Question

Gravitational force acts on all objects in proportion to their masses. Why then a heavy object does not fall faster than a light object.

Answer 1

Hint
Every object in the universe attracts every other object with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. Also, force is the product of mass and acceleration. We have to use the two definitions to solve this problem.

Complete step by step answer
We know that the gravitational force acting on all bodies will be proportional to their masses. But heavy objects are not observed to fall faster than lighter objects. This can be explained as shown below.
For an object on the surface of the earth, weight can be written as
$\Rightarrow W = mg$ (Where ${\text{ m }}$stands for the mass of the body and ${\text{ g }}$stands for the acceleration due to gravity)
The gravitational force can be written as, $F = G\dfrac{{Mm}}{{{r^2}}}$
Where${\text{ G }}$stands for the gravitational constant, ${\text{ M }}$stands for the mass of the Earth, ${\text{ m }}$stands for the mass of the body and ${\text{ r }}$stands for the distance between the body and earth. That is the radius of the earth (because the object is on the surface of the earth)
The weight of the object can also be considered as the force of gravity acting on it.
$\therefore F = G\dfrac{{Mm}}{{{r^2}}} = mg$
From this equation we get,
$\Rightarrow g = \dfrac{{GM}}{{{r^2}}}$
From this, we can understand that the acceleration due to gravity does not depend on the mass of the object falling. It will be the same for all objects regardless of the mass of the object.

Note
If we have considered two masses \[{\text{ }}{{\text{m}}_1}\] and ${\text{ }}{{\text{m}}_2}{\text{ }}$separated by a distance ${\text{ r}}{\text{. }}$According to Newton’s law of gravitation the force between the two masses is directly proportional to the product of their masses ${\text{ }}\left( {{m_1} \times {m_2}} \right){\text{ }}$ and it is inversely proportional to the square of the distance ${\text{ }}\left( {{r^2}} \right){\text{ }}$
i.e.
$\Rightarrow F \propto \dfrac{{{m_1} \times {m_2}}}{{{r^2}}}$
$\Rightarrow F = - G\dfrac{{{m_1} \times {m_2}}}{{{r^2}}}$
Where ${\text{ G }}$is a constant called the gravitational constant.
There is a negative sign in the expression of the gravitational field. This shows that the gravitational field is attractive in nature. Gravitational force is the weakest force out of all the four fundamental forces. Though gravity is more clearly explained by Einstein's general theory of relativity. For many practical applications, we use Newton's law of universal gravitation. The gravitational field of the earth depends on the position on the surface of the earth. The acceleration rate of freely falling bodies varies with the variation in latitude.