Re: Astrophysics question

Posted by Mike Kremer on Feb 01, 2004 at 20:14

Re: Astrophysics question (Amaranth Rose)

Without doing any maths, I am very certain that your 1/10 size moon satellite would need to be quite close to your 'earth' type water world, in order to raise an appreciable tide.
In being close enough to raise tides, it would have to be circling around your 'earth' fast enough to obey Newton's inverse square law.
Your 1/10 diam moon might even tend to break up?

Thinking of our own Earth and Moon, and taking their diameters to be approx 4000 miles and 1000 miles diameters respectively, they are locked into a synchronous orbit, based upon our Moon rotating around our Earth in 24 hours. Meaning that the centipetal force upon
(m)oon, should equal the mass of the Earth X the centipetal acceleration, or m2 velocity squared divided by the radius.

To keep it simple assume your hypothetical Earth diam is 1000 miles then your 1/10 size moon is just 100miles in diam.

Their Densitys are, D=mass/vol

While a spheres volume is 4/3 pi r cubed

Hmmmm, sorry Ive lost it for now, im not quite sure how one would calculate the attraction of your 'earth' water surface towards
your small moon. But it does depend upon how close its orbit is.

I think i need to find the surface acceleration of both your bodies Not sure how though.
Think Ill sleep on it. Sorry

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