just for fun, let's try the worst-case scenario: $\theta=180^\circ$, $v_\text{i}=72\,\text{km/s}$: $$ \begin{align} u_{T,\oplus} &= 23.239\;\text{km}\,\text{s}^{-1},\\ u_{r,\oplus} &= 0\;\text{km}\,\text{s}^{-1},\\ E' &= -617.10\;\text{km}^2\,\text{s}^{-2},\\ h'^2 &= 1.2086\times 10^{19}\;\text{km}^4\,\text{s}^{-2},\\ a' = -\frac{\mu}{2E'} &= 1.07530\times 10^8\;\text{km},\\ e' = \big[1- h'^2/(\mu a')\big]^{1/2} &= 0.391,\\ r_\text{p} = a'(1-e') &= 0.65462\times 10^8\;\text{km},\\ r_\text{a} = a'(1+e') &= 1.49598\times 10^8\;\text{km}, \end{align} $$ so that the radius at impact has become the aphelion, and the change in eccentricity is highest. Although I wonder how much would be left of the Earth after such an apocalyptic event...
If the collision isn't central, then part of the energy will be transferred to the axial rotation of the Earth, which should reduce the effect on the orbit. But that will be more difficult to quantify.
The orbital plane of the Moon around the Earth is at an angle to the orbital plane of the Earth about the Sun.
This diagram was drawn to show why we do not have eclipses more often but is also shows how it is that a full Moon occurs.
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Image downloaded from Taylor Science Geeks website
Best Answer
Here is a picture that may help in describing this.
First row: Sun (red), Earth (blue) and Moon (black) to scale (axes - in km). You can see just how far away the sun is... and on this scale the Earth and Moon are essentially invisible (they are inside the "zoom box").
Second row: zooming in (50x), you can barely make out the Earth and Moon, Zooming in all the way to 300x, you can finally see just the Earth and the Moon - still to scale.
Now in order for the Moon to be in full sunlight, it needs to shift out of the direct line from the Sun to the Earth. The distance it has to shift to be in "full sunlight" is approximately the radius of the Earth plus the radius of the Moon: the actual shadow (umbra, where you get a total lunar eclipse) has a diameter of just 9000 km at the distance of the moon. This means that the Moon needs to be just 1.3° below the ecliptic to be out of the shadow (the ecliptic is the name of the plane containing the Sun and the Earth ... so called because when the Moon is on the ecliptic, you get an eclipse). The lunar orbit is in fact inclined by 5.145° relative to the ecliptic - so it spends most of its time away from the shadow of the earth. And that's why most of the time, we can see the full moon.
Bottom row: the difference between what a full moon looks like at 0°, 1°, and 5° "off perfect" illumination. As you can see, "full" looks pretty full for all of these - so a casual observation of the shape of the Moon along won't allow you to tell how far off the ecliptic it is. (The "flattening" is happening at the bottom of these plots - if you look closely, you can just see a few missing pixels on the right-most image).
And that is why you can see the full moon. And the earth is not flat.
You can find a nice animation (not to scale) on the earthsky.org site