Question

If the symbols have their usual meanings then $\dfrac{{g{R^2}}}{M}$ represents.
A. $G$
B. ${G^2}$
C. $\dfrac{1}{G}$
D. $\dfrac{1}{{{G^2}}}$

Answer 1

Hint:In order to this question, to know the exact option which is represented as $\dfrac{{g{R^2}}}{M}$ , we will go through the acceleration due to gravity in terms of mass and the gravitational force. That’s how we will find the correct option which is represented.

Complete step by step answer:
The acceleration gained by an object due to gravitational force is known as acceleration due to gravity. The SI unit for it is $m.{s^{ - 2}}$ . It is a vector quantity since it has both magnitude and direction. The acceleration owing to gravity is denoted by the letter $g$ . $G$ has a standard value of \[9.8{\text{ }}m.{s^{ - 2}}\] on the earth's surface at sea level.

The acceleration of an object in free fall within a vacuum is known as gravitational acceleration in physics (and thus without experiencing drag). This is the gradual increase in speed induced only by gravitational attraction.So, as we know that, acceleration due to gravity,
$g = \dfrac{{GM}}{{{R^2}}}$
where, $M$ is the mass of any object, $G$ is the gravitational force applied and $R$ is the radius or length of the object.
$ \Rightarrow \dfrac{{g{R^2}}}{M} = G$ …..eq(1)
Therefore, in the above equation(1), $\dfrac{{g{R^2}}}{M}$ represents $G$ .

Hence, the correct option is A.

Note:This formula can be extended to any pair of objects in which one is significantly more massive than the other, such as a planet compared to any man-scale item, using the integral form of Gauss's Law. The distances between planets, as well as the distances between planets and the Sun, are many orders of magnitude greater than the sizes of the sun and planets.