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Write all constant values in gravitation.
Answer
472.5k+ 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.
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