If large mass causes a curvature around spacetime, then why don't we see a gravity lens around our planets?

  • 25
    $\begingroup$ Did you try to calculate the magnitude of the effect? $\endgroup$ Jul 20, 2020 at 11:59
  • 1
    $\begingroup$ I don't know about physics and calculations. It's just an open question to learn more. $\endgroup$
    – user6579
    Jul 20, 2020 at 12:07
  • 3
    $\begingroup$ I've removed a number of comments that were attempting to answer the question and/or responses to them. Please keep in mind that comments should be used for suggesting improvements and requesting clarification on the question, not for answering. $\endgroup$
    – David Z
    Jul 21, 2020 at 2:20
  • 1
    $\begingroup$ Historical side-note: Einstein originally wanted experimentalists to measure light lensing around Jupiter to support GR. But, then when he did the calculations he found out that the amount of lensing would be too small to be detected then. $\endgroup$
    – PNS
    Jul 21, 2020 at 4:26
  • 3
    $\begingroup$ @OscarBravo: Answers should not rewrite questions, it defeats the purpose of keeping the question around. Question correction is independent of whether answers have been given and what those answers may contain. $\endgroup$
    – Flater
    Jul 21, 2020 at 10:58

4 Answers 4


Contrary to the other answers, I will point out that gravitational lensing due to planets in the solar system is a significant and measurable effect. The measured positions of stars, as seen from a point in the solar system near the Earth are altered by the gravitational deflection due to the fields of the Sun and then, in order of decreasing effect, Earth Jupiter, Saturn, Venus, Uranus, Neptune.

The size of the effect depends on the angular separation of a star from the solar system object and can be anything from 0 to 70 microarcseconds.

This sounds very small, but is easily within the precision of positional measurements by the Gaia spacecraft. In fact, the effects of these deflections have to be taken into account in order to provide accurate positions as intermediate data for the calculation of the parallaxes and proper motions of stars.

A brief information sheet on the topic has been produced by ESA.

However, this lensing and these kinds of deflections are not strong enough to produce multiple images or Einstein rings. The angular size of a perfect Einstein ring is given by $$\theta_E \simeq \left[ \left(\frac{4GM}{c^2}\right) \frac{D_{LS}}{D_L D_S}\right]^{1/2} \ ,$$ where $M$ is the mass of the lens, $D_L$ is the distance from observer to lens, $D_S$ is the distance to the source and $D_{LS}$ is the distance between the lens and the source. Since in this case $D_{S} \gg D_{L}$ and $D_{S} \simeq D_{LS}$, then $$\theta_E \simeq \left(\frac{4GM}{c^2D_L}\right)^{1/2} = 0.071 \left(\frac{M}{M_{\rm Earth}}\right)^{1/2} \left( \frac{D_L}{1{\rm AU}}\right)^{-1/2}\ {\rm arcsec}$$

Taking Jupiter as an example, this yields $\theta_E \sim 0.6$ arcsec. But this is a lot smaller than the angular diameter of the planet ($\sim 50$ arcsec). In other words, to see multiple images or rings you would need to be much further away from Jupiter than a few AU and a star that is directly behind Jupiter is completely hidden when viewed from the Earth.

  • 9
    $\begingroup$ It might be worth adding for OP that not only do planets produce lensing effects, this is one of the ways we detect exoplanets (Wikipedia claims we've found 89 exoplanets this way). $\endgroup$
    – user107153
    Jul 20, 2020 at 17:57
  • 7
    $\begingroup$ @tfb This method of detecting exoplanets does not use lensing effects of the exoplanet, but instead looks for aberrations that an exoplanet causes on lensing by its parent star of a more distant star or galaxy. $\endgroup$
    – gandalf61
    Jul 20, 2020 at 19:03
  • 3
    $\begingroup$ The point is that you are not directly detecting the [IMPOSSIBLY TINY] lensing of the planet. You are detecting a fluctuation in an (BARELY ACCESSIBLE TINY) lensing effect of the parent star. $\endgroup$ Jul 20, 2020 at 21:21
  • 7
    $\begingroup$ @RossPresser the perturbation in the lensing signature (in this case is a magnification and positional shift) is due to the exoplanet. However this question was about "our planets". $\endgroup$
    – ProfRob
    Jul 20, 2020 at 21:27
  • 4
    $\begingroup$ @gandalf61: Detecting a perturbation in the lensing caused by a star is detecting the lensing caused by a planet. Saying otherwise is like saying that detecting small perturbations to Newtonian predictions of the orbit of Mercury is not detecting effects due to GR: it is. However, as Rob Jeffries says this is a question about Solar system planets, so perhaps not relevant to it. $\endgroup$
    – user107153
    Jul 21, 2020 at 9:25

The deflection angle for a light ray that is just grazing the surface of a planet or star (so the maximum observable deflection) is

$\displaystyle \theta = 2 \left( \frac {v_e} {c} \right)^2$

where $\theta$ is in radians, $v_e$ is the escape velocity of the planet or star and $c$ is the speed of light (see this Wikipedia article). For a planet, $\frac {v_e}{c}$ is too small for this deflection to be detected. You need an escape velocity of hundreds of km/s to produce a measurable deflection of light, so this requires something as massive as a star or as compact as a black hole.


Ok, so deflections of light rays by planets in the Solar System can be detected by space based instruments such as Gaia. But to get a lens effect (i.e. multiple images) you need to be far enough away from the planet that the planet's angular diameter is less than twice the maximum deflection. So we are too close to any planet in the Solar System to see it cause multiple images of a distant object due to gravitational lensing.

  • 5
    $\begingroup$ I would just like to note that it is possible to see even small deflection angles, but you need to be very far away. That is how e.g. galaxies end up being effective lenses. $\endgroup$
    – Void
    Jul 20, 2020 at 14:26
  • 2
    $\begingroup$ The angle of deflection is detected and is routinely measured and corrected for. Neither is this the reason specifically that gravitationally lensed images are not seen. $\endgroup$
    – ProfRob
    Jul 20, 2020 at 17:45

The effect is proportional to mass and it takes a big galaxy to see it. A big galaxy has typically a mass of $10^{11} $ solar masses so about $3\cdot10^{17} $ Earth masses. The angle of deflection is incredibly small for a planet.

  • 1
    $\begingroup$ I do love the typo "soNar" masses :-) $\endgroup$ Jul 20, 2020 at 13:50
  • 6
    $\begingroup$ The angle of deflection is small but can be, and is, measured. $\endgroup$
    – ProfRob
    Jul 20, 2020 at 17:46
  • 1
    $\begingroup$ It's not just the mass but also the density that matters. A black hole a tiny fraction of a galaxy's size can cause more lensing. $\endgroup$ Jul 20, 2020 at 23:00

Your question is a red herring. There is this effect around all bodies. It is probably very weak and therefore not likely to make a significant impact of local observations. "Lensing" is the result of distorted ray paths and ALL sources of gravity will distort ray paths. The issue is whether this will be noticeable over short distances.

  • $\begingroup$ Based on false assumptions? Yes. But a red herring? To my understanding, a red herring deliberately misleads. Anyway, it's well possible that there some nuances of the real-world use of the term I just don't get, for I'm not an English native speaker. $\endgroup$ Jul 22, 2020 at 4:13
  • $\begingroup$ The term "red herring" can and is often applied to cases where false info is inadvertently provided but you are correct that perhaps the original and more prevalent use is in describing an intentionally misleading piece of info. I use the term in the more loose sense (presence of false info even inadvertent). Was not my intent to accuse the OP of intentional misleading. $\endgroup$
    – user196418
    Jul 22, 2020 at 19:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.