Question about gravity
-
Topic authorDanial
- Posts: 33
- Joined: 31.03.2008
- Age: 45
- With us: 16 years 8 months
- Location: Gold Coast, Australia
Question about gravity
With the Jovian planets being so large and most of their moons being tidally locked, are the moons' gravity less on the planet side due to the planet's pull and greater on the far-side? If so, does anyone know how much?
-
Arenamontanus
- Posts: 10
- Joined: 07.04.2008
- With us: 16 years 8 months
Re: Question about gravity
I think tidal forces are the answer. All points on the surface have to be at the same gravitational potential (otherwise they would slide around until they had reached a minimum), but that does not mean the local gravity has to be at the same strength. Tidal forces would make local gravity at the subjovian point and the antijovian point weaker, while strengthening the local gravity along the "equator". So the difference ought to be about the same size as a tide, which is around F = 2MmGr/R^3 where M and m are the masses of the jovian and moon, G the gravitational constant, r the radius of the moon, R the distance to the jovian.
Once that force starts to become of the order of local gravity the moon is going to start to deform rather severely and when they are equal you reach the (rigid body) Roche limit and it breaks up. (the fluid body limit is about 2.44 times further out). So if you are finding yourself almost weightless at the subjovian point, you better get away from there soon!
Once that force starts to become of the order of local gravity the moon is going to start to deform rather severely and when they are equal you reach the (rigid body) Roche limit and it breaks up. (the fluid body limit is about 2.44 times further out). So if you are finding yourself almost weightless at the subjovian point, you better get away from there soon!