3
$\begingroup$

I have a planetary system consisting of an oceanic world [7% Land] with a large precious metal moon [extremely dense with valuable alloys]. The planet has 110% earth mass, with a size of 1.4 Earth radii.

  1. I want the moon and earth to be tidally locked with one another, with the orbital period of the moon equal to the rotation of the earth. Essentially placing this smaller but denser moon in geostationary orbit much closer than our own.

  2. Would this create any strange phenomenon around to sub/anti-lunar points? My first concern was a permanent mountain of boiling water, or at the weirdest a shared-atmospheric system. I know the tides will absolutely bizarre relative to earth.

  3. I had a fun thought that there would be massive dust storms of valuable metals that scoured the moon's surface, and I thought that if the atmospheres were joined, some of this might leak into the upper atmosphere and fall in a seasonal heavy metal snow {highly toxic of course}.

Is this feasible without the moon crashing apocalyptically into the surface of the host planet instantly? And if so, would the host world be rendered uninhabitable by the lunar tidal forces?

  • You may generate as much atmosphere as needed to make things even.
Improve this question
$\endgroup$
2
  • $\begingroup$ The first thought that comes to mind is that heavier elements tend to sink to the center of a given system rather than the periphery. This would make the planet a likelier place to find precious metals than its moon, unless during planet formation the moon collided with a bunch of asteroids made of heavier metals? $\endgroup$
    – Adam Coville
    Commented Jan 2, 2022 at 22:30
  • $\begingroup$ @AdamCoville not too sure of the orbital dynamics of it, but a capture of a planet from a closer orbit could account for the heavier elemental mixture $\endgroup$
    – Sonvar
    Commented Jan 3, 2022 at 17:11

1 Answer 1

Reset to default
2
$\begingroup$

The planet has 110% earth mass, with a size of 1.4 Earth radii.

This would mean that the average density of the planet would be 2.5 times smaller than the average density of Earth. Being the average density of Earth 5.5 $5.5\ \text{g}/\text{dm}^3$, this planet would be at about $2.2\ \text{g}/\text{dm}^3$, which is roughly the density of silicon. Somehow this planet has very little content of heavy elements. This alone would make very difficult for this planet to be hospitable for life as we know it.

The Roche limit for the planet would be $d=R_M(2$$\rho_M\over \rho_m$$)^{1/3}=1.71R_M$, considering that the Earth radius is 6360 km, the Roche limit for your planet would be at about 10800 km from its center, or about 1900 from its surface.

What would be the period of the geosynchronous orbit?

$T=2\pi \sqrt{a^3/(GM_M)}$

You haven't specified a value for $a$, but if you want the satellite to not be destroyed by tidal forces, it has to be greater than those 10800 km calculated above. Once you decide it, you will also derive the duration of the rotation.

You haven't specified neither the radius of the satellite nor the duration of the rotation, so I am not able to proceed further with the estimates.

Improve this answer
$\endgroup$
2
  • $\begingroup$ In your TeX (math markup) you formatted units as if they were variables. $\endgroup$
    – JDługosz
    Commented Jan 3, 2022 at 16:47
  • $\begingroup$ Upon initial reading to the question I know the density would be off. thanks for doing the math. However, assuming a relatively similar Earth radii and gravity parameters, geosynchronous orbit would be ~32000 km, outside your calculated Roche Limit. $\endgroup$
    – Sonvar
    Commented Jan 3, 2022 at 17:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .