Tag Info

New answers tagged

0

I have an answer, but I was hoping to see confirmation from another source. I used the centripetal force formula to come up with: $$ v = \sqrt{Gm/r-ar} $$ Where v is the tangential velocity of the orbit and a is the radial acceleration away from Earth. Is this right?


1

Since the force is radial, you are not changing the angular momentum, but you are adding potential energy: this tells you what must happen to the tangential velocity (decreases) and radial velocity (increases). I will leave it up to you to figure out by how much.


1

No, that's not possible. Even if the two bodies could be compressed to be just larger than their Schwarzschild radius (they can't really, without collapsing further to black holes), their combined Schwarzschild radius, which grows linearly with mass, is twice their individual Schwarzschild radii. That means that even if they effectively rolled on each other ...


2

If we take some object approaching a star along a geodesic then the angular momentum of that object about the centre of the star picks out a natural axis: If we consider a Schwarzschild metric (NB this applies only if the star is not rotating) we can choose this axis to be the $\theta$ axis then the symmetry means the geodesic must lie in the plane ...


0

Here's a summary of what Malcolm Longair discusses in Theoretical Concepts in Physics. Kepler was assigned by Tycho to analyze the huge amounts of data he had collected from about 20 years of observations. Tycho was interested in whether his hybrid model would fit the data better than the Ptolemaian (sp?) and the Copernican models. they were notoriously ...


4

Let's consider a circular orbit in Schwarzschild coordinates, taken to be in the equatorial plane for simplicity. The test particle's position has components $x^\mu = (t, r, \pi/2, \phi)$, where $t$ and $\phi$ vary linearly with time/proper time and $r$ is constant. Then the $4$-velocity is $u^\mu = (\dot{t}, 0, 0, \dot{\phi})$, where dots denote derivatives ...


1

The answer to your question is that to a first approximation the direction of the spaceship will not change, so the upper diagram is the correct one. However the direction of the spaceship will change very slightly due to a phenomenon called the geodetic effect. The easiest way to see this is to replace the spaceship by a gyroscope, and make the gyroscope ...


1

First, you have to prove that the equations of motion of the satellite with mass $m$ relative to the planet with mass $M$ is given by $$ {\bf\ddot{r}} = -{\frac{\mu}{r^3}} {\bf r} $$ where ${\bf r}$ (with magnitude $r=\|{\bf r}\|$) is the position vector of the satellite relative to the center of mass of the planet, and where $\mu=G(M+m)\approx GM$ is the ...


7

Basically the rings don't fall into Saturn for the same reason the Moon doesn't fall into the earth. The rings are billions of little moons, each in it's own stable, or largely stable orbit. The rings are also likely resupplied with new ring material from Enseladus, Saturn's 2nd closest moon. (ice volcanoes due to strong tidal forces that can shoot ice ...


1

Let me assume that the object has spherical symmetry, however, for solving the present problem it is painted on its surface with different colors. So, imagining a plane section that contains the orbit of the object around the earth. The section of the plane through the object is a circle and we will see different points of the circumference painted in ...


1

Yes, the moon does rotate. In a tidally locked system, the angular speed of the rotation of the satellite about its own axis is equal to the angular speed of the rotation of the satellite about its primary. Thus, the moon has an angular speed about the Earth of 1 cycle / 28 days, and an angular speed of 1 rotation / 28 days about its own axis; this is why ...


0

Since there are no extra rules for black holes they should follow the same laws. In general relativity this effects are called the de Sitter- and the Lense Thirring-effect which have been verified with big trumpets call by the famous Gravity Probe B.


1

Assuming earth mass, procession would be reduced practically to zero because tidal effects would be practically zero. Procession, at least according to this site, has to do with tidal effects and a planet being mailable. A black hole - 2/3rds of an inch in diameter would experience essentially zero tidal effects and it would be far less prone to bulging ...


1

Another, secondary question that might stem from this is whether, at any point, this (or the fact that my engines' exhaust would appear to have a lower velocity to the external observer) would significantly affect the gains from the Oberth Effect, and if so, what the effect would be and at what point it would start to matter. When time dilation factor ...


0

Angular momentum $\mathbf{L}=\mathbf{q}\times m\mathbf{\dot{q}}$ is a vector which is always perpendicular to $\mathbf{q}$ (as it comes from a cross product). So, since angular momentum is conserved, $\mathbf{q}$ is always perpendicular to the same unit vector $\hat{\mathbf{L}}$ which can be taken as the unit normal of a plane. In general, the magnitude of ...


1

It may be easier to consider a change in orbit caused by a pair of separate short thruster burns, to split the process into its various parts. Consider a satellite in a perfectly circular orbit around the Earth. Its speed is constant, say $v_1$. To go to a lower orbit, the satellite fires a thruster that opposes its motion. The satellite slows down, to a ...


2

As said by lemon, the reason is that the satellite drops to a lower orbit. This can be seen in a very simple way. Quantitatively The orbit can be approximated with a circular motion with a very slowly decaying radius. This implies we can write down equations for circular orbital motion: $$F = \frac{GMm}{r^2} = m\frac{v^2}{r} \Rightarrow$$ $$v = ...


0

Objects at both L4 and L5 have a bit of an "orbit" around the central point. To simulate this, you'll need to use somewhat complicated rotating reference frame physics. Here is an example of the Jupiter Trojan asteroids. You can see that they all dance around each other in the vicinity of the Jupiter-Sun L4 and L5. http://youtu.be/yt1qPCiOq-8?t=43s You ...


1

According to this source, (and this one), it's the "wobble" one can detect in nearby stars if they have an orbiting exoplanet - either detected directly or via Doppler shift of the star's spectrum.



Top 50 recent answers are included