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Why do we use gravitational force in earth by relating just the mass of an object with the acceleration produced by the gravitational field:
$F_g=m\cdot <!--\; \cdot \; -->\stackrel{\to <!--\; \to \; -->}{g}$
And when we're dealing with planets, we use a relation defined by the masses of two planets, distance squared and gravitational constant:
$F_g=G\cdot <!--\; \cdot \; -->\frac{M_1\cdot <!--\; \cdot \; -->M_2}{d^2}$
I really don't get why we use just the first relation here on earth, because we're dealing with a interction between two objects... It's because our mass is irrelevant??

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.

Why do we integrate along the whole length while finding the gravitational force between a object of mass (m) and a rod of length L and mass M? Can't we simply use $F=\frac{GmM}{r^2}?$?

Two objects are attracted to each other with 36 N of gravitational force. What would the force between them be if the distance between them were doubled?

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?

A book is lifted upward a vertical distance of 0.800 m. Duringthis displacement, does the gravitational force acting on the book do positivework or negative work? Does the gravitational potential energy of thebook increase or decrease?

1. Why is gravitational force always an attractive force?
2. And is the Newtonian formula of gravitational force applicable for very small particles like electrons and protons etc.? From Formula of Gravitational Force, I'm referring to:
$F_G=\frac{GMm}{R^2},$
where M and m are the masses of objects.
A logical explanation will be much appreciated.

Compare the gravitational attraction between objects on earth and interaction of celestial bodies in space. Which gravitational force is almost negligible? Why?

Earth is the third planet in our solar system. If Earth became the fifth planet in our solar system, how would that affect the gravitationallforce? A. The gravitational force would increase. B. The gravitational force would decrease. C. The gravitational force would remain the same. D. The gravitational force would not change .

I have heard that no one escape from the intense gravitational field of a black hole (obviously that's why it is black). And gravitational force is due to the mass having the body, no mass no gravity, as
$F=GMm/r^2$
Putting 0 in one of the two bodies' masses, gives no gravitational force. But light cannot escape from black hole due to its strong gravity, does that mean that light has mass? please correct me.

What is the ratio of Earth’s gravitational force on a satellite when it is on the ground to the gravitational force exerted when the satellite is orbiting at an altitude of 890 km?

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.

The gravitational force between two objects is 2400 N. What will be the gravitational force between the objects if the mass of one object is doubled?