
Write all constant values in gravitation.
Answer
476.1k+ views
Hint: Here we need to know about Newton's law of gravitation. This law gives the amount of gravitational force acting between the two masses which are separated by a distance. From the mathematical form of the law we will find a constant. We need to write about the different properties of this constant.
Formula used: $F=\dfrac{G{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}$
Complete step by step answer:
Newton’s law of gravitation states that the gravitational force of attraction acting between the two bodies is proportional to the product of their masses and inversely proportional to the square of the distance between them. Thus if we have two masses ${{m}_{1}}$ and ${{m}_{2}}$ separated by a distance $r$ and the gravitational force of attraction $F$ is acting between them then we can write
$\begin{align}
& F\propto \dfrac{{{m}_{1}}{{m}_{2}}}{{{r}^{2}}} \\
& orF=\dfrac{G{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}.........(1) \\
\end{align}$
Here $G$ is the constant of proportionality and is called universal gravitational constant.
It has the value of $6.67\times {{10}^{-11}}N{{m}^{2}}k{{g}^{-2}}$ , So its value is very small. Now from (1)we can write
$G=\dfrac{F{{r}^{2}}}{{{m}_{1}}{{m}_{2}}}$
So its dimensions are
$\dfrac{[ML{{T}^{-2}}][{{L}^{2}}]}{[M][M]}=[{{M}^{-1}}{{L}^{3}}{{T}^{-2}}]$
In the law of gravitation, the value of constant of proportionality $G$ is supposed to be the same for all pairs of bodies at all places ,at all times, at all distances of separation, and also independent of the presence of other bodies or the properties of the intervening medium.
Additional Information: In Newton’s law of gravitation the only constant is $G.$ For any planet the acceleration due to gravity $g$ is related to G by the relation $g=\dfrac{GM}{{{R}^{2}}}$ where $M$ is the mass of the planet and$R$ is the radius of the planet.
Note: In Newton's law of gravitation the masses and the separation between them are variables and $G$ is constant. The $G$ is really a universal constant and the calculations made on this assumption to describe different planetary motions have come out to be correct. There is so far no evidence against this assumption.
Formula used: $F=\dfrac{G{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}$
Complete step by step answer:
Newton’s law of gravitation states that the gravitational force of attraction acting between the two bodies is proportional to the product of their masses and inversely proportional to the square of the distance between them. Thus if we have two masses ${{m}_{1}}$ and ${{m}_{2}}$ separated by a distance $r$ and the gravitational force of attraction $F$ is acting between them then we can write
$\begin{align}
& F\propto \dfrac{{{m}_{1}}{{m}_{2}}}{{{r}^{2}}} \\
& orF=\dfrac{G{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}.........(1) \\
\end{align}$
Here $G$ is the constant of proportionality and is called universal gravitational constant.
It has the value of $6.67\times {{10}^{-11}}N{{m}^{2}}k{{g}^{-2}}$ , So its value is very small. Now from (1)we can write
$G=\dfrac{F{{r}^{2}}}{{{m}_{1}}{{m}_{2}}}$
So its dimensions are
$\dfrac{[ML{{T}^{-2}}][{{L}^{2}}]}{[M][M]}=[{{M}^{-1}}{{L}^{3}}{{T}^{-2}}]$
In the law of gravitation, the value of constant of proportionality $G$ is supposed to be the same for all pairs of bodies at all places ,at all times, at all distances of separation, and also independent of the presence of other bodies or the properties of the intervening medium.
Additional Information: In Newton’s law of gravitation the only constant is $G.$ For any planet the acceleration due to gravity $g$ is related to G by the relation $g=\dfrac{GM}{{{R}^{2}}}$ where $M$ is the mass of the planet and$R$ is the radius of the planet.
Note: In Newton's law of gravitation the masses and the separation between them are variables and $G$ is constant. The $G$ is really a universal constant and the calculations made on this assumption to describe different planetary motions have come out to be correct. There is so far no evidence against this assumption.
Recently Updated Pages
The correct geometry and hybridization for XeF4 are class 11 chemistry CBSE

Water softening by Clarks process uses ACalcium bicarbonate class 11 chemistry CBSE

With reference to graphite and diamond which of the class 11 chemistry CBSE

A certain household has consumed 250 units of energy class 11 physics CBSE

The lightest metal known is A beryllium B lithium C class 11 chemistry CBSE

What is the formula mass of the iodine molecule class 11 chemistry CBSE

Trending doubts
Why was the Vernacular Press Act passed by British class 11 social science CBSE

Arrange Water ethanol and phenol in increasing order class 11 chemistry CBSE

Name the nuclear plant located in Uttar Pradesh class 11 social science CBSE

What steps did the French revolutionaries take to create class 11 social science CBSE

How did silk routes link the world Explain with three class 11 social science CBSE

What are the various challenges faced by political class 11 social science CBSE
