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1. Is material on Earth's surface not in free fall around Earth's center? No. Material on the Earth's surface -- or inside it -- is not in orbit, and so is not in free fall. You can temporarily put yourself into an orbit (and thus into free fall) by jumping up into the air, or jumping off a higher surface. When you do this, you are briefly in a very ...


11

Assuming a circular orbit of a planet of mass $m_p$ around a star of mass $m_*$, with an orbital period of $P$ and we can assume that $m_* \gg m_p$ for the case of an Earth-like planet. Newton's second law tells us that $$ m_p a \frac{4\pi^2}{P^2} = G\frac{m_* m_p}{a^2}\ , \tag*{(1)}$$ where $a$ is the orbital radius. The observed radial velocity amplitude ...


10

Masers tend to be extremely bright, compact sources with line emission at cm wavelengths (usually from OH or H$_{2}$O molecules; this is a technique for gas clouds, albeit one that only works in special conditions). This makes them ideal for very long baseline interferometry (VLBI) with radio telescopes, which allows you to get extremely precise and accurate ...


6

From its RV graph alone, with no other information, you cannot calculate inclination. This is why RV velocity measurements are typically reported as "$v \sin{i}$", because what you're actually measuring is the orbital velocity projected along the line of sight. Without other information, you cannot disentangle the orbital velocity from the viewing angle. ...


6

What's happening is that (i) the star is spinning and (ii) it is being eclipsed (by an exoplanet perhaps?) As the exoplanet (or whatever dark object eclipses the star, the same explanation would apply to a brown dwarf or very low-mass companion) traverses across the face of the star it will eclipse parts of the stellar surface that gave their own specific ...


5

I don't know of a source for the CSV data directly but if you are OK with a little bit of Python, this can be done with the NASA Exoplanet Archive. Looking at one famous example (HD 189733b), if you do a search for this object on the front page it should bring you to a page of detailed information and links to datasets. Expanding the 'Ancillary Information' ...


5

Imagine you are in orbit around the earth, several 100 km upwards. What happens when you slow down? That's right, you fall down until some force stops your fall. That force is the pushback from the ground. So next imagine: What happens when you throw a ball in the air? It falls back down to the ground. So it follows, that the ball is too slow to be in orbit....


4

If you are referring to the Earth... If the number is positive, the star is moving away from us. When the number is negative, the star is moving towards us. Hope that helps! Source with more info: https://cseligman.com/text/stars/radialvelocity.htm


4

It depends what lines you are measuring and in what kind of star. When you measure the RV from a spectral line, you are measuring an intensity-weighted average RV over the region where the line is formed. For a star like the Sun, the photospheric lines are all formed within a layer no thicker than about a 1000 km and the differences in RV with depth in the ...


3

An interesting corollary to this question: if the ground is not in orbit, how does it move (roughly) in a circle? If we model a section of ground as an isolated particle, it's clear that in order to move in a circle despite having a relatively low tangential velocity, it would need to have an ongoing force being applied to counteract the direction the ...


3

@NeutronStar's excellent answer sums up the situation nicely: From its RV (radial velocity) graph alone, with no other information, you cannot calculate inclination. This is why RV velocity measurements are typically reported as "𝑣sin𝑖", because what you're actually measuring is the orbital velocity projected along the line of sight. Without other ...


3

I am not sure what you meant by "introduce another one." The system of equations is derived simply from Newtonian mechanics, and it is already completed in itself. So, my guess answer is, no that is not possible. I think Figure 2 here is a good one to have in mind when thinking about this. The nature of RV data is that you have a bunch of data points ...


2

The number -5.5 km/s in this context means that the relative line of sight velocity between Sirius and the Sun is 5.5 km/s in the sense that Sirius and the Sun are getting closer to each other. Note that the line of sight velocity of Sirius as observed from the Earth will vary throughout the year, since the Earth moves with an orbital speed of 30 km/s ...


2

There are many ways to measure a star's motion, radial velocity (wobble) can be measured using doppler spectroscopy. The first exoplanet discovered by this method was 51 Pegasi b by Michel Mayor and Didier Queloz, who discovered the planet on December 1995. A drawback of this method is that it can only detect the movement of a star towards or away from the ...


