Understand the Acceleration Due to Gravity Formula and More!

Acceleration due to gravity during its journey up and down
When we throw an object up into the air, ignoring air resistance, etc, we define acceleration to be -9.8 m/s^2. When it goes down after its journey up, like a parabola, do we define the acceleration as 9.8 m/S^2, OR still the same as -9.8m/s^2?
I know that acceleration due to gravity is always constant, but would it be wrong to say it like the question i have above?

How do you calculate the acceleration due to gravity of a massive object (black hole)
Is there an online calculator to find the acceleration of gravity a massive object exerts from a specific distance? I know the effects of gravity kind of just fade out with distance but I'm asking for mass-specifics.
For example, you plug in the mass of the object and the distance from the object to get the m/s^2 (and maybe even the escape velocity)???
(I'm also thinking this would be some sort of exponential graph, like "acceleration" on the y axis and "distance" on the x axis)

Is the acceleration due to gravity the same no matter the mass of the object?
The University of Illinois Department of Physics article Q & A: Heavy and Light - Both Fall the Same states that the acceleration due to gravity is the same no matter the mass of the object.
However if an object, say for example the moon, were falling to the Earth, gravity would have to pull harder to speed it up the same. Would the acceleration of gravity be the same?
My question is not the same as Don't heavier objects actually fall faster … because I'm asking about acceleration of gravity, I'm asking if heavier objects would fall slower, not faster. So, wouldn't they fall slower because gravity has to pull harder?
Side note: Let's just say we also ignore the effects the other question mentions.

Need help in understanding accelaration due to gravity
I've been having some trouble in understanding acceleration due to gravity.
On earth, the acceleration due to gravity is an average of $9.80$$m/s^2$. The mass of the earth is approximately $5.972\times <!--\; \times \; -->{10}^{24}kg$. The acceleration due to gravity on the surface of the sun is $273.7$$m/s^2$ and its mass is about $1.989\times <!--\; \times \; -->{10}^{30}$
So
$\frac{\text{Mass of Sun}}{\text{Accelaration at Sun surface}}=\frac{\text{Mass of Earth}}{\text{Accelaration at Earth surface}}$
Why don't the above numbers equal each other? Is it because I am doing mass divided by acceleration?

How to expand this equation considering acceleration due to gravity into 3D vector space?
How can we expand this following equation into 3D vector space? I learned this equation from this answer: Don't heavier objects actually fall faster because they exert their own gravity?
The answer shows how we can find time as function of distance (or single dimension) when there's force due to gravity accelerating both pieces of mass as follows:
$t=\frac{1}{\sqrt{2G(m_1+m_2)}}(\sqrt{r_i{r}_f(r_i-<!--\; -\; -->r_f)}+r_i^{3/2}{\cos }^{-<!--\; -\; -->1}<!--\; \; -->\sqrt{\frac{r_f}{r_i}})$
Where $m_1$ and $m_2$ are the masses of two bodies, $r_i$ is initial distance from mass to another and $r_f$ is the final distance from mass to another.
If this equation had just the force caused by gravity within it, I could easily divide it into components, and use this equation for each dimension, but as seen, this equation only has masses and distances in it.
How can I apply this kind of equation into 3d vector space, ie. get the position as 3d coordinates as function of time when initial position and velocity are known?
If this equation cannot be applied in to 3d space, then how could we derive another equation, which applies same kind of relations into 3d vector space? (I checked the answer I linked above, but it goes somewhat above my understanding, as differential equations are used.)

Acceleration due to gravity on Coil vs Insulator
Consider I have a coil with a length of $5m$ and made up of copper wire with a diameter of cross-section $d$ and wrapped around an imaginary axis with radius $R$ with mass $M$
Now consider an insulator made up of a glass of mass $M$ of $5m$ of diameter $D$
If both were taken very high in the atmosphere but not away from the gravity of the earth. Then it is dropped from that point at the same level without providing any external force.
My question is that which object will move faster
Conductor. OR
Insulator
I have been taught that every object of any mass will have same acceleration due to gravity but electromagnetism course taught us that a coil will produce induction and it will slow down or speed up the coil.
Which theory is true?

Does GR imply a fundamental difference between gravitational and non-gravitational acceleration?
1.Does the equivalence principle imply that there is some fundamental difference between acceleration due to gravity and acceleration by other means (because there is no way to 'feel' free fall acceleration for a uniform gravitational field)?
2.Does General Relativity allow you to describe the acceleration due to gravity without Newton's second law (because every other source of 'push or pull' outside the nucleus involves the electromagnetic field)?
3.Is the acceleration due to gravity a result of changes in time dilation/length contraction as opposed to an actual push or pull?\

What is the jerk due to gravity with ascent?
Acceleration is defined as the rate of change of velocity with time. Jerk is defined as the rate of change of acceleration with time. What is the jerk due to gravity with ascent?

When an object, say a ball, is attracted by the black hole it gets acceleration due to gravity. Suppose light is moving towards the black hole vertical to it... then does it gain acceleration due to gravity? If yes then won't be the speed of light increase and get violated?

Is acceleration due to gravity constant?
I was taught in school that acceleration due to gravity is constant. But recently, when I checked Physics textbook, I noted that
$F={\displaystyle \frac{G{m}_1{m}_2}{r^2}}.$
So, as the body falls down, $r$ must be changing, so should acceleration due to gravity.

