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26

You've used the gravitational constant with only three significant digits. So it's no surprise that your answer isn't accurate to five significant digits. Instead of $G$ and $M_\odot$ separately, you should use the product $GM_\odot$, known as the standard gravitational parameter. Its value is known very accurately: in the link, you'll find $$ GM_\odot = ...


12

Dark matter would affect planetary motion, but the influence of dark matter on planets in our solar system is too small to detect even currenlty, due to the low concentration of dark matter compare to ordinary matter in our solar system. See Constraints on Dark Matter in the Solar System. The density of dark matter is very low, $ <~10^{-19} grams/cm^3$, ...


11

In a two body system the orbit can have any eccentricity you want. The key point is that the eccentricity cannot change. In a three or many body system the bodies perturb each others orbits and the eccentricity can change with time. For example this graph shows the eccentricities of the rocky planets as a function of time. These changes are mainly caused by ...


11

To expand on Prahar's answer, let me run some numbers to try and convince you this is reasonable. Your answer is correct to within one part in 104: $$ \frac{365.256363004}{365.2075}\approx 1.000133795. $$ The main perturbing influence on Earth's orbit is the gravitational pull of Jupiter, whose mass is about 1000 times smaller than the Sun, and which orbits ...


10

Kepler's 3rd law assumes that the Earth travels in a perfect ellipse with the only gravitational force on it being from the Sun. Further, Kepler's laws are derived from Newtonian gravitation. In reality, the orbit of the Earth is affected by the gravitational pull of other planets, and by the effects of General Relativity and is therefore not quite ...


8

The answer is because dark-matter has relatively constant density, as has been given explicitly in another answer. Then, it logically follows that the impact on the Milky Way due to this low density. To show this step, I will establish a figure of merit. $$ FOM = \frac{M_{dark}}{M_{normal}} $$ That is, the ratio of dark matter within the area of ...


6

As you have said, the general solution for the Kepler problem is an elliptic orbit. The shape of an ellipse is determined by its semi-major axis and the eccentricity. The semi-major axis is determined entirely by the energy, but the eccentricity depends also on the angular momentum. Both energy and angular momentum are conserved in the two-body problem, so ...


4

Earth has few and relatively tiny sattellites other than the moon specifically because of this large moon. Note the mass ratio of our moon to our planet. It is the highest in the solar system by a large amount. This one large sattellite will over time sweep up and aggregate other smaller sattellites. Put another way, we probably did have other smaller ...


3

As indicated in the comments, this is probably a question for Computational Science. But I will try to answer your question in such a manner that it may be on-topic enough to stay here. I think you can solve this by using Python's atan2(y,x) function, rather than defining $z = y/x$ and using atan(z). Let me address this concern first: ...except where ...


3

No. The n-body problem is a problem of calculation, not of determinism. There is nothing non-deterministic about Newtonian mechanics, it is just hard to calculate. By contrast, a quantum system is generally accepted to be non-deterministic by its nature; Einstein disagreed with this idea, hence his statement that 'God does not play dice'.


3

Orbits are pretty complicated. Most texts on this deal in terms of predicting positions at a specific time rather than just a simple ellipse, because that model while correct is too basic. As others mentioned position estimation makes it more complex because there are more parameters involved. I'll just pull some stuff for you on the basics. For standard ...


2

The planets are attracted towards the sun, as you would expect from the gravitational force. The planets don't fall into the sun, though, because their velocities are at right angles to that force. The planets end up being pulled by the sun into a circle. A planet's speed is constant, but its direction changes. I think it's easiest with a picture: ...


2

Dark matter collects in larger quantities (thus a higher proportion relative to matter) in the centre of galaxies compared to in the centre of stellar systems such as the solar system. galaxies are not very dense, as stellar systems are sparsly spaced. So even though on a galactic scale the dark matter is in high ratios, on a stellar scale the ratio is ...


2

A Lagrange point is a position relative to a system of two-body gravitational objects at which a third negligible small object, if its velocity is correct too, would not move relative to the two other objects. The two gravitational objects will orbit around the center of mass of both objects together (also called the barycenter). I am not sure, but I also ...


2

As the formula you wrote suggests, it depends on $r$, the distance between the satellite and the centre of mass. $3.1\:\mathrm{km/s}$ is for geostationary orbit, and $8000\:\mathrm{m/s} = 8\:\mathrm{km/s}$ is for low Earth orbit. Please see Wikipedia for more details.


1

Let's assume the light from the Sun is parallel, then the shadow of Earth looks like this: The dotted line is the orbit of the satellite at a height $h$ (I've exaggerated the height a bit to make the diagram clearer). All we have to do is calculate the angle $\theta$, because the time the satellite is in the Earth's shadow is simply: $$ t = \tau ...


1

Yes, the Sun and our whole Solar System are revolving around the centre of our galaxy, the Milky Way. Milky Way is a spiral galaxy and hence has four major spiral arms and a central buldge. The Sun (and, of course, the rest of our solar system) is located near the Orion arm, between two major arms (Perseus and Sagittarius).


1

In an atom, the electron is not just spread out evenly and motionless around the nucleus. The electron is still moving, however it is moving in a very special way such that the wave that it forms around the nucleus keeps the shape of the orbital. In some sense the orbital is constantly rotating. To understand precisely what is happening you need to use ...


1

$\theta$ is the angle made between the positive x-axis and the spacecraft's velocity vector (all at closest approach). Don't you need $\theta$ to be the angle of the spacecraft's velocity relative to Mars? Try subtracting the Mars velocity vector from the spacecraft's velocity vector, calculating $\theta$ for the resultant vector and using that angle ...


1

For your first question: Yes gravity is mediated in all directions equally. As for your second question: I assume you are asking about elliptic orbits. If this is the case then, from the law of conservation of angular momentum, when the planet is closer to the sun it has greater velocity, and when its further out it travels slower because $\vec{L}$ = ...


1

The acceleration due to the Sun's gravity at the orbit of the Earth is given by: $$ a = \frac{GM}{r^2} $$ where $M$ is the mass of the Sun ($1.9891 \times 10^{30} \mathrm{kg}$), $r$ is the Earth-Sun distance ($\approx 1.5 \times 10^{11} \mathrm{m}$) and $G$ is Newton's constant ($6.6738 \times 10^{-11} \mathrm{Nm^2kg^{-2}}$). This gives: $$ a \approx ...



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