Why no gravity lens around planets?

Question

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

Answer 1

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.

Answer 2

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.

Correction:

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.

Answer 3

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.

Answer 4

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.