
What is the S.I. unit of the gravitational constant $G$?
Answer
554.4k+ 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
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
10 examples of friction in our daily life

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

Difference Between Prokaryotic Cells and Eukaryotic Cells

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

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

