# Tag Info

4

It's easier to dispel this with biology and geology than with physics. That our circadian rhythms are 25 hour long is based on research done in 1962 that was later found to be faulty. For more recent research, see, for example, Czeisler, et al. (1999), "Stability, precision, and near-24-hour period of the human circadian pacemaker," Science 284.5423 : ...

1

SciShow covered this and their sources are in the video description. It probably can't happen, but it depends on your tolerances. Planetary alignments have a "quality" which is how far apart in the sky the planets are allowed to be and still considered in "alignment". All planets lined up in a nice straight line from the Earth to the Sun? Almost ...

3

There should be no preferred reference frame and from an observer on the sun, the sun will rotate around the Earth. Aside: You have the sense backwards. From the perspective of an observer on the Earth, it's the Sun that appears to rotate around the Earth, and at three frequencies, once a day, once a year, and once every 26,000 years. An observer on the ...

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I prefer to think of it that the Earth and Sun actually orbit around their combined center of mass, which just so happens to be very deep inside the sun. The same can be said for the Earth-Moon system.

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People say the Earth rotates around the sun and not vice versa because a reference frame attached to the center of the Sun more closely approximates an inertial reference frame than a reference frame attached to the center of the Earth. Yes, there is the issue of the path of the other planets, but even if the Earth were the only planet, we would still say ...

1

If I understand your problem correctly, the diagram looks like this: Where I am drawing on the unit sphere, so distances $a$ and $b$ scale from 0 to $2\pi$. The sine rule for spherical triangles is $$\frac{\sin{A}}{\sin{a}}=\frac{\sin{B}}{\sin{b}}$$ Now since $B=90°$, it follows that $$a = \sin^{-1}(\sin{A} \sin{b})$$ Please see whether that makes ...

0

You seem to be neglecting the change in the moment of inertia of the planet in the definition of angular momentum: $$\mathbf L=I\boldsymbol\omega$$ Since $\boldsymbol\omega$ increases, then $I$ must decrease to satisfy angular momentum conservation. There isn't any need for angular momentum to somehow "reach out" and take it from a nearby planet.

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You say: But for the observer on the planet, since the total angular momentum of the star about its axis is zero it should remain zero. But the observer on the planet does not occupy an inertial frame. An observer in a rotating frame measures fictitious forces. So there is no reason why angular momentum should be conserved.

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according to some theories (like big bang) the planets are parts of greater bulks and as gradually they collide and teared apart (after becoming cold) they have gotten a speed due to the explosion and then they have stuck in the gravitational field of bigger planets and stars and rotating in vacuum around them so they don't lose their velocity due to some ...

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Let me answer another component: where the initial energy for their movement came from. Imagine two bodies separated by a large distance. In this case, the gravitational pull is small and the gravitational potential is low. Their relative velocities are just about zero. For all intensive purposes, our energy accounting is zeroed out. KE=0 GE=0 (kinetic and ...

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Unfortunately the answer to "is the structure gravitationally stable?" is "most definitely not." Anything planet-sized pretty much has to be close to a sphere, unless it's spinning very rapidly, because the gravitational forces increase with the body's size, whereas the electromagnetic forces holding atoms together don't, so the material's strength will ...

4

As explained in this article by Neill DeGrasse Tyson, the tidal forces between the Earth and the moon do indeed slow down the rotation of the Earth each year, the same process that caused the moon's rotation to become tidally locked with its orbit of the Earth. This effect would eventually cause the Earth's rotation to be tidally locked with the moon as ...

2

This basically is a specific case of Lambert's problem. I will cover the maths involved solving the problem in your case. When looking at the velocity of mass C, 10000 m/s radial outwards relative to the sun, it can be noted that its movement therefore will basically be one dimensional along the radial direction. This means that its angular moment is zero ...

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Oke this link is a nice reference sheet for elliptical orbits. we are interested in two things the speed at perihelion The time it will take to reach aphelion The initial velocity is what OP wants to now. The time it will take to reach aphelion will be compared with the position at time t of object C So lets call $r_c(t)$ the distance of Object C to ...

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There seem to be three kinds of slingshot manoeuver. You can bleed some kinetic energy off a moving body, sort of like an ancient slingshot; that allows a spacecraft to either increase or decrease its own kinetic energy. Or you can get more "bang for the buck" with the assistance of a gravity well, be it moving or fixed, by expelling mass after having ...

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If you are in a circular orbit what you need is a Hohmann transfer, from Wikipedia: In orbital mechanics, the Hohmann transfer orbit /ˈhoʊ.mʌn/ is an elliptical orbit used to transfer between two circular orbits of different radii in the same plane. It works like this assuming the planet is in a circular orbit. Then the amount of delta v needed to ...

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The gravitational tides generate atmospheric tides ( http://en.wikipedia.org/wiki/Atmospheric_tide ), earth tides and water tides. The water tides can be found in oceans, groundwater or rivers (about the last 2 types you can read on http://www.nature.com/srep/2014/140226/srep04193/full/srep04193.html ). The moon is the principal cause of the tides on Earth; ...

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