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Has there been an experiment to measure the (my vocabulary may be incorrect), intensity, strength of gravity emitted from a body, when approaching and retreating from a measuring device? If so, was there a difference between the two measurements at the same distance? Is the gravitational field stronger from a body when coming towards you than it is when its moving away from you?

Paul
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It is an important principle in physics that all motion is relative. That means the situation where you are stationary and an object is coming towards you at some speed $v$ is physically equivalent to the body being stationary and you approaching it at the same speed $v$. We can use this to restate your question as:

Is the gravitational field from a body stronger when you are approaching it than it is when you are receding from it?

And this is easy to answer because many of the objects in the Solar System are in elliptical orbits so the alternately head towards the body the are orbiting then head away from it. For example the Earth is in an elliptical orbit around the Sun, though the Earth's orbit is only slightly elliptical. That means if the strength of the gravitational field was different in the approach and retreat parts of the orbit the orbits would be changed in a way that would be very easy to detect.

And no such perturbation of the orbits has ever been detected. So we can be pretty sure that the strength of a gravitational field is the same whether your are approaching it, receding from it or stationary with respect to it.

Our current best theory of gravity is general relativity, and this theory has been tested to high accuracy. General relativity predicts no difference in the gravitational acceleration when approaching or receding from a body, so this reinforces our confidence that no such difference exists.

John Rennie
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  • Thank you, that makes sense even though its counter intuitive. – Paul Jan 03 '20 at 15:15
  • @Paul Perhaps this will make it more intuitive for you. Make a movie of some bodies interacting gravitationally. Now play the movie backwards. The reversed motions in the backwards movie are physically correct. So if you see such a movie, it's impossible to tell if it's playing forwards or backwards. – PM 2Ring Jan 03 '20 at 16:11
  • That is my very stumbling block. The movie requires a function of time to exist. If I sit on a seesaw, the opposite end goes up as I go down. It doesn't require time to react. If gravity was an alteration to space, the effects should be instantaneous. But with the detection of gravity waves, came the speed of gravity's effects. To my mind, which it seems is wrong, indicates that gravity is emitted, and travels. Anything which travels, should have an alteration in its effects depending on whether the distance is reducing as opposed to increasing. I accept the answer, but don't like it. – Paul Jan 03 '20 at 20:09
  • @Paul Ok. In Newtonian theory, gravity acts instantaneously, so Earth orbiting the Sun is attracted towards the direction the Sun is right now, not the direction where the Sun was 8 minutes ago. But in general relativity, gravitational waves travel at the speed of light, so it may seem that GR's orbit predictions would strongly clash with Newton's. But they don't - GR corrections to planetary orbits are tiny. This phenomenon is discussed here: http://math.ucr.edu/home/baez/physics/Relativity/GR/grav_speed.html and elaborated on here: https://physics.stackexchange.com/a/75124/123208 – PM 2Ring Jan 03 '20 at 21:01
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John Rennie's empirical argument is correct. The Earth (non-spinning) causes a gravitational acceleration of a test mass that is independent of the relative velocity between them. Here is one way to show that mathematically.

There is an analog called GravitoElectroMagnetism (GEM) between weak field gravity and electromagnetism. In our case the gravitoelectic field of the Earth is

$$\vec{E_{grav}}=-G\frac{M_{Earth}\hat{e_r}}{r^2}$$

The gravitomagnetic field of the Earth is

$$\vec{B_{grav}}=\frac{G}{2c^2}\frac{\vec{S}-3(\vec{S}\cdot \hat{e_r})\hat{e_r}}{r^3}$$

where $\vec{S}$ is the angular momentum of the Earth. The gravitoelectromagnetic analog of the Lorentz force in electromagnetism is

$$ \vec{F_{grav}}=m(\vec{E_{grav}}+4(\vec{v} \times \vec{B_{grav}}))$$

where $m$ is the mass of our test particle. If the angular momentum (spin) of the Earth is zero, then the force of gravity has no dependence on $\vec{v}$ as answered by John Rennie. If the spin of the Earth is non-zero, then the force on the test particle will depend on $\vec{v}$ and the path of the particle will bend around the $\vec{B_{grav}}$ field lines just like a charged particle does in electromagnetism. The effect of the $\vec{B_{grav}}$ of the Sun and planets in the solar system on the orbits of other planets is so small that it is normally neglected.

Gary Godfrey
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