
The SI unit of acceleration due to gravity is:
\[A.\,m{{s}^{-2}}\]
\[B.\,N{{m}^{2}}k{{g}^{-2}}\]
\[C.\,m{{s}^{-1}}\]
\[D.\,N{{m}^{2}}k{{g}^{+2}}\]
Answer
487.8k+ views
Hint: This is a direct question. The formulae for computing the force should be used. The general equation of the force and the universal law of gravitation should be equated and then, the acceleration should be replaced with the acceleration due to gravity to obtain the SI unit of the acceleration due to gravity.
Formula used:
\[\begin{align}
& F=ma \\
& F=\dfrac{GMm}{{{r}^{2}}} \\
\end{align}\]
Complete step by step solution:
The formulae for computing the force are discussed as follows.
The general formula of the force is defined as the product of the mass and acceleration. The mathematical representation of the force is given as follows.
\[F=ma\]…… (1)
Where m is the mass and a is the acceleration.
The formula for the universal law of gravitation is given as follows.
\[F=\dfrac{GMm}{{{r}^{2}}}\]…… (2)
Where G is the gravitational constant, M and m are the masses and r is the radius of the earth.
We can equate the equations (1) and (2), as the LHS of both the equations is same, that is, the force. So we get,
\[ma=\dfrac{GMm}{{{r}^{2}}}\]
Cancel out the common terms.
\[a=\dfrac{GM}{{{r}^{2}}}\]
This acceleration is the same as the acceleration due to gravity. So, we can write as follows.
\[g=\dfrac{GM}{{{r}^{2}}}\]…… (3)
Now, we will substitute the units of the above terms in the RHS part of the equation.
The unit of the gravitational constant G is, \[G={{m}^{3}}k{{g}^{-1}}{{s}^{-2}}\]
The unit of the mass M is, \[M=kg\]
The unit of the radius of the earth r is, \[r=m\]
Now, substitute these values of the units in the above equation (3), so, we get,
\[g=\dfrac{{{m}^{3}}k{{g}^{-1}}{{s}^{-2}}\times kg}{{{m}^{2}}}\]
Cancel out the common terms.
\[g=m{{s}^{-2}}\]
Therefore, the SI unit of the acceleration due to gravity is \[m{{s}^{-2}}\].
So, the correct answer is “Option A”.
Note: The formulae relating the acceleration and the acceleration due to gravity should be known to solve such problems. Even the units of each parameter used in the formulae should be known. The acceleration due to gravity ‘g’ is different from the gravitational constant ‘G’.
Formula used:
\[\begin{align}
& F=ma \\
& F=\dfrac{GMm}{{{r}^{2}}} \\
\end{align}\]
Complete step by step solution:
The formulae for computing the force are discussed as follows.
The general formula of the force is defined as the product of the mass and acceleration. The mathematical representation of the force is given as follows.
\[F=ma\]…… (1)
Where m is the mass and a is the acceleration.
The formula for the universal law of gravitation is given as follows.
\[F=\dfrac{GMm}{{{r}^{2}}}\]…… (2)
Where G is the gravitational constant, M and m are the masses and r is the radius of the earth.
We can equate the equations (1) and (2), as the LHS of both the equations is same, that is, the force. So we get,
\[ma=\dfrac{GMm}{{{r}^{2}}}\]
Cancel out the common terms.
\[a=\dfrac{GM}{{{r}^{2}}}\]
This acceleration is the same as the acceleration due to gravity. So, we can write as follows.
\[g=\dfrac{GM}{{{r}^{2}}}\]…… (3)
Now, we will substitute the units of the above terms in the RHS part of the equation.
The unit of the gravitational constant G is, \[G={{m}^{3}}k{{g}^{-1}}{{s}^{-2}}\]
The unit of the mass M is, \[M=kg\]
The unit of the radius of the earth r is, \[r=m\]
Now, substitute these values of the units in the above equation (3), so, we get,
\[g=\dfrac{{{m}^{3}}k{{g}^{-1}}{{s}^{-2}}\times kg}{{{m}^{2}}}\]
Cancel out the common terms.
\[g=m{{s}^{-2}}\]
Therefore, the SI unit of the acceleration due to gravity is \[m{{s}^{-2}}\].
So, the correct answer is “Option A”.
Note: The formulae relating the acceleration and the acceleration due to gravity should be known to solve such problems. Even the units of each parameter used in the formulae should be known. The acceleration due to gravity ‘g’ is different from the gravitational constant ‘G’.
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