Here a way to understand why apples fall because of quantum mechanics.
First you need to know that apples are accelerated toward earth center because they follow longest timelike spacetime worldline, and because time is longer at the tree level than at the earth level. We will consider that the gravity acceleration g is the same at both altitudes.
For a small difference of altitude δh, the relative difference of time indicated by two clocks after a delay t is
δt/t=gδh/c2.
For g=9.8m/s2, the relative difference is only
1.1⋅10−16 for δh = 1 m. This is very small, but it is sufficient for understanding the falling apple. You just need to compute and maximize the length of its worldline in spacetime.
Now the problem is to evaluate this 1.1⋅10−16 relative time difference using quanta.
First I rewrite the equation as δt=gtδh/c2. And you should ask «what is t, exactly? Is it the time at apple level?». The response is no, it is the time at infinity, it is the variable t used in Schwarzschild metric. The difference will be not negligible for black-holes apple-trees.
That’s a pity to measure this t time at infinity. Why not to measure it on the mass worldline, as its proper time? Problematic with black holes. Less problematic for an homogeneous earth. Nevertheless a formula using the time at earth center will be awful. Quantum orientated formula will be much simple and beautiful.
We know that the approximate equation is
δt=GMtR2δh/c2
The product Mt is the action of the mass worldline. It is natural to measure it as number nq of h quanta: nq=Mc2/h. Earth action: 8.103⋅1074 quanta per second. A huge value for a so small δt.
The radius R seems also problematic. In fact this radius should be not measured as the distance from M but as the ratio of the circonference on 2π or better R2=dSdΩ, with horizontal dS and horizontal surface delimited by non parallel vertical lines defining a solid angle dΩ.
Now we can write the exact formula:
δt=σPnqdΩdSδh
The constant σP=2πlPtP can be called the unreduced Planck space-time surface, having the extraordinary small value σP=5.475⋅10−78m⋅s. The constants lP and tP are respectively the (both reduced) Planck length and Planck time.
Eventually we compute the expected value for the gravity acceleration at the surface of earth as the product of these extraordinary small and big quantum values. First in natural units:
δttδh=5.475⋅10−78×8.103⋅1074×(6.31⋅106)−2=1.1⋅10−16m−1,
Then in SI units:
g=c2δttδh=9.8m⋅s−2
Originally Answered: Why would gravity have quantum properties?
Because the universe is inherently quantum. The definitions of time and space are incompatible between gravity and quantum mechanics. Our theory of gravity only works because we apply it at scales where the incompatibilities are negligible, and don’t affect the results.
To properly model a situation where both gravity and quantum mechanics have significant effects, you’ll need to harmonize the definitions of space and time. That will inherently be quantum-mechanical; there is no way to account for quantum mechanical effects with classical space-time.
Most likely, the resulting theory won’t really be identical to quantum mechanical space-time but will instead be yet another thing. That “other thing” is going to have to have a lot of the properties that we think of as quantum mechanical, so it’s easiest to say that gravity will have quantum properties.