2

Let's assume we had a mathematical method that could separate the planet's mass from the inclination of the planet-star system. Then we could derive the mass of the planet and the inclination independently from each other. In particular we would be able to determine the mass of the planet in the case of inclination $i = 0$. But this cannot be true for the ...


2

If the orbital period is $P$ (in units of days) and the time of periastron passage (a specific date when the stars are closest together) is $T_0$ (generally given as a Julian Date, or JD), then the conversion from orbital phase $\phi$ to JD (days) will be: $$ JD = T_0 + \phi P $$ In other words, $\phi$ is just the fraction of the orbital period, so ...


2

I hope that someone more knowledgeable than me will answer this question, but this is what I understand it is happening. The bootstrap method gives an estimate of the probability distribution of the period, given the data. If the probability distribution were a Gaussian centered on the true period, you would probably see a Gaussian. But when estimating the ...


2

No, it is correct. The inclination of an eclipsing binary can be estimated from the light curve (and is likely to be close to 90 degrees in order to produce an eclipse). The radial velocity curves can then give $M_{1,2}\sin(i)$ and hence the component masses.


2

$$v_r = c\left( \frac{\lambda - \lambda_0}{\lambda_0}\right),$$ where $\lambda$ is the observed wavelength and $\lambda_0$ is the wavelength at rest. This gives an equivalent velocity with respect to the central wavelength of the line. In your plot, what is shown is a line profile where the wavelength separation from $\lambda_0$ for that line (or from the ...


2

A line profile may be formed in material that has a range of line of sight velocities. The answer to your question depends on what radial velocity you are trying to measure. For example, if the line broadening is dominated by macroscopic plasma motions, then the wavelength of any particular point in the line profile corresponds to a different line of sight ...


1

I tried to find detailed derivation online after my fail. Finally I found it in The Exoplanet Handbook by Michael Perryman: Actrually K, or the so called radial velocity semi-amplitude, is not the peak radial velocity (which I understand before) but some kind of amplitude of the fluctuation of the radial velocity. And it must be considered in 3D space. let z ...


1

There are several corrections here, and as you suspect, all of them can be done in astropy. You'll need to do something like this: Transform your observed velocity to the solar system barycenter (thus correcting for Earth's rotation on its axis, and its orbit around the Sun. There is an example here in the astropy docs that shows how to do that. If you ...


1

Why not fit the orders you are interested in separately and then use the standard error of the mean (possibly weighted by the signal-to-noise in each CCF) as the precision in the final, averaged RV. In terms of what to fit, I don't see why a Gaussian is so bad? You probably need to limit the fit to the inner $\sim \pm 1 \sigma$ to avoid noise outside the ...


1

Proper motions are not velocities. They are angle traversed per unit time. They can be converted to tangential velocities (i.e. tangential to a line joining the star and the Sun) if the distance to the star is known, but you need additional information to get the line of sight (radial) velocity and hence a 3D velocity. Such velocities are with respect to ...


1

An excellent page to get most or all names for a star is Simbad which also happens to know the Kepler IDs (KID). Ont the other hand, every of the RV data files contains info on the star's identifier it belongs to in the header, e.g: \STAR_ID='HD 4628' So using Simbad you should be able to do a cross-matching. SimBad allows scripted query and includes ...


1

The first you need to determine if you're looking for a zebra or a unicorn. That is, if there's a herd of horses before you, you expect to see all horses. Maybe there's a wayward zebra in there. Unlikely but possible, but you're not going to find a unicorn because, as far as we know, unicorns don't exist. We know how planets form and we expect to find ...


1

To use the Doppler shift, you need to know by how much the light is blue/red-shifted. Suppose you have a light curve where the wavelength varies between $\lambda_1$ and $\lambda_2$. Using these wavelengths, the Doppler shift will give you an interval of radial velocities. This gives you a value for K that you can plug into your last equation to derive $M_p$...


1

You don't need to collect data for one cycle in order to determine the period. From Newtonian's physics, we know precisely how a trajectory of one particle orbitting around another one would be. Since the instantaneous velocity (v_inst) at any point on the trajectory is tangential to the point on the trajectory, we can decompose the v_inst into the radial ...


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