Answer
Verified
437.4k+ views
Hint: The proportionality constant used in Newton’s universal law of gravitation is termed as the gravitational constant that we can write in terms of force, mass, and distance. We will figure out our desired answer from the S.I. units of the other quantities associated with Newton’s law of gravitation.
Formula used:
$F \propto \dfrac{{{m_1}.{m_2}}}{{{r^2}}}$
Complete solution:
The gravitational constant, also known as the universal gravitational constant, is an empirical constant that is involved with calculating the gravitational force of an object. According to Sir Isaac Newton’s law of universal gravitation, the gravitational force acting between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance connecting the centers of the two bodies. Therefore we have-
$F \propto \dfrac{{{m_1}.{m_2}}}{{{r^2}}}$
${m_1},{m_2}$ are the masses of the two bodies
$r$ is the distance between the two bodies
Hence we have-
$F = K.\dfrac{{{m_1}.{m_2}}}{{{r^2}}}$
Where $K$ is a proportionality constant, and this constant is termed as the universal gravitational constant, denoted as $G$. Therefore, the above equation is written as-
$F = G.\dfrac{{{m_1}.{m_2}}}{{{r^2}}}$
The S.I. unit of $F$ is Newton$(N)$
The S.I. unit of ${m_1}$ and ${m_2}$ is Kilogram$(Kg)$
The S.I. unit of $r$ is meter$(m)$
From the above equation, we have-
$\Rightarrow G = F.\dfrac{{{r^2}}}{{{m_1}.{m_2}}}$
We put the S.I. units of the other quantities in the right-hand side of the equation-
$\Rightarrow G = N.\dfrac{{{{\left( m \right)}^2}}}{{{{\left( {Kg} \right)}^2}}}$
$ \Rightarrow G = \left( {Kg.m.{s^{ - 2}}} \right).\dfrac{{{{\left( m \right)}^2}}}{{{{\left( {Kg} \right)}^2}}}$
Hence, $G = {m^3}.K{g^{ - 1}}.{s^{ - 2}}$
Therefore the S.I. unit of the universal gravitational constant, i.e., $G$ is ${m^3}.K{g^{ - 1}}.{s^{ - 2}}$.
Note: The universal gravitational constant is very significant in the world of physics. The magnitude of the universal gravitational constant is $6.674 \times {10^{ - 11}}{m^3}.K{g^{ - 1}}.{s^{ - 2}}$. It is also useful in Einstein’s General Theory of Relativity. In Einstein’s field equations, this constant quantifies the relation between the energy-momentum tensor and space-time geometry.
Formula used:
$F \propto \dfrac{{{m_1}.{m_2}}}{{{r^2}}}$
Complete solution:
The gravitational constant, also known as the universal gravitational constant, is an empirical constant that is involved with calculating the gravitational force of an object. According to Sir Isaac Newton’s law of universal gravitation, the gravitational force acting between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance connecting the centers of the two bodies. Therefore we have-
$F \propto \dfrac{{{m_1}.{m_2}}}{{{r^2}}}$
${m_1},{m_2}$ are the masses of the two bodies
$r$ is the distance between the two bodies
Hence we have-
$F = K.\dfrac{{{m_1}.{m_2}}}{{{r^2}}}$
Where $K$ is a proportionality constant, and this constant is termed as the universal gravitational constant, denoted as $G$. Therefore, the above equation is written as-
$F = G.\dfrac{{{m_1}.{m_2}}}{{{r^2}}}$
The S.I. unit of $F$ is Newton$(N)$
The S.I. unit of ${m_1}$ and ${m_2}$ is Kilogram$(Kg)$
The S.I. unit of $r$ is meter$(m)$
From the above equation, we have-
$\Rightarrow G = F.\dfrac{{{r^2}}}{{{m_1}.{m_2}}}$
We put the S.I. units of the other quantities in the right-hand side of the equation-
$\Rightarrow G = N.\dfrac{{{{\left( m \right)}^2}}}{{{{\left( {Kg} \right)}^2}}}$
$ \Rightarrow G = \left( {Kg.m.{s^{ - 2}}} \right).\dfrac{{{{\left( m \right)}^2}}}{{{{\left( {Kg} \right)}^2}}}$
Hence, $G = {m^3}.K{g^{ - 1}}.{s^{ - 2}}$
Therefore the S.I. unit of the universal gravitational constant, i.e., $G$ is ${m^3}.K{g^{ - 1}}.{s^{ - 2}}$.
Note: The universal gravitational constant is very significant in the world of physics. The magnitude of the universal gravitational constant is $6.674 \times {10^{ - 11}}{m^3}.K{g^{ - 1}}.{s^{ - 2}}$. It is also useful in Einstein’s General Theory of Relativity. In Einstein’s field equations, this constant quantifies the relation between the energy-momentum tensor and space-time geometry.
Recently Updated Pages
Identify the feminine gender noun from the given sentence class 10 english CBSE
Your club organized a blood donation camp in your city class 10 english CBSE
Choose the correct meaning of the idiomphrase from class 10 english CBSE
Identify the neuter gender noun from the given sentence class 10 english CBSE
Choose the word which best expresses the meaning of class 10 english CBSE
Choose the word which is closest to the opposite in class 10 english CBSE
Trending doubts
Which are the Top 10 Largest Countries of the World?
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How do you graph the function fx 4x class 9 maths CBSE
A rainbow has circular shape because A The earth is class 11 physics CBSE
The male gender of Mare is Horse class 11 biology CBSE
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths