Question

Give the relation and difference between 'g' and 'G'. ( related with class 9 physics)​

Answer 1

Answer:

we know that

$F = \frac{GMm}{ {R}^{2} } - - - - (1) \\ g = \frac{F}{m} - - - - - -(2) \\ substituting \: equation \: 1 \: and \: 2 \\ g= \frac{GMm}{ {R}^{2} m} \\ g = \frac{GM}{ {R}^{2} }$

hope it will help!!!!!@!

Answer 2

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/s

2

.

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=

r

2

GMm

..........(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=

r

2

GMm

or g=

r

2

GM

...............(3)

∵ r=(R+h)

g=

(R+h)

2

GM

................(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=

R

2

GM

.............(4)

If Earth's average density is ρ then;

Mass of the Earth , M = Volume × Density

M=

3

4

πR

3

ρ

From the equation (4)

g=

R

2

G

3

4

πR

3

ρ

or g=

3

4

πGRρ .............(5)

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

Hope it will helps......❤

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