8

The gravitational force does do work on a satellite in elliptical orbit because there is a component of the force in the direction the satellite moves. One easy way to verify this is conservation of angular moment which states that v×r = constant so as the distance varies so does the velocity Note that as the satellite is moving away from the Sun, the ...


6

Let me try to make an analogy: Suppose you and I both start at town A and we drive to town B. We take different routes and it turns out that your route is 20 metres shorter than mine, so although we start at the same place and end at the same place you have travelled 20 metres less than me and your car is 20 metres less worn out than mine. This is travelling ...


5

“The square of the velocity minus some constants divided by radius”. Let’s write down the law of conservation of energy: kinetic plus potential energy is constant. $$K.E. = \frac12 m v^2$$ $$P.E. = -\frac{G m M}{r}$$ Where $m$ is the mass of the planet, $M$ the mass of the sun it is orbiting, $G$ the gravitational constant, and $r$ the distance. Add them, ...


5

In the three-body problem, the Lagrange points are those points in space where two bodies with large mass (Earth and Sun), through the interaction of the respective gravitational force, allow a third body with a much lower mass to maintain a stable position relative to them. In a planetary system it implies that a small object, such as a satellite or an ...


5

We were taught that lagrangian points are where the resultant gravitational intensity is zero. No this is not the case. Lagrange points are the points where the net gravitational force of Sun and Earth provides the necessary centripetal force to make an object go around their common center of mass.


4

Yes, a satellite is deformed by the gravity of its primary. More precisely, this deformation is due to the tidal effects caused by the difference in the gravitational attraction of the primary from one side of the satellite to the other. The effect is greater the larger the satellite and the closer the satellite is to the primary. Tidal forces have several ...


4

Let go of a stone while high up in the sky. It comes crashing vertically down. Gravity pulls along with its velocity, so it speeds up and up and up. Repeat but this time push it a bit sideways also. It still comes crashing down but lands a bit further to the side. The sideways speed and the downwards speed caused by gravity combine into a slightly tilted ...


4

In a circular orbit the speed of the orbiting body does not change because the acceleration vector is always perpendicular to the velocity vector - so the direction of the velocity vector changes over time but its magnitude is constant. In an elliptical orbit the acceleration vector is generally not perpendicular to the velocity vector (except at the apsides ...


4

Decades ago I learned there were five Lagrangian points around the orbit of a planet correction: with or without a moon but I understood that these were just simple solutions for the three body problem (digression - 'Three Body Problem' is an excellent novel by Liu Cixin). At these points gravity acts on the third body in a manner that maintains stable ...


3

In this answer I will name the celestial bodies as follows: the Primary, the Secondary, and the Tertiary. For example, in the case of the Sun, Jupiter, and the Trojans the Sun is the Primary, Jupiter is the Secondary, and any member of the Trojans is the Tertiary. For a Tertiary orbiting at a Lagrange point the following applies: at the position of the ...


3

In an axisymmetric system, the gravitational force on a particle is going to be radial. It therefore exerts zero torque about the origin, so angular momentum about the origin is conserved: $$L = mr^2\dot{\varphi} = \text{const.} $$ Now differentiate both sides with respect to time to get $$ 2mr\dot{r}\dot{\varphi} + mr^2\ddot{\varphi} = 0,$$ which upon ...


3

Are orbiting masses in a uniform disc affected by masses outside its orbit? Yes. The gravitational potential inside a massive ring or annulus (in the plane of the ring or annulus) is not uniform. There is no “Ring Theorem” similar to the Shell Theorem. Since the potential is not uniform, there is a gravitational field from the outer mass. The potential ...


2

I don't know what you're trying to achieve, but even in a two body problem, the center of revolution is always between the centers of masses of two bodies. This phenomenon, is so intense between Jupiter and Sun that their center of revolution is actually outside the sun. And for your simulation, check out https://prappleizer.github.io/Tutorials/RK4/...


2

Does time period of orbit loses it's meaning in general relativity? The period of an orbit is the time that it takes to go 360 degrees around the barycenter. It doesn’t matter if perihelion moves a bit or if the distance to the barycenter is slightly off.


2

I think the choice of a specific set of coordinates is somewhat blurring the answer to your question. Let start from the beginning and let us follow the main steps required to arrive the Kepler's first law in the form of the equation (29) in your figure (the polar representation of a conic). The usual strategy is to write down the equation of motions for two ...


