# Tag Info

## New answers tagged celestial-mechanics

0

Is there a step in your code that takes an inverse trig function? Each inverse trig function has two answers, a positive and a negative. Calculators and presumably program functions probably just take the positive automatically. I made this mistake once in calculating a gravity assist trajectory for a class problem and your description reminded me of my ...

2

I will simplify this problem by assuming that the only forces come from Newtonian gravity and by limiting the masses of the moons to much smaller values than that of the planets, such that the moons exert a much smaller gravitational forces on all other bodies and therefore can be neglected; so the only sources of gravitational forces are the two planets. ...

0

I suspect that they were lucky that their predictions agreed with reality so closely, but any prediction was going to have Neptune roughly (perhaps very roughly) in the same direction as Uranus, during the times when it affects Uranus the most. So I suspect their calculations meaningfully ruled out large swathes of sky, which improved odds of finding it.

4

Although this may not be what you're looking for... They weren't "simply lucky." In fact, they didn't use Bode's law at all- they used calculations based on Neptune's supposed gravitational effect on Uranus. In fact, had the two used Bode's law, they would never have found Neptune, as the Bode "law" would predict a completely different location. (This is ...

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The only factor of $\pi$ in the problem comes from the final integral you have to do, which in dimensionless form is $$\int_0^1 \sqrt{\frac{u}{1-u}} du = \frac{\pi}{2}$$ This integral is pretty hard, so if you did it quickly (which I think you did, since you didn't mention any painful integrals here), then you made a mistake here.

5

How “large” is a Lagrange point? L1, L2 and L3 are essentially zero size cause they're never stable. They're still useful cause an orbital near L1, L2 or L3 doesn't require a lot of energy to stay in that general area. so we can use L1, 2 or 3 for not quite stable orbits that don't require much energy adjustment. L1 Halo orbits are used too, not ...

1

Actually, the orbit of the Earth around the Sun is influenced by all the planets and every other gravitating object in the solar system. But their gravitational influences are relatively small compared to the Sun's, and it becomes computationally unwieldily and even impossible to account for the orbital motions of more than two and at most three mutually ...

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