# Tag Info

18

This is actually pretty difficult to do, because it depends on where from due to uneven distribution of matter (local parameters), how far from the galactic center due to radial velocity, the direction in which you want to reach escape velocity (how much of the radial velocity can be used), and that it's hard to estimate mass of the Milky Way (global ...

17

One of your instincts was correct; it is indeed the influence of the Moon. Wikipedia notes: Over millions of years, the rotation is significantly slowed by gravitational interactions with the Moon; both rotational energy and angular momentum are being slowly transferred to the Moon: see tidal acceleration. And here is the general case of a satellite ...

14

Yes, well, kind of; JAXA's Kaguya (SELENE) took images of the Earth during the February 10, 2009 penumbral lunar eclipse from lunar orbit of roughly 50 km altitude, using its HDTV camera:    Japan's Kaguya probe caught this sight of Earth's diamond ring as the planet passed in front of the sun as seen from the lunar orbit. The ring appears ...

13

Getting hard numbers about how accurate measures we can get from current systems, adapted to the Sun instead of far away stars is difficult, bordering to impossible. But we can get data about the relative difficulty of the solar system planets. First off, we can do some cheating for Mercury and Venus, as they occasionally go in front of the Sun. Given your ...

12

The Earth's rotation rate is very gradually slowing down. Over the short term, the Earth's rotation rate changes a bit chaotically, sometimes speeding up, sometimes slowing down. These short term changes are due to transfers of angular momentum amongst various components of the Earth. The long term secular variations in the Earth's rotation rate are due to ...

11

Planets are much, much bigger than you think. The total mass of all the bodies in the main asteroid belt is about half of 1% of Mars's present mass. There's no realistic way to round up enough loose matter to appreciably increase the mass of a planet. 38% of Earth's gravity is pretty substantial. Humans have managed to survive a year in microgravity ...

11

Rearranging the lifeless rocky planets might make terraforming and transportation easier. No, it mightn't, because the amounts of energy it would involve are so ridiculously gigantic that terraforming a planet is a very easy job in comparison. The kinetic energy of an orbiting body is $\epsilon_k = G\cdot \frac{m\cdot M}{2\cdot r}$ where $G$ is the ...

11

This is not a complete answer. It is instead an extended comment to the following: I understand I will have to retard the gravity from each source by its particular light-time, as well as correct for the light time for the HST images. While you do want to correct for light time travel with regard to seeing a moving remote object, you definitely do not ...

10

The moon weighs 7.34 × 10$^{22}$ kg. For an asteroid to have a noticable effect, its mass must be in that ballpark. the asteroid mass needed also depends on its speed (kinetic energy). what timeframe do you want to consider? On a timescale of a billion years, a tiny asteroid will have measurable effects: the change in orbit will be tiny, but over ...

10

As many of the comments have already mentioned, there are several different reasons people might recommend the use of Fortran over Matlab. One of the most straightforward answers is that a lot of legacy (read: validated) code is written in Fortran, and depending on your job function, learning to use Fortran might make you more productive - for instance, if ...

8

This is to do with Frames of Reference. The video below gets across nicely that all motion is relative, your perception of it depends on your point of view: Frames of Reference - Hume and Ivey, 1960, Educational Services Inc. If we chose the Sun as the center point of the Universe, we could say everything moves around it - ...

8

Setting aside the fact that this would be a very bad idea and that there are far less energy intensive ways to terraform the surface of a planet, the most energy efficient way to do this I can think of is by using Oort Cloud objects. It takes very little $\Delta V$ to make an Oort cloud object dive towards the Sun. In fact, it is the source of most comets. ...

7

The most up-to-date is from the work in this paper, due to be published in July 2016. The actual numbers can be found in this large file, for which the number you seek is the second number in the file, and its uncertainty is the third number in the file. (The rest of the numbers are the higher order terms of the gravity field. You can find a detailed ...

7

For reasons of convenience, departure excess energy is plotted as $C_3$, and arrival excess energy is plotted as $v_\infty$, where $C_3=v_\infty^2$. The reason is that launch vehicle performance is always quoted in terms of $C_3$, whereas computing things like orbit insertion $\Delta V$ or entry velocity is more straightforward with the arrival $v_\infty$. ...

6

callhorizons is depricated now and refers to the python library astroquery which now seems to be the way to go. astroquery (GitHub, readthedocs) is "an astropy affiliated package that contains a collection of tools to access online Astronomical data. Each web service has its own sub-package.", where making Horizons queries is just one of many options. It's ...

