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6

This is a fun question, so I spend some time thinking about it. This is what I came up with. Is this possible? Yes, this is certainly possible. Just consider the Solar System itself. We have a massive central body, the Sun, and several tiny subsystems (planets with their moons) orbiting it. Is it stable? Again, from the simple observation that the ...

-2

I would say no. The gas giant's gravity (I think) would rip the smaller asteroid bodies out of the orbit of the moon, and inevitably destroy it. This explains in part what I am referring to. I will continue to find my other resources of Jupiter's moons being 'doomed'... eventually causing Jupiter to have rings similar to Saturn. ...

1

You are correct that an hyperbola's outgoing speed is the same as the incoming speed with regard to the body lieing at hyperbola's focus. The direction is changed. But with regard to another body, the change of direction can mean a change of speed. Here is a diagram of how the moon might be employed in the capture of an asteroid to reduce it's hyperbolic ...

3

Here's an intuitive understanding without math or physics explanations (others will provide that stuff here): You are right that approaching and leaving the vicinity of a planet in itself adds up to zero effect. Gravity assist is the effect of being "dragged along" with the movement of the planet. If a spaceship approaches the planet from behind in its ...

2

The diagram is in the rest frame of the planet. Now suppose a spacecraft is slowing down in the frame of the solar system. A planet is nearby, so it now starts accelerating due to its gravity and gains speed. Now, this speed increase is added to some component of the speed of the planet's motion when it comes out on the other side (this added component can ...

6

Depends on the interpretation of your question... The best places not to observe the moon are the north and south pole. On the north pole you will only be able to see objects above the celestial equator. As the moon orbits the Earth in one month its orbit is inclined from the celestial equator. This inclination is almost the same as the inclination of the ...

2

He knew the orbital period of Earth. You can do this by observing the sun's motion in the sky over a year. (And then you know that the orbital period of the earth is a year.) Now from earth, you can observe Mars from one Opposition (https://en.wikipedia.org/wiki/Opposition_(planets)) to another. You can find out the time ($T_\text{relative}$ it takes for ...

1

Kepler did know the orbital period of the Earth, Mars, Venus, Mercury, Jupiter, and Saturn. He had access to observations going back centuries, including Brahe's most excellent data. What couldn't know was the absolute size of the Earth's orbit, so the best he could do was give the size of the plantets' orbits in terms of the Earth-Sun distance, 1 A.U. ...

1

If you know the apoapsis and periapsis, you can find the eccentricity, using $r_a = a (1+e), r_p = a (1-e)$ (https://en.wikipedia.org/wiki/Apsis#Mathematical_formulae). Using the eccentricity, you can find the radius at any angle $\theta$ from the periapsis (this angle is also called the true anomaly). \$r(\theta) = \frac{a(1-e^2)}{1+e.\cos ...

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