Question

a1) What is acceleration due to gravity!​

Answer 1

Answer:-

Acceleration due to Gravity:- The acceleration is caused due to the force of gravity is called acceleration due to gravity.

✧ Acceleration due to gravity is denoted by g .

✧ The acceleration due to gravity near the earth's surface is 9.8m/s²

To Calculate Acceleration due to gravity on any planets in the universe , we use

⟹ a = F/m = GM/R²

here

F is the force

a is the acceleration

m is the mass

G is the universal constant.

M is the mass of the body

R is the distance between them

The value of Universal Gravitational Constant is 6.67×10^-11 Nm²/Kg².

The value of Universal Gravitational Constant was discovered by Henry Cavendish .

✧ Universal Law Of Gravitation : The gravitational force of attraction between any two particle is directly proportional to the product of the masses of the particles and is inversely proportional to the square of the distance between the particles.

Answer 2

AnswEr:

$\underline{\bigstar\:\sf{Acceleration \: due \: to \: gravity :}}$

⋆ The uniform acceleration producedproduced in a free falling body due to the gravitational force of earth is known as "Acceleration due to gravity".

⋆ When an object is dropped from some height its velocity increases at a constant rate or we can say when an object is dropped from some height a uniform acceleration is produced innate by the gravitational pull of the earth.This acceleration does not depend on the mass of the falling object.

⋆ Acceleration due to gravity is denoted by letter "g" .

⋆ The value of Acceleration due to gravity(g) is $\bf\ 9.8m\setminus\ s^{2}$

⋆ The point that should we should remember is that " when a body is dropped freely it falls with an acceleration but when a body is thrown vertically upwards it undergoes retardation.

$\underline{\bigstar\:\sf{Calculation \: of \: Acceleration \: due \: to \: gravity \: \left(g \right) :}}$

Suppose, we drop a ball of mass "m" from a distance of "R" from the centre of Earth at mass "M", then force exerted is :

$\dashrightarrow\sf\ F \: = \: G \times\ \frac{M \times\ m}{R^{2}} \quad\sim\ eq.1$

$\scriptsize\sf{\quad\dag\ G \: = \: Gravitational \: constant}$

We also know that, Force is the product of mass and acceleration; Hence :

$\dashrightarrow\sf\ F \: = \: m \times\ a$

$\dashrightarrow\sf\ a \: = \: \frac{F}{m} \quad\sim\ eq.2$

Put the values of eq.1 & 2 in Formula of Force;

$\dashrightarrow\sf\ a \: = \: \frac{G \times\ M \times\ m}{R^{2} \times\ m}$

$\dashrightarrow\sf\ a \: = \: \frac{G \times\ M \times\ \: \cancel{m}}{R^{2} \times\ \: \cancel{m}}$

$\dashrightarrow\sf\ a \: = \: G \times\ \: \frac{M}{R^{2} }$

Here;

⋆ a = acceleration due to gravity

⋆ G = Gravitational constant

⋆ M = Mass of earth

⋆ R = Radius of earth

Now; put the value of G, M and R ;

⋆ Value of G =$\sf\ 6.7 \times\ 10^{-11}Nm^{2}\setminus\ kg^{2}$

⋆ Value of M =$\sf\ 6 \times\ 10^{24}kg$

⋆Value of R =$\sf\ 6.4 \times\ 10^{6}m$

$\dashrightarrow\sf\ a \: = \: \frac{6.7 \times\ 10^{-11} \times\ 6 \times\ 10^{24}}{\left(6.4 \times\ 10^{6} \right)}$

$\dashrightarrow\sf\ a \: = 9.8m\setminus\ s^{2}$

$\dashrightarrow\normalsize{\underline{\boxed{\mathsf \red{ Value \: of \: g \: = \: 9.8m\setminus\ m^{2} }}}}$