Question

Derive the relation between Acceleration due to gravity and Universal Gravitational Constant.​

Answer 1

Answer:

Explanation:

Gravitational force is given by

F = GMm/r^2

Here F = ma = mg

Therefore,

mg = GMm/r^2

Divide both sides by m

g = GM / r^2

In above g and G relation

g = acceleration due to gravity of planet

G = universal gravitational constant

M = mass of planet

r = radius of planet

Answer 2

The Force (F) of Gravitational attraction on a body mass "m" due to earth of mass "M" and Radius "r" is given by

• $\large\rm{F=G\times{\frac{Mm}{r^2}}}$......1

We know that Newton's second law of motion that the force is Product of Mass and acceleration.

• $\rm{F=ma}$

But the acceleration due to gravity is represented by the symbol "g". Therefore, We can write

• $\rm{F=mg}$..........2.

From the eq 1 and eq 2, we get

$\implies\rm{mg=G\dfrac{Mm}{r^2}}$.

$\implies\rm{\cancel{m}g=G\dfrac{M\times{\cancel{m}}}{r^2}}$

Hence,

$\implies\rm{g=\dfrac{GM}{r^2}}$.......3

Subsitituting the value of G,M,and r kn equation 3 "g" may be estimated. These values are:-

• $\rm{Mass\:of\:earth=6\times{10^24}}$

• $\rm{Radius\:of\:earth=6.4\times{10^6}}$.

$\implies\rm{g=\dfrac{6.7\times{10^{-11}}\times{6}\times{10^{24}}}{(6.4\times{10^6)^2}}}$

$\implies\rm{g=9.8m/s^-2}$.

• Since the radius if the earth does not change much over it's entire surface ,g is more or less constant on or near the earth.

• The value if acceleration due to gravity is independent of mass,shape and size of the body but depends upon mass and radius of the earth.