Understanding Gravitational Pull and Its Effects on Humans

[yuganes warman]

Everything that has mass , posseses gravitational pull. We as humans do have mass, and thus posses gravitational pull too ? If so , do we posses tiny amount of gravitational pull ?

 

[Rithikha]

Yes, that is true. Every object that has mass exerts a gravitation force.
You can quantify this using the formula,
g=(G.m1.m2)/r^2
g is the gravitational force
G is the gravitational constant = 6.7x10^-11
m1 and m2 are the masses of the objects you are considering
r is the distance between them
If you calculate the gravitational force (g) for everyday objects, the masses, distance are extremely small, not to forget the smaller constant. Hence, the resultant g is also very small and thus negligible. For objects with greater masses like that of the Earth, it is more prominent and the force is significant.

 

[yuganes warman]

Thank you very much , now my curiosity is solved

 

[A.T.]

g=(G.m1.m2)/r^2

If you calculate the gravitational force (g) for everyday objects, the masses, distance are extremely small, not to forget the smaller constant. Hence, the resultant g is also very small and thus negligible.

Reducing the distance increases g. And why would the constant be smaller? Doesn't the term "constant" give you a hint?

The only part of your explanation that is correct, are the small masses. Also note that "g" usually refers to gravitation acceleration, not the force.

 

[Rithikha]

Reducing the distance increases g. And why would the constant be smaller? Doesn't the term "constant" give you a hint?

The only part of your explanation that is correct, are the small masses. Also note that "g" usually refers to gravitation acceleration, not the force.

I meant, the constant is smaller compared to the masses. Why do you think I mentioned the value if I didn't know that?
And the gravitational acceleration formula is different. This is the gravitational force formula.
Yes, but the whole point was to say that the force is negligible, which it is due to the small numerator.

 

[A.T.]

I meant, the constant is smaller compared to the masses.

The constant has different units than mass. It doesn't even make sense to compare them.

This is the gravitational force formula.

Then you should use "F" for force, not "g".

 

[yuganes warman]

yes gravitational force should be denoted with Fg

 

[quincy harman]

I thought that Force = mass * acceleration. So if something has very little mass and is accelerating towards an object the size of Earth at 9.8 meters per second² then wouldn't you be able to say that the force acting on the much smaller object is = to acceleration?

 

[SteamKing]

I thought that Force = mass * acceleration. So if something has very little mass and is accelerating towards an object the size of Earth at 9.8 meters per second² then wouldn't you be able to say that the force acting on the much smaller object is = to acceleration?

It's not clear how you can say "the force acting on the much smaller object is = to acceleration" when you also say "Force = mass * acceleration".

 

[quincy harman]

It's not clear how you can say "the force acting on the much smaller object is = to acceleration" when you also say "Force = mass * acceleration".

You're right it makes no sense bahaha. Just thought about it.

 

[HallsofIvy]

I thought that Force = mass * acceleration. So if something has very little mass and is accelerating towards an object the size of Earth at 9.8 meters per second² then wouldn't you be able to say that the force acting on the much smaller object is = to acceleration?

F= ma and, for gravitational force, F= GmM/r^2 where "m" and "M" are the masses of the two objects. If we take m to be the "little mass" and M to be the "larger mass" then, for the smaller mass, ma= GmM/r^2 so a= GM/r^2. For the larger mass, Ma= GmM/r^2 so a= Gm/r^2. That tells us, first, that all objects, attracted by the earth, accelerate toward the Earth with the same acceleration, GM/r^2. At the same time, the Earth is accelerating toward the object with acceleration Gm/r^2 which is, of course, far smaller than GM/r^2.

However, force is NEVER "equal to acceleration". They are different kinds of "things" with different units so never "equal". (For mass, say, 1 kg, the acceleration and force, in the MKS system, will have the same numerical value but still are not "equal". "2 meters per second" is NOT the same as "2 kilogram meters per second".)

 

[quincy harman]

F= ma and, for gravitational force, F= GmM/r^2 where "m" and "M" are the masses of the two objects. If we take m to be the "little mass" and M to be the "larger mass" then, for the smaller mass, ma= GmM/r^2 so a= GM/r^2. For the larger mass, Ma= GmM/r^2 so a= Gm/r^2. That tells us, first, that all objects, attracted by the earth, accelerate toward the Earth with the same acceleration, GM/r^2. At the same time, the Earth is accelerating toward the object with acceleration Gm/r^2 which is, of course, far smaller than GM/r^2.

However, force is NEVER "equal to acceleration". They are different kinds of "things" with different units so never "equal". (For mass, say, 1 kg, the acceleration and force, in the MKS system, will have the same numerical value but still are not "equal". "2 meters per second" is NOT the same as "2 kilogram meters per second".)

What about Einsteins thought experiment in which he said that if you're in a closed box and accelerating at 9.8 meters per second per second in space that you would not know the difference from standing on Earths surface?

 

[A.T.]

What about Einsteins thought experiment in which he said that if you're in a closed box and accelerating at 9.8 meters per second per second in space that you would not know the difference from standing on Earths surface?

You would observe the same forces and accelerations in both cases. That doesn't make force equal to acceleration