What is the value of Gravitational constant, $G$
(i) on earth and
(ii) on the moon?
Answer
559.2k+ views
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.
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.
Recently Updated Pages
Which will be the least stable resonating structure class 11 chemistry CBSE

How many 5 digit telephone numbers can be construc-class-11-maths-CBSE

How do you find the angle of the resultant vector class 11 physics CBSE

Draw labelled diagram of the following i Gram seed class 11 biology CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

What is the need and importance of classification class 11 biology CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

The teeth used for biting and cutting food are called class 11 biology CBSE

