Tag Info

New answers tagged

5

Orbits in Schwarzschild spacetime can be described by the effective potential $$V_\text{eff} = -\frac{GM}{r} + \frac{\mathfrak{l}^2}{2r^2} - \frac{GM\mathfrak{l}^2}{c^2r^3}\text{,}$$ where $\mathfrak{l} = r^2\dot{\phi}$ is the specific angular momentum of the orbit, which is a conserved quantity. The first two terms match the form of the Newtonian effective ...


2

The two terms are used in answering different questions. Hill Sphere: given a large mass (eg Sun) and a small mass (eg Earth), can a tiny mass (eg Moon) find a stable orbit around the small mass? (If the tiny mass goes outside the Hill Sphere of the small mass, no.) SOI: given two large mass objects and a small object between them, (eg sending a probe ...


0

There is a much easier way to do this. 1) Look up the length of the solar year in earth days 2) multiply the length of the years like this: Mercury year * Venus year * Earth year * Martian year * Jovian year * Saturn year * Uranus year * Neptune year 3) Divide by 365 to get earth years. And you have a time when they will align again ...


4

Very cool question. I want to get into a little bit of detail here because otherwise there would be a one-paragraph answer, and I don't think that would cut it. So here goes. The planets in the solar systems have orbits with pretty low eccentricities (see this for more eccentricity values). At the upper end is Mercury, with an eccentricity of 0.2056. At the ...


5

Back in 1246 CE, the perihelion coincided with the December solstice. The Earth was at perihelion at 1 Jan in 2005 and will be on 6 Jan in 2096, GMT time. The drift is not monotonic, but it does progress overall throughout the centuries. Eventually, it will drift towards the March equinox and beyond, to come back to the winter solstice in around twenty ...


1

The beginning of the year (1st of January) has been chosen for historical reasons (near the winter solstice) and has absolutely nothing to do with Earth's perihelion passage. Earth's perihelion could as well have occurred on the 10th of October. That it is near the 1st of January is pure chance (see for instance a post by EarthSky). It is, therefore, not a ...


3

For really low orbits over an atmosphereless body, the body needs to have a uniform density. Otherwise the gravity field is not symmetric, the orbit changes shape over time, and you end up with a crater. NASA had trouble with this, lunar Mascons, during the Apollo era. In one case: "The Moon has no atmosphere to cause drag or heating on a spacecraft, so ...


0

This isn't necessarily a full answer to your question, but this could cover at least part of it. Have you ever heard of the Roche limit? It's the distance inside which an object cannot orbit another object because tidal forces will tear it apart. That's thought to be the reason behind the rings of the gas giants: In each case, a moon could have wandered too ...


0

I use this app for it. Thanks to it I could be able to take a picture of it. https://play.google.com/store/apps/details?id=com.runar.issdetector&hl=es


0

Centrifugal force is too small to cancel gravity. The ISS, as all the satellites, are actually "always falling", but the horizontal component of their speed will always take them over the horizon line, keeping the distance to Earth constant. You have a very good explanation here http://www.lasalle.edu/~smithsc/Astronomy/Orbits/orbits.html


0

i use this site and it works perfectly. You just enter your coordinates and you can get times ( UTC ) and cardinal directions ( i.e. from SW to NW ) for all passes of ISS and a number of satellites. http://heavens-above.com/



Top 50 recent answers are included