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16

For simplicity, consider a perfectly circular orbit; the gravitational acceleration is always at a right angle to the velocity vector. This means that the speed cannot change despite the fact that there is constant acceleration. Note that for the speed to change, there must be a non-zero component of acceleration parallel (or anti-parallel) to the velocity ...


6

$$1- 0.01672*\cos(0.9856*(\text{day}-4))$$ This is an approximate expression. Term by term, $1$ The mean distance between the Earth and the Sun is about one astronomical unit. $0.01672$ This is the eccentricity of the Earth's about about the Sun. $\cos$ This is of course the cosine function, but with argument in degrees rather than radians. $0.9856$ This ...


4

No. The shape of the orbit, i.e. how elliptical it is, does not depend on the relative masses of the two bodies. All objects in the solar system orbit around the centre of mass of the solar system. For obvious reasons, namely that the Sun contain far and away most of the mass of the solar system, the centre of mass of the solar system is quite close to the ...


4

First, let us look at the current rate at which the moon slows down. I have a few different sources, and they don't all give me the same answer. First, there is this claim that Earth slows down at a rate of about 0.005 seconds per year per year. A year has approximately $365.25 \cdot 24 \cdot 3600 = 3.15\cdot 10^7 \mathrm{sec}$, so 0.005 seconds change ...


3

To say that the orbit becomes more circular the greater the Sun's mass is not true. Instead, the eccentricity (i.e. how much the shape of an orbit varies from being circular) is governed by a couple of factors. If you have a planet orbiting about the Sun with a mass much less than that of the Sun, and you know the following for an instantaneous point in the ...


3

By the sounds of it you have made a mistake with the units. In fact, you should not be using SI units at all in your simulation; astronomical values in SI units vary by such huge orders of magnitude that they are often a source of floating point errors that can destroy trajectories. You should instead use the astronomical system of units. Specifically, ...


3

One has to realize that Kepler's laws are a mere approximation. The motion of planets around the sun is a two-body problem. In case of such two-body problems, both the bodies revolve around the center of mass. But it turns out that Sun is much much heavier than the planets. So the center of mass of the system is very close to the Sun and Hence it is a good ...


3

If the rope is "radially directed" it means every point has the same angular velocity $\omega$. Assume a length $2\ell$, then you can integrate the force on the rope from $r-\ell$ to $r+\ell$ - gravitational force must equal centripetal force. This gives you an equation for $\ell$ as a function of $\omega$ and $r$. See if that gets you going.


3

The rotation period $T$ is given by $$T=2\pi \sqrt{\dfrac{a^3}{G(M_\text{Sun}+M_\text{planet})}}$$ where $a$ is the sum of the half axes of the ellipse. Routhly: $M_\text{Sun}=2\times 10^{30}$ kg $M_\text{Earth}=6\times 10^{24}$ kg $M_\text{Jupiter}=2\times 10^{27}$ kg If you assume both Earth and Jupiter are orbiting around the Sun (and neglect the ...


3

I was wondering that if they were orbiting in same orbit then will they both have same time period? If yes, then why because as they both have different angular momentum and both have so much of differences. I'll break this down into two parts, first looking at the period of individual objects orbiting the Sun at a distance of one astronomical unit (but ...


2

A collapsing gas cloud is an open system. It loses mass, energy and angular momentum as it collapses. Even if the net angular momentum of the cloud is zero, after the collapse the final planetary disk can have a significant net angular momentum, and the ejected material will have the opposite angular momentum. What can not happen, and that's where your ...


2

Yes, objects with mass all attract to each other and move each other of course, except that the star doesnt change it's theoretical orbital shape depending on it's mass, it probably just experiences small tidal forces that aren't much bigger than it's own centrifugal forces. The oscillation of the star position is a complex dynamic based on it's surrounding ...


2

The Moon rotates around the Earth slower than the rotation of the Earth itself. That's why, from a fix point on the Earth, the Moon appears to be moving. The Moon creates the tide on Earth. So the tide "follows" the Moon. However as the Earth rotates faster than the Moon it will tend to carry the tide with itself "forward". The Moon pulls the tide toward ...


1

If I understand the question right, we suppose we want to prove to someone that the earth orbits the sun. I'm not quite sure that' the case from a scientific point of view. Literally speaking, we can choose any reference frame we like and thus prove a heliocentric system or a or a geocentric. Quoting Einstein:" The struggle, so violent in the early days ...


1

A presentation on the SETI Weekly Seminar series (available on Youtube) points out that tidal locking (e.g. expected of a planet in the habitable zone of a red dwarf) can involve higher muliples than same-face-shows, and in fact an eccentric orbit favors an odd half multiple (3:2 like Mercury). There is also orbital inclination to consider. The pattern of ...


1

Acceleration is any change in velocity, whether it's the direction of the velocity or its magnitude. In the case of a perfect orbit it is the direction, the tangential velocity of the satellite should keep a constant magnitude and just change direction as the satellite orbits. When the velocity is slightly less than what is needed for a perfect orbit, it ...


1

Yes, both special relativity and general relativity have to be taken into account. The total time dilation is given by $$ \frac{\text{d}\tau}{\text{d}t} = \sqrt{ \left(1 - \frac{2GM}{rc^2}\right) - \left(1 - \frac{2GM}{rc^2}\right)^{-1}\frac{v^2}{c^2}}, $$ where $\text{d}\tau$ is the time measured by a moving clock at radius $r$, and $\text{d}t$ is the ...


1

Its almost impossible to have an n body system not create gravitational waves. Natural systems of more than one body will always have a quadropole moment of some sort, along with some angular motion. Then if the entire system is somehow finely balanced, there would certainly be regions in the system radiating GR waves. Three body (or more) systems can't be ...


1

The answer based on accepted physics is trivial: none. That's why they call it science FICTION. One can theoretically reduce the amount of propellant, but then the required power these ships would have to develop would go trough the roof. Keep in mind that a medium sized rocket develops on the order of 10GW of power for a few minutes during first stage burn ...


1

The answers are yes and no. Special relativity does make ellipses precess, but it only accounts for 7" out of 43" per century of Mercury's anomalous precession. I wonder if Einstein and/or Sommerfeld knew that. To first order, incorporating special relativity results in a small inverse cube correction to the gravitational force, which is well known to ...



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