What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is 6 × 1024 kg and radius of the earth is 6.4 × 106m.) a ep by step explanation of last step of substituting F = gm1m2/r²
Answers
Given:-
- Mass of Object (m1) = 1kg
- Mass of Earth (m2) = 6×10²⁴ kg
- Radius of Earth (r) = 6.4×10⁶m
- Gravitational Constant = 6.67×10^-11
To Find :-
- Gravitational Force between Earth and Object.
Solution:-
By using this Formula we get
→ F = Gm₁m₂/r²
Substitute the value we get
By solving this We get the Final answer
→ F = 9.8N (approx)
∴ The Force Of Gravitation between object and Earth is 9.8 Newton (Approx) .
Additional Information!!
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.
mass of body =m
m=1kg
mass of earth =M
M=6×10^6
Radius of earth =R
R=6.4×10^6
magnitude of the gravitation force (F) between the earth and the body can be given as -
Explanation:
F=GMm/ R^2
F=6.67×10^-11×6×10^24×1/(6.4×10^6)^2
F=6.67×6×10/6.4×6.4
ANS=9.8N