
Write the numerical value of gravitational constant G with its SI unit.
Answer
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Hint:: From the very name of the constant you may realize that the constant is directly related to the gravitational force. You could then recall the expression in which you have come across this constant. You could easily deduce the constant’s SI unit from there and you may also recall its value and hence answer the question.
Complete answer:
In the question, we are asked to find the numerical value of gravitational constant G and also give its SI unit. In order to answer this, we have to recall where we have seen this constant.
You may recall the gravitational law. This law states that the gravitational force between any two bodies is found to be proportional to the product of their masses and inversely proportional to the square of the distance between them. Mathematically,
$F\propto \dfrac{{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}$
$\therefore F=G\dfrac{{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}$
So the gravitational constant G is the proportionality constant introduced to the gravitational law. The gravitational constant is known to have a value given by,
$G=6.67\times {{10}^{-11}}{{m}^{3}}k{{g}^{-1}}{{s}^{-2}}$
The value above is expressed in the SI unit and this unit can be derived from the above given expression.
$G=\dfrac{F{{r}^{2}}}{{{m}_{1}}{{m}_{2}}}$
Substituting the SI units of the given quantities respectively, we get,
$\left( G \right)=\dfrac{\left( kgm{{s}^{-2}} \right){{m}^{2}}}{k{{g}^{2}}}$
$\therefore \left( G \right)=k{{g}^{-1}}{{m}^{3}}{{s}^{-2}}$
Thus, we have found the value of gravitational constant G in SI units to be,
$G=6.67\times {{10}^{-11}}{{m}^{3}}k{{g}^{-1}}{{s}^{-2}}$
Note:
As physicists, we should understand that any measurement made without the knowledge of uncertainty is meaningless. We know that the gravitational force is a very weak force when compared to other fundamental forces that we know. Thus, quite obviously the measurement of this constant with high accuracy is a very difficult task. So, the above found value of this constant is with an uncertainty of 22ppm.
Complete answer:
In the question, we are asked to find the numerical value of gravitational constant G and also give its SI unit. In order to answer this, we have to recall where we have seen this constant.
You may recall the gravitational law. This law states that the gravitational force between any two bodies is found to be proportional to the product of their masses and inversely proportional to the square of the distance between them. Mathematically,
$F\propto \dfrac{{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}$
$\therefore F=G\dfrac{{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}$
So the gravitational constant G is the proportionality constant introduced to the gravitational law. The gravitational constant is known to have a value given by,
$G=6.67\times {{10}^{-11}}{{m}^{3}}k{{g}^{-1}}{{s}^{-2}}$
The value above is expressed in the SI unit and this unit can be derived from the above given expression.
$G=\dfrac{F{{r}^{2}}}{{{m}_{1}}{{m}_{2}}}$
Substituting the SI units of the given quantities respectively, we get,
$\left( G \right)=\dfrac{\left( kgm{{s}^{-2}} \right){{m}^{2}}}{k{{g}^{2}}}$
$\therefore \left( G \right)=k{{g}^{-1}}{{m}^{3}}{{s}^{-2}}$
Thus, we have found the value of gravitational constant G in SI units to be,
$G=6.67\times {{10}^{-11}}{{m}^{3}}k{{g}^{-1}}{{s}^{-2}}$
Note:
As physicists, we should understand that any measurement made without the knowledge of uncertainty is meaningless. We know that the gravitational force is a very weak force when compared to other fundamental forces that we know. Thus, quite obviously the measurement of this constant with high accuracy is a very difficult task. So, the above found value of this constant is with an uncertainty of 22ppm.
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