5

Let's see what it would take to move Mars to an Earth orbit. It's not a pretty picture. The most efficient way to move from one orbit to another is via a Hohmann transfer. We'll apply a delta-V to Mars to slow it down and put the planet into an elliptical transfer orbit that just intersects Earth's orbit, then another delta-V once Mars reaches perihelion. ...

5

One method of calculating the angle involves using the ellipse reflection law. Light from one focus reflects off the ellipse into the other focus. Thus in the picture below (by the author), the radial vector from the focus $F_1$ is reflected at $P$ onto the second focus $F_2$, forming a triangle whose third side is the line between the foci. Your flight ...

4

Your encounter must by definition occur somewhere around the MOID line, if you do not consider to significantly change the trajectory. A solution can be obtained using Keplerian analysis, but that is only an approximation, due to the hard-to-restrict three body nature of the problem. If I understand your question correctly, you want a higher degree of ...

4

As Russell Borogove pointed out, that plan is quite impossible because you need to get that mass from somewhere, and all the other mass is already concentrated on even larger planets. And even when we would have enough asteroids, the plan would still not be even remotely feasible due to the huge energy requirements it would take to change the orbits of such ...

4

I massaged some raw numbers from https://nssdc.gsfc.nasa.gov/planetary/factsheet/ For each body-Sun pair the velocity of the Sun is the velocity of the planet times the ratio of the masses since they orbit around their center of mass. Eclipse depth is just the ratio of diameters. Jupiter results in the largest velocity by far, thought the amplitude of the ...

4

The rings became flat over time as the trillions of particles in them collided over and over, slowly causing their vectors (direction of motion) to average out until they were all aligned in the same direction. See this short video from Minute Physics. A flat disk rotating in one direction is the only arrangement of these particles that is stable. The ...

4

Within Known Physics There are three vaguely plausible methods for changing a planet's orbit that come to mind without breaking the accepted laws of physics: direct application of thrust Consistent close passes of a series of massive objects targeted solar flares Direct application of thrust is going to be incredibly disruptive - it's either by impact or ...

4

Think of the earth as a 12,800-km diameter gyroscope. It takes an enormous amount of energy to change the axis of rotation. The blips that are there are due to gyroscopic precession, primarily due to gravity gradient torques from the sun and the moon.

4

The laws of planetary motion and orbits are underpinned by Newtonian Physics and Kepler's Laws. These physical laws apply to everything in the universe and, as such, apply equally to the motion of planets and the motion of artificial satellites. The major difference when calculating orbits for an artificial satellite is that it's very common to ignore its ...

4

If your ellipse is a circle, the Flight Path Angle is 0. You’re done. Otherwise, for an elliptical orbit, start with the polar equation that relates radial distance $r$, true anomaly $\theta$, semimajor axis $a$, and orbital eccentricity $e$: $$r=\frac{a(1-e^2)}{1+e\cos\theta}$$ Solving for $\theta$ gives us the following: \theta = \arccos\left({\frac{-...

3

First of all you need to specify the bodies you are considering. There are 5 Lagrangian points in the Sun-Earth system and there are other 5 Lagrangian points in the Earth-Moon system. Basically you can find 5 Lagrangian points (which are equilibrium points int the synodic reference frame) for each system made of 3 bodies where $m_3 << m_1,m_2$. In ...

3

It's not possible! At least not if you want to keep Earth safe in its orbit. And while I like some suggestions previously mentioned here because playing celestial game of pool sure is fun, they're inherently not possible because of the timescale at which all these orbital changes would have to be applied. So forget about Hohmann transfer on a planetary scale,...

3

It depends on where and when the probe is captured and read. If it's within a few thousand years and within our own galaxy, photos as suggested by BAR might be enough. Redshift would also work. NASA put a map on the Voyager spacecraft using the position of pulsars. If the timescale is much longer (millions of years), using a photo of the night sky ...

3

I just ran it, and mine look pretty much like those in the paper. See some coordinates at the bottom. Here are some {x,y} coordinates at the times in the left column: 0. {1.,3.} {-2.,-1.} {1.,-1.} 5. {2.46917,-1.22782} {-2.2782,-0.20545} {0.34106,0.901049} 10. {0.77848,0.141392} {-2.02509,0....

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