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I wonder whether we know at how much gravitation a gravitational lensing around a celestial body becomes observable (visible in the sense of that we see stars on the wrong places, stars behind the body etc, similar like around this neutron star, but of course by far not as strong, I'd like to know where the boundary lies). Such visible gravitational lensing occurs around neutron stars, black holes and white dwarfs, right? What about red and brown dwarfs? They're also quite dense and have a high surface gravity (though not as high as the other mentioned bodies of course). Do they curve the space around them visibly?

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    $\begingroup$ Depends on what wavelength you want the effect to be observable at. For pulsar timing and radio astronomy, the extra time delay (the Shapiro delay) caused by the gravitational lensing is included in the standard TEMPO2 analysis code for everything more massive than Venus (see Table 2 of Hobbs, Edwards and Manchester 2006) as it has effects on the signals if they are passing close enough to one or more of the planets/Sun. So you may want to edit and clarify the question for what you mean by "observable" $\endgroup$ – astrosnapper Feb 5 at 18:25
  • $\begingroup$ @astrosnapper Sorry, I don't understand a word. :-) I mean "visible to our eyes", I 'll clarify my question. $\endgroup$ – user30007 Feb 5 at 19:20
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    $\begingroup$ Sorry let me try again. The effects of gravitational lensing from the Sun and planets are "observable" to radio astronomers looking at the pulses of radio "light" coming from distant pulsars. The extra distance the radio pulses have to travel as they bend round the Sun, Jupiter or Saturn have to be accounted for when they look at their data. Although this is "observable" and "light" in the broad electromagnetic spectrum sense, it's probably not what you meant so the clarification to mean "visible light" is helpful. $\endgroup$ – astrosnapper Feb 5 at 19:56
  • $\begingroup$ @astrosnapper Now I understand. You can't see the Sun bending visible light at all. I'll improve my question again. $\endgroup$ – user30007 Feb 6 at 6:06
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    $\begingroup$ As that Wikipedia article says, "For light grazing the surface of the sun, the approximate angular deflection is roughly 1.75 arcseconds". Sure, you won't notice that by naked eye observation, but it is observable by careful observation with a telescope. $\endgroup$ – PM 2Ring Feb 6 at 9:50
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The article 'A practical relativistic model for microarcsecond astrometry in space' (Klioner 2003; caution, very heavy duty maths...) describes the framework which was developed to support the data processing for the ESA Gaia astrometric space mission. This mission has the goal to measure the position (astrometry) of stars and other objects to approx. 20 microarcseconds (one millionth of an arcsecond, which is 1/3600th of a degree). In order to do this, you need to account for all the sources of error and deflection, such as light bending near massive objects, to a level that is well below you expected accuracy (this level of systematic error was chosen to be 1 microarcsecond, hence that part of title).

Section 6 of the paper discusses gravitational light bending and Table 1 gives the size of the effect for various bodies in the Solar System. The column headed by $\delta_{pN}$ gives the size of the effect in microarcseconds ($\mu$as). As expected the Sun has the largest effect at $1.75 \times10^6\,\mu$as or 1.75", with Jupiter second etc. The next two columns gives the angular separation a light ray has to pass to the body to cause a $1 \mu as$ or $10 \mu as$ effect. So for example, any light ray from a distant object passing within $11.3^\circ$ of Jupiter will cause at least a $10 \mu as$ effect. The smallest bodies that have an effect larger than $1 \mu as$ are the dwarf planet Ceres (mass $8.958 \times 10^{20}$ kg), and the moons Dione ($1.05 \times 10^{21}$ kg) and Umbriel ($1.27 \times 10^{21}$ kg) - these masses are several thousand times smaller than the Earth for example.

Now the level of precision needed by Gaia is far more precise than and the deflections are tens of thousands times smaller than you would see in an optical image with a regular telescope e.g. Eddington's confirmation of general relativity during the 1919 total solar eclipse (space.com story). The Klioner (2003) paper includes a formula to estimate the size of the effect of the light deflection as a function of the density but you need to know what level of deflection you care about to get an answer. The formula (number 35 on page 11 of the paper) for the radius $L$ of a body producing a light deflection than $\delta$ is: $$ L\geq \left(\frac{\rho}{1\,\textrm{g cm}^{-3}}\right)^{-1/2}\left(\frac{\delta}{1\,\mu\textrm{as}}\right)^{1/2}\times 624\,\textrm{km} $$ where $\rho$ is the mean density of the body. Once you have chosen a level of deflection/lensing you want ($\delta$), you can chose suitable densities (e.g. asteroids are $\sim2\,\textrm{g cm}^{-3}$, Earth and other rocky planets are $\sim5.5\,\textrm{g cm}^{-3}$)

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