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Tidal effects are due to the difference in the gravitational acceleration from another object at two different points on Earth. If $P$ is the point directly below the object and $R$ is its antipode, …

John Baez's article How Many Fundamental Constants Are There? is a pretty good explication of how physicists think about fundamental constants. It's not quite written at the layman's level, though — …

It emerges from Einstein's general relativity, rather than Newtonian gravity (which, as has been noted in other answers, is definitely not Lorentz-invariant.) … This effect was finally confirmed a few years ago (after many, many years of trials and tribulations) by the Gravity Probe B experiment. …

Answer: yes, so long as you use the word "reflect" broadly enough. Einstein's equation says that the curvature of the metric is related to the stress-energy tensor by $$ G_{\mu \nu} = 8 \pi G T_{\mu …

In the original $f(R)$ gravity, the action is $$ S = \int \sqrt{-g} \left[ \frac{f(R)}{16 \pi G} + \mathcal{L}_\text{mat}(g^{ab}, \psi) \right] \, d^4x. $$ Here, $\psi$ represents the collection of matter … So in that sense, $T_{ab} = 0$ in the absence of matter ($\psi = 0$) in $f(R)$ gravity if and only if the same statement holds in conventional gravity. However, there's one small ambiguity here. …

Yes, it will fall back to Earth (given the initial velocity you mentioned.) But that's to be expected, since you can't get a satellite into orbit by launching it vertically. Presumably this is a sim …

As the balls fall, gravity will pull them both the same distance downward relative to the positions they would have had if there was no gravity. … And so no matter how strong gravity is, the ball will hit the target. …

The Moon has a "tidal bulge" of about 50 cm due to the Earth's gravity. This was measured in 2014 via data from the Lunar Reconnaissance Orbiter. …

The first law would still be valid. The Binet equation can be used to show that any central force of the form $\vec{F} = -k/r^2 \hat{r}$ (with $k > 0$) will lead to orbits that are conic sections. Ho …

If you allow for non-Newtonian gravity (i.e., general relativity), then an extended body can "swim" through spacetime using cyclic deformations. … Even in Newtonian gravity, it appears to be possible. The second paper above cited "Reactionless orbital propulsion using tether deployment" (Acta Astronautica, v. 26, p. 307 (1992).) …

If the aircraft is not actually moving up or down, then there is zero vertical displacement, and so the energy imparted by the lift force is zero. (And so is the power.) This is a standard result th …

The kinetic energy of a relativistic object is given by $K = (\gamma - 1) m c^2$, where $\gamma = 1/\sqrt{ 1-v^2/c^2}$. Since $\gamma \to \infty$ as $v \to c$, the kinetic energy of an object diverge …

Systems, and Kleppner & Kolenkow's An Introduction to Mechanics, among many others. 1 As noted in the comments, I think what you're calling "standard GR" is what almost everyone else calls "Newtonian gravity …

Tidal forces find a very natural expression in the language of spacetime curvature, via the concept of geodesic deviation. Here's how: First, we need the concept of a geodesic. This is the equivalen …

The parametrized post-Newtonian (PPN) formalism is a generalized way of exploring gravity theories, including general relativity. … In other words, we're pretty sure that kinetic energy and pressure create the same amount of gravity that mass do to within a few percent. …