# What would happen if the Earth's orbit around the sun were faster (shorter year)? [closed]

I'm writing a book and I need to know if it would be possible to have a four month long year. I hope you can help!

• You should start by understanding Kepler’s Law Dec 28 '16 at 6:21
• Note that you can't have just any speed for any given orbit. In order to remain in orbit, the distance from the Sun (or whatever primary) and the speed are related by a formula. The formula also involves the mass of the primary. If you travel too slowly for your orbit, then you will fall toward the Sun. Travel too fast and you will fly off into space. See, e.g. en.wikipedia.org/wiki/Orbital_speed.
– Jay
Dec 28 '16 at 6:48

The Earth's orbit would be well inside that of Venus. There'd be no liquid water on the Earth, and all life more complicated than bacteria would be dead.

However, if you made the sun dimmer, you could have a 4-month orbit and still keep the Earth inside the 'Goldilocks zone'.

The 30-day month originally derives from the lunar orbit or about 30 days. If you moved the moon further away from the Earth you could increase its orbital period thereby causing the inhabitants to invent 91-day months. This would result in a dimmer moon (unless you made the moon larger to compensate), darker nights (because there's no moon), and much smaller tides.

Having a shorter orbital period of four months (assuming Earth months) will require you to solve a set of equations. First, you have the generalised form of Kepler's law:

$$p^2 = \frac{4\pi^2 r^3}{G M}$$

Where $p$ is the period, $r$ is the distance from the centre of gravity, $G$ is the gravitational constant, and to simplify from $M_1 + M_2$, I will use $M$ because planetary masses are negligible compared to stars.

Then, I have the equations for extraterrestrial solar irradiance, holding the Earth value of 1367 $\textrm{Wm}^{-2}$ constant.

$$1367 = \frac{L}{4 \pi r^2}$$

Where $L$ is the luminosity of the light source. Now, because I can relate luminosity to mass, I am going to use the mass-luminosity relation with the general approximate value for low-mass main sequence stars of around 4:

$$\frac{L}{L_{\odot}} = \left(\frac{M}{M_{\odot}}\right)^{4}$$

Substituting all of these equations together, starting with the solar radiation equation, then substituting the mass-uminosity relation for $L$ and then Kepler's law, solved for $r$, I get:

$$1367 = \frac{\left(\frac{M}{M_{\odot}}\right)^{4} L_{\odot}}{4\pi \left(\sqrt[3]{\frac{p^2 G M}{4\pi ^2}}\right)^2}$$

I then substitute in the appropriate value for your four month year and come up with some mass value $M$ necessary for that star. Using Wolfram Alpha (because I really don't want to solve that by hand), with this query,

1367 = {(\frac{M}{1.988e30})^{4} * 3.848e26}/{4 * \pi * ({(1.051e+7)^2 * 6.67e-11 * M}/{4 * \pi^2})^(2/3)}, solve for M

I get a value for the mass around $1.28 \cdot 10^{30}$ kg, or, 0.643722 $M_{\odot}$. The distance from the star, therefore by Kepler's law, would be somewhere around (using the expression substituted for $r$ and the query string below)

(((1.051e+7)^2 * 6.67e-11 *(1.28e+30) )/ (4pi^2))^(1/3)

some 0.418 astronomical units from the star, which would itself be a K3 or K4 class star and exhibit a slightly orange glow.

Note: The original version of this answer calculated the distance from the Sun, without having accounted for a dimmer and therefore, less massive, star. This version has been updated to constrain all variables onto the mass. Other answers already show the relation given that the Sun would stay in place. Mathematically, the orbital radius for a 4 month orbit would be somewhere around 0.48075 AU, slightly higher than the orbit of Mercury, where the Sun's heat can melt lead. The Earth, without accounting for a dimmer star, would be sterilised.