Physics, asked by sasik3534, 3 months ago

derive the relation between g and G​

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Answered by niharikagurjar2005
3

 Answer

The acceleration on an object due to the gravity of any massive body is represented by g (small g). The force of attraction between any two unit masses separated by unit distance is called universal gravitational constant denoted by G (capital G). The relation between G and g is not proportional.

Answered by Rizakhan678540
3

Answer:

Relation between Acceleration due to Gravity (g) and Universal Gravitational Constant (G)

The acceleration produced by the gravitational force of the Earth is called acceleration due to gravity (g)

" The acceleration gained by an object freely falling the Earth because of gravitational force is called acceleration due to gravity".

It is represented by 'g'. This value is independent of the shape and mass of the body. Its S.I unit is m/s².

Suppose, that the Earth's radius is R and its mass is M. The value of the gravitational force acting on a particle of mass 'm' at a distance 'r' from the center.

F → GMm/r².....(1)

By Newton's second law of motion.

f = ma

∴ If the acceleration due to gravity is g at point P.

f = mg......(2)

From the equation (1) and (2) we get.

→ mg = GMm/r²

→ Or g = GM/r².....(3)

r = (r + h)

g =  \frac{gm}{(r + h) {}^{2} } ......(3)

where ' h' is the height of the particle above the Earth's surface. If point P is close to the surface of the Earth then the value of h will be negligible with reference to the radius R of the Earth , i.e.,

→ h < < R

→ h + R ≈ R

From the equation (3)

g =  \frac{gm}{r {}^{2} } .....(4)

If Earth's average density is ρ then;

Mass of the Earth , M = Volume × Density

m =  \frac{4}{3} \pi \: r {}^{3} p \\

From the equation (4)

→ g = G 4/3 πR³ p/R²

or \: g  \:  =  \frac{4}{3} \pi \: grp.....(5)

This equation (5) shows the relationship between (g) and density (ρ).

 \:  \:

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