What is the S.I. unit of the gravitational constant $G$?
Answer
583.8k+ 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
Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

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

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

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

State and prove Bernoullis theorem class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

Which among the following are examples of coming together class 11 social science CBSE

