New answers tagged orbit
0
This isn't exactly an answer to the question as asked, but it's close enough to one that you might find it interesting and hopefully helpful in some way.
The gravitational force of the Sun on the Earth is given by
$$F_{Grav} = \frac{GM_{Sun} M_{Earth}}{a^2}$$
where $a$ is the semimajor axis of Earth's orbit and is about 1.5E+11 meters and $GM_{Sun}$ is ...
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A planet's mass does not affect a planet's orbit around the Sun.
2
Humans don't have this capacity. The Earth has a kinetic energy of about $10^{33}$ joules, relative to the sun. Even if we put the total energy that humans use per year into the Earth's orbit (about 10^{24} joules) we still only have one-billionth of the kinetic energy of the Earth.
Moreover, the annual changes in temperature have very little to do with ...
0
Yes,
As the moon more or less follows the same path as the sun in the sky, from the poles the moon will be above the horizon for two weeks then below the horizon for two weeks, so the answer to this question is yes.
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Many red and brown dwarfs must have planets that are in full tidal lock with their host star, that means that the same side of the planet is always facing the star so there is always day while on the other side there is always night. Our Moon and many other moons are also in such a tidal lock with their planet but not with the Sun so that the often-used term ...
2
The rings of Saturn are only about 10 meters thick. What phenomenon pulls the particles, over time, into a thin disk?
Consider a particle that is in a circular orbit inside the ring. Its velocity can be resolved into an in-disk component, Vd, and a component, Vn, that is normal to the disk. Since the orbit is circular, the radial component of velocity is ...
4
I can reproduce your JPL Z coordinate plot if I use Earth's heliocentric position relative to the J2000 ecliptic.
Naturally the amplitude has a minimum around the year 2000:
Relative to the ecliptic of date, which accounts for precession, the heliocentric Z coordinate of the Earth-Moon barycenter is much smaller:
The IAU 2006 precession model has two ...
3
If you know the length of a sideral day and the length of a year, you can determine the length of a solar day.
From Wikipedia:
After one year, there will have been one more solar day than sideral day. Therefore :
solar day = sideral day / (1 - (sideral day/orbital period))
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And we know that many of astronomers know that mercury orbit does not precess ...
Mercury's orbit does precess, by a good amount. The greatest amount is explained by Newtonian mechanics. Venus, Jupiter, and to a lesser extent, all the other planets, make Mercury's orbit precess by over 500 arcseconds per century. A key problem of the latter half of the 19th ...
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I'm assuming this is Pluto's equatorial plane.
That is incorrect. Charon's right ascension of ascending node with respect to Pluto's equator is undefined. Seemingly paradoxically, it is well defined with respect to Earth's equatorial plane, and since that has become the universal plane of reference, that is what is used.
Charon and Pluto are tidally locked ...
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Talking about instantly teleporting planets isn't very meaningful if you want to get scientific answers, as PM 2Ring pointed out. But is is almost certainly not physically impossible for them to have trajectories that let them all pass between the earth and moon at the same time. But the earth and moon would have to start out farther apart than they are now ...
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