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0

There are several effect that could explain retrograde captures being easier than prograde. As Nickolai noted, aside from atmospheric effects and non-uniform gravitational fields, there's not way for an orbiting object to know about the spin of the primary. However, in general, planet's do have non-uniform gravitational fields- they bulge due to rotation, ...


3

There are three possible forms of motion for a small object in a $\propto \frac{1}{|r|}$ potential: hyperbolic motion parabolic motion elliptic motion The first and second possibilities are unbound, so the object comes from infinity and will go back to infinity afterwards. The third one is bound, which means that the object cannot escape the ...


3

Let $\Delta_S$ and $\Delta_G$ be the time dilation effects due to General Relativity (gravity) and Special Relativity (motion) respectively (i.e. the clock rate on the satellite due to SR and GR is $1 - \Delta_S + \Delta_G$, signs chosen for simplicity). If these are small, they can be approximated as : \begin{eqnarray} \Delta_S &=& 1 - \sqrt{1 - ...


1

For a satellite to orbit around the earth, we need to ensure the following: (a) Satellites orbital plane must pass through the center of the earth (b) It must have sufficient CPF centripetal force, to continue on its orbit without getting pulled into by the gravity of the earth (c) At a constant orbital speed the satellite must cover same length of the track ...


0

$ r^2 \dot{\theta} $ is known as the specific angular momentum. Also, the correct formula for the 2 body Lagrangian is actually: $$ \mathcal{L} = \frac{\mu}{2} (\dot{r}^2+r^2 \dot{\theta}^2) + \frac{GMm}{r} = \frac{Mm}{M+m} \frac{\dot{r}^2+r^2 \dot{\theta}^2}{2} + \frac{GMm}{r} $$ where $ \mu = \frac{Mm}{M+m} $ is the reduced mass.


2

Phil Frost's argument in his answer (v4) is correct. Assuming a spherical Earth with constant density $\rho$ (and assuming for simplicity that the object for some reason can move freely$^1$ through Earth so that there is no air drag, and so that we can skip all the tunnel drilling and not worry about that the Earth's rotation could press the object up ...


8

Phil's answer, while beautifully illustrated, is a little incomplete. It relies on the fact that in the case of the tunnel you're solving the one dimensional projection of the low earth orbit satellite, but doesn't prove this. I do this below. The force applied on the object, for a sphere of uniform density, is actually : \begin{eqnarray} F &=& - ...


1

My question is: why are these two periods (oscillating through the earth and a LEO) the same? I am sure that there is some fundamental physical reason that I am missing here. Help. It's a result of the (flawed) assumption of a uniform density Earth. The Earth is anything but a constant density object. The Earth's core is five times more dense than ...


5

An alternate explanation (which really is the same as the answer from @Phil): as per Kepler's laws, an orbit is an ellipse, and the orbiting period is proportional to the semi-major axis of the ellipse. A satellite in the lowest orbit will try to follow a special kind of ellipse (namely, a circle), whose semi-major axis is really the Earth radius (this is ...


4

Following the same orbit at greater velocity would mean that the satellite has greater acceleration (its velocity is changing at a greater rate than if it follows the same route more slowly). But the only force acting on it is earth's gravity, which creates a particular acceleration at any given altitude above the earth. It can't go any faster or slower ...


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Intuitive explanation Suppose you drill two, perpendicular holes through the center of the Earth. You drop an object through one, then drop an object through the other at precisely the time the first object passes through the center. What you have now are two objects oscillating in just one dimension, but they do so in quadrature. That is, if we were to ...


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In my opinion "description of trajectory" is not a topic of physics it is a topic of kinematics (geometry of motion, see http://en.wikipedia.org/wiki/Kinematics). Whereas explaining the mechanism which causes an object to follow a particular trajectory IS a matter for physics. To say that "B goes around C" is to describe a trajectory in space and time. A ...


2

This is not quite a full answer. I need some sleep before I start on a long drive at 3 AM. Porkchop plots are a way to envision solutions to Lambert's problem, which addresses a class of orbital mechanics problems. In it's simplest form, Lambert's problem addresses a vehicle orbiting a central mass. The orbiting vehicle has an initial position and velocity ...


1

Even though your integration method seems wrong, like David Hammen pointed out, the results look do not look wrong. The problem rely lies with the way you define your ellipse. Your definition for the semi-major axis, $a$, and eccentricity, $e$, are correct. The way you define the semi-minor axis, $b$, probably is not, which is defined as: $$ b = a \sqrt{1 - ...


1

Here's your key error: var dx = Add(Mul(gv , dt*dt) , Mul(o.v, dt)); // position changes In math, that's $\Delta \vec x = \vec g \Delta t^2 + \vec v \Delta t$. That's incorrect. Assuming a constant acceleration $\vec g$, the change in position over some time interval $\Delta t$ is given by $\Delta x = \frac 1 2 \vec g \Delta t^2 + \vec v \Delta t$. That ...


1

I think that a disc of orbiting material is something that arises naturally when you have enough objects to do statistics with, but not necessarily when you have only a handful of masses as in your example. I happen to be playing with an orbit simulator lately, so I put two Jupiter-mass objects in perpendicular circular orbits 1 AU from a solar-mass central ...


0

If we draw in the acceleration and velocity vectors of an orbiting body, you'll see why it acceleration due to gravity does not cause an orbiting body to fall towards an object. First, assume the orbit is circular. In any motion of an object, the velocity vector is tangent to the object's path. Because acceleration due to gravity is straight towards the ...


1

You're right - the objects are always accelerating while orbiting, and this acceleration is due to gravitational influence. However, force/acceleration don't always cause an increase in speed. For example, a force acting opposite to the velocity will slow an object, and a force acting perpendicular to the velocity will merely change its direction. In an ...


1

It does keep accelerating. It's velocity in the direction of the object being orbited keeps increasing. But this direction keeps changing. The reason the satellite's total speed doesn't increase, at least in the case of a circular orbit, is that while the it's velocity towards the object increases, it's tangential motion moves it forward so that that ...


2

Note that circular orbits never exist in nature, and that the scenario that you described would be very rare. The reason why most solar systems have planets with coplanar orbits is because these planets were formed from an accretion disk, which as the name implies, is a very thin (but dense) cloud of dust and debris orbiting a star. That being said, it ...


2

You can solve this problem by using energy conservation. It holds \begin{equation} -\frac{GMm}{P}+\frac{1}{2}mV^2 = - \frac{GMm}{2a}, \end{equation} where $a$ is the semimajor axis and related to $P$ by $P=a(1-\epsilon)$. You can understand the RHS from Virial's theorem, for instance. You may also want to check out the vis-viva equation. psm.


-2

Earth rotation is slowing dowm through ebb and flow and the resulting friction. It will have found it's equilibrium as soon the day on earth will be one month (and the tides will have ended). The moon has finished this earlier, since it's moment of inertia is much lower


1

My suggestion would be to simply show him a picture or better yet animation of how the earth moves around the sun. There are also a number of interactive gravity simulations like "Universe Sandbox", that let you play around with gravity. If the question is "why?" rather than "how?", then I would go with: No one really knows, but physicists (or some more age ...



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