DaveC426913 said:
TL;DR Summary: Is this as much nonsense as I think it is?
I know, I know, I'm committing the cardinal sin of putting any stock in the technobabble of a TV show, but this one seems egregious.
Lewis (English WhoDunnit-in-Oxford on Britbox) Season 8 Episode 2 "The Lions of Nemea" had an astro physics professor who said this:
"... and so we see that one AU cubed is equal to G times the mass of the Sun times one year squared times four pi squared...."
Not only can I not imagine what this could possibly be referring to but the units don't even match.
m3 = N · kg · s2
m3 = kg2 · m
(Did I get that right? Also: Sorry, I just cannot manage LaTeX)
I'm not (presently) a subscriber of BritBox, so I can't watch the episode myself, but going by your quote, it almost makes sense if
- He's talking about the orbit of the Earth around the Sun,
- with the assumption that the orbit is circular,
- The mass of Earth is relatively small compared to the Sun, such that we can assume their common barycenter is pretty much the same location as the Sun.
- ignoring any gravitational effects of other bodies,
- and also assuming that both the Earth and the Sun are spherical (i.e., tidal effects can be ignored).
The centripetal force on the Earth in such a model is:
[itex]F = m_🜨 \frac{v^2}{R}[/itex],
where,
- [itex]m_🜨[/itex] is the mass of the Earth,
- [itex]v[/itex] is Earth's orbital speed (again, with the simplistic model discussed above)
- [itex]R[/itex] the distance between the Earth and the Sun.
We can re-write this equation, knowing that the orbital period [itex]T[/itex] is one year, and Earth traverses [itex]2 \pi R[/itex] per year,
[itex]F = m_🜨 \frac{ \left( \frac{2 \pi R}{T} \right)^2}{R} =m_🜨 \frac{\frac{4 \pi^2 R^2}{T^2}}{R} = m_🜨 \frac{4 \pi^2 R}{T^2}[/itex],
Along with all that, we also have Newton's law of universal gravitation,
[itex]F = G \frac{m_🜨 m_☉}{R^2},[/itex]
where.
- [itex]G[/itex] is Newton's gravitational constant. This is usually verbalized/pronounced "big gee." Note that this is different than "little" [itex]g[/itex], the typical acceleration due to Earths gravity at its surface. They're different. Little [itex]g[/itex] plays no role here.
- [itex]m_☉[/itex] is the mass of the sun.
Equating the forces gives:
[itex]m_🜨 \frac{4 \pi^2 R}{T^2} = G \frac{m_🜨 m_☉}{R^2}[/itex]
We can see that the mass of the Earth, [itex]m_🜨[/itex] , cancells out from both sides, leaving
[itex]\frac{4 \pi^2 R}{T^2} = G \frac{m_☉}{R^2}[/itex].
Rearanging variables gives us,
[itex]R^3 = \frac{ G m_☉ T^2}{4 \pi^2}[/itex]
So it's pretty close to what was quoted. The only mistake I see is instead of saying "...
times 4 pi squared ..." should have been "...
divided by 4 pi squared."