Question regarding calculating acceleration due to gravity on planet Mercury
I was asked to calculate the acceleration due to gravity on planet Mercury, if the mas of Mercury is $2,99\times <!--\; \times \; -->{10}^{22}kg$ and its radius is $2,42\times <!--\; \times \; -->{10}^{3}km$. The mass of the object is $10kg$ and the mass of Earth is $6\times <!--\; \times \; -->{10}^{24}kg$ and the Radius of the Earth is $3,82\times <!--\; \times \; -->{10}^{3}km$
This question rather puzzled me because I was not sure if my answer is correct or not but let me proceed :
$\stackrel{\to <!--\; \to \; -->}{F}=m\stackrel{\to <!--\; \to \; -->}{a}=F_g=G\frac{m_1{m}_2}{r^2}$
$m_{1}g=G\frac{m_1{m}_2}{r^2}$
(Note that $m_2$ = mass of mercury)
$g=\frac{(6.673\times <!--\; \times \; -->{10}^{-<!--\; -\; -->11}\frac{m^2}{k{g}^2})(2,99\times <!--\; \times \; -->{10}^{22}kg)}{(2,42\times <!--\; \times \; -->{10}^{6}m{)}^2}$
I compute my answer to be $0.34\frac{m}{s^2}$
What really is confusing me is that when I look at my textbook, it shows me the gravitational acceleration due to gravity on mercury to be $3.59\frac{m}{s^2}$
Can someone please explain to me what the answer that I am getting is giving me? My computation was marked correct in a test but I do not understand what this value is giving me.

Why do two bodies of different masses fall at the same rate (in the absence of air resistance)?
I'm far from being a physics expert and figured this would be a good place to ask a beginner question that has been confusing me for some time.
According to Galileo, two bodies of different masses, dropped from the same height, will touch the floor at the same time in the absence of air resistance.
BUT Newton's second law states that $a=F/m$, with $a$ the acceleration of a particle, $m$ its mass and $F$ the sum of forces applied to it.
I understand that acceleration represents a variation of velocity and velocity represents a variation of position. I don't comprehend why the mass, which is seemingly affecting the acceleration, does not affect the "time of impact".
Can someone explain this to me? I feel pretty dumb right now :)

Does Acceleration due to Gravity take into consideration the centrifugal acceleration due to Earth's spin?
Is the gravitational acceleration we consider only the attraction due to the Earth's gravity or is it that of gravity plus the attraction due to Earth spinning?
We know that earth produces an acceleration towards the centre on any body near it due to gravitational attraction.
We denote this acceleration as $g$
But we also know that any body on Earth is also undergoing rotational motion due to earth's Spin. So every body should experience an outward centrifugal acceleration.
Does acceleration due to gravity $g$ take the centrifugal acceleration into consideration?

Is time dilation due to relative velocity equivalent in principle to time dilation due to relative gravitational strength?
Is time dilation due to relative velocity and relative gravitational strength equivalent? That is, similar to Einstein's thought experiment where an observer in an enclosed capsule with no windows cannot tell the difference between acceleration due to gravity or due to some applied force, is there any observable difference between time dilation due to velocity versus that due to gravity?

Given Angle, Initial Velocity, and Acceleration due to Gravity, plot parabolic trajectory for every $x$?
Given any Angle -> 0-90
Given any Initial Velocity -> 1-100
Given Acceleration due to Gravity -> 9.8
Plot every x,y coordinate (the parabolic trajectory) with cartesian coordinates and screen pixels (not time)
This should only be one equation as far as I can tell, a "y=" type equation, telling you the height in y coordinates based on the current x coordinate which is increasing by 1 as you plot across a computer screen.
I've come up with an equation that works for the angle 45, but doesn't seem to be entirely accurate for any other angle, I suspect it has something to do with the angle 45 having 1 solution and every other angle having 2 solutions (two different initial velocities land on the same spot), but I'm stumped beyond that.
$y=y_0-<!--\; -\; -->(x-<!--\; -\; -->x_0)\cdot <!--\; \cdot \; -->\tan <!--\; \; -->\frac{\frac{1}{2}\cdot <!--\; \cdot \; -->\frac{9.8}{v_0^2}\cdot <!--\; \cdot \; -->x-<!--\; -\; -->x_0^2}{\cos <!--\; \; -->{\mathrm{\theta <!--\; \theta \; -->}}^2}$

Time period related to acceleration due to gravity
The period of a pendulum is given by
$T=2\mathrm{\pi <!--\; \pi \; -->}\sqrt{\frac{L}{g}}.$
If we take a pendulum where there is no gravitational field, then $g=0$, therefore the period should become infinity. In such a condition what will our time relative to that of a person on earth will be? I believe that the time of a person will become too slow as it takes infinite time to complete one oscillation. Please tell me whether I am right or wrong, and if I am wrong please help me understand why.

Acceleration due to gravity at Centre
Why is the acceleration due to gravity at the center of the earth $0$?
My Attempt
actually at the center distance from the surface is equal to the radius so the formula
$g^{\prime }=(1-<!--\; -\; -->\frac{d}{R})\times <!--\; \times \; -->g$
yields $0$.
But can anyone give me the other reason for this?

Acceleration due to gravity
Does acceleration due to gravity increase when an object at a said radius is doubled? Like for instance, if I said the radius of the earth is $5\mathrm{k}\mathrm{m}$ (not to scale obviously) if I doubled that, so $10\mathrm{k}\mathrm{m}$ will acceleration due to gravity double? Or maybe $g/2$? Or would the acceleration stay the same no matter distance?

Acceleration due to gravitational force
I need some help with below. So according to Newton's 2nd law, $a=F/m$, for a given mass, the acceleration depends on mass.
But acceleration due to gravity is independent of mass. There seems some contradiction; I am sure I am missing something, but I can't pin-point what that is.