Question

If M the mass of the earth and R its radius, the ratio of the gravitational acceleration and the gravitationalconstant is​

Answer 1

$\huge{\underline{\underline{\red{\mathfrak{AnSwEr :}}}}}$

${\underline{\boxed{\sf{\dfrac{g}{G} \: = \: \dfrac{M}{R^2}}}}}$

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$\large{\underline{\underline{\red{\mathfrak{Step-By-Step-Explanation :}}}}}$

As we know that :

$\large{\boxed{\sf{g \: \propto \: M}}}$

And

$\large{\boxed{\sf{g \: \propto \: \dfrac{1}{R^2}}}}$

Combining both the above equations we get,

$\implies {\sf{g \: \propto \: \dfrac{M}{R^2}}} \\ \\ \implies {\sf{g \: = \: G \dfrac{M}{R^2}}} \\ \\ \implies {\boxed{\boxed{\sf{\dfrac{g}{G} \: = \: \dfrac{M}{R^2}}}}}$

On removing proportionality symbol there comes a constant known as Gravitational Constant.

Where,

• Mass is mass of earth
• R is radius of earth
• G is Gravitational Constant
• g is Gravitational force or acceleration due to gravity.

Answer 2

Answer:

Given:

Mass of Earth = M

Radius of Earth = R

To find:

Ratio of Gravitational acceleration and gravitational constant.

Concept:

Gravitational acceleration is a Gravitational Field Intensity directed towards the centre of the Earth.

Universal Gravitational Constant is numerically equal to the Gravitational force in between unit masses separated by unit distance.

Let gravitational acceleration be g , and gravitational constant be G

Calculation:

As per Newton's Law of Gravitation , we can say that :

$\sf{Force = G\dfrac{Mm}{ {R}^{2} }}$

Now gravitational acceleration be g

$\sf{g = G\dfrac{M}{ {R}^{2} }}$

$\sf{ \implies \: \dfrac{g}{G} = \dfrac{M}{ {R}^{2} }}$

So final answer :

$\boxed{ \huge{ \blue{\sf{ \dfrac{g}{G} = \dfrac{M}{ {R}^{2} }}}}}$