2

Interesting excercize. Kepler's Third Law for the orbit of the Earth gives us $$ T^2_\text{e} = \frac{4\pi^2}{k}R^3\tag{1}, $$ with $k=GM_\odot$ (ignoring the mass of the Earth). The Kepler orbit of a parabola in polar coordinates is $$ r = \frac{h^2}{k}\frac{1}{1 + \cos\theta},\tag{2} $$ where $h$ is the specific angular momentum. The closest approach ...


2

The orbital velocity (object at $r=2R$) can be calculated if we set $$\frac{1}{2} mv^2 - \frac{GMm}{2R} = 0$$ so that as you point out the velocity is $$\tag 1 v= \sqrt{\frac{GM}{R}}$$ and also $$g = \frac{GM}{R^2}$$ meaning $$GM = gR^2$$ or $$gR = \frac{GM}{R}$$ and from equation (1) this means $$v = \sqrt{gR}$$ Since escape velocity is given by $$v_e = \...


1

Consider the angular momentum difference of the two orbits w.r.t. the planet. Obviously, the larger orbit has a greater angular momentum. This means there must be a net angular impulse when the thruster fires at point A; that is, the torque should be nonzero and should act to increase the angular momentum. The answer should be evident from here.


1

You've added a new planet to the solar system. You have to add the kinetic and potential energies of that object to the rest of the solar system. If you consider that enlarged solar system, you would find that KE + PE remains constant. One troubling word in your question is "all of a sudden". One interpretation of that is the new planet ...


1

I don't have a full solution but would attempt the following in Cartesian coordinates. determine the coordinates of points $A$ and $B$, from: $$x^2+y^2=R^2$$ and: $$y=cx^2+\beta R$$ So that: $$\frac{y-\beta R}{c}+y^2=R^2\tag{1}$$ The two roots of $(1)$ give the coordinates of points $A$ and $B$ determine the path length between $A$ and $B$ (with ...


1

The amount of radiation absorbed by a surface is proportional to the intensity of the beam of light (from the Sun in this case) multiplied by the cosine of the angle between the beam direction and a normal to the surface. In other words, when a surface is tilted at an angle to the light source it absorbs less energy. This is what is happening with the Earth. ...


1

You are correct in expecting to need two coordinates to specify position in the orbital plane. Distance from the central body supplies one of them. The other is simply the angle $\theta$. The two numbers give you the position vector in the polar coordinates $(r,\theta)$. In order to convert them to the Cartesian coordinates $(x, y)$ you can use these ...


1

You can only have a gravitational collapse if the gas cloud continuously loses energy through inelastic collisions (which is then radiated off). Otherwise the collapse would soon come to a halt as the lost potential energy would be turned into kinetic energy and a steady state would be reached once the kinetic energy is -1/2 x the potential energy (virial ...


1

To my understanding the physics of the process of contraction of a gas/dust cloud to a star is a highly specialized branch of astrophysics. I assume that the specialists that create simulations for that know what they are doing, but I haven't encountered articles in which the author sets out to present the findings in this branch of physics to a wider ...


1

It's already the case that below a certain size of the satellite the probability of a sizable chunk reaching the surface is very low. I used a search engine, using the search terms "satellite" "wood" to find your information source. According to that information source the consideration is that some portion of the aluminium that goes into ...


1

You can find it in Chandrasekhar's textbook ("The mathematical theory of black holes" Chap 3, sec. 19). Only you need is to show the conservation laws $r^2 d\phi/d\tau$ of particle in Schwarzschild metric. Edit 1: How to derive approximate Kepler's second law from GR The static metric on $\theta=\pi/2$ $$ ds^2=\frac{1}{1-\frac{1}{r}}dr^2+r^2 d\phi $...


1

Any two bodies orbiting each other will both orbit the center of their combined masses, the barycenter. If they are of equal mass the barycenter will be of equal distance from each. If one has greater mass the barycenter will be closer to it.


1

If you want to accurately calculate intersections, you need much more information about the orbits. But the perihelion of the object is sufficient to rule out an intersection entirely. Earth's orbit is slightly eccentric, but both perihelion and aphelion are close to 1AU. Your asteroid has a perihelion beyond 3AU. Intersection is not possible.


1

When you integrated $$\frac{d}{dt}(\vec{v} \times \vec{L}) = GMm\frac{d}{dt}\hat{r}$$ it appears that you assumed the constant vector, $\vec{B}$, was in the same direction as the unit vector, $\hat{r}$. However, $\hat{r}$ is not constant. With a constant vector of integration the result is $$(\vec{v} \times \vec{L}) = GMm\hat{r}+\vec{B}$$ Now taking the dot ...


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