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What is the value of Gravitational constant, $G$
(i) on earth and
(ii) on the moon?

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Answer
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Hint: While solving the question one must be very careful to differentiate between acceleration due to gravity, $g$ and the Universal Gravitational constant, $G$. Newton’s Law of gravitation will further help to understand the concept of gravity and the formula involved in it.

Complete step by step solution:
It is a well-known fact that Sir Isaac Newton discovered gravity and formulated the law of gravitation. According to Newton’s law of Gravitation, “Everybody in the universe attracts every other body with a force which is directly proportional to the product of the masses and inversely proportional to the square of the distance between them.” In mathematical form it can be written as:
$\begin{align}
& \,F\propto \dfrac{{{m}_{1}}\times {{m}_{2}}}{{{r}^{2}}} \\
 & \Rightarrow F=G.\dfrac{{{m}_{1}}\times {{m}_{2}}}{{{r}^{2}}} \\
\end{align}$
Where
$F$ = Gravitational force in Newton
${{m}_{1}}$ and ${{m}_{2}}$ = masses of the bodies in kilograms
$r$ = distance between the masses in meters
$G$ = Gravitational constant
Here $G$ is a proportionality constant and the definition of constant is that its value never changes. So, be it earth, moon or any other planet, the value of $G$ will always be$6.67\times {{10}^{-11}}N.{{m}^{2}}.k{{g}^{-2}}$ .
On the other hand, when an object falls freely, its acceleration changes due to gravity. Such acceleration which is gained by the object due to gravitational force is called acceleration due to gravity, $g$. The formula for $g$ near the surface of the earth is given by:
$g=\dfrac{GM}{{{r}^{2}}}$
Where $g$ = acceleration due to gravity in $m/{{s}^{2}}$
$M$= mass of the earth
$r$ = radius of the earth
Therefore, acceleration due to gravity on the surface of the earth depends on the mass and radius of the earth. So, if the question asks about acceleration due to gravity on the moon, then it can be easily calculated if the mass and radius of the moon are known. Acceleration due to gravity also depends on the height of the object from the surface of the earth. When the height of the object increases from the surface of the earth, the value of $g$ decreases.

Note:
The acceleration due to gravity on the earth is $9.8m/{{s}^{2}}$ and on the moon is $\dfrac{1}{6}th$ that of earth. The value of $g$ differs for different altitudes and depths from the surface of the earth whereas $G$ is a physical constant that is invariant and does not depend on any factors whatsoever. Gravitational force is a weak force and hence the accuracy with which $G$ is measured is less.