what is the weight of a 70kg body on the surface of a planet whose mass is 1/7th that of earth and radius is half of earth
Answers
GIVEN
The Mass of a body = 70kg
Mass of the planet = 1/7th of Earth
Radius of the planet = 1/2th of Earth
TO FIND
The Weight of the body.
SOLUTION
》Method - 01
We know that
Weight= Mass × Acceleration due to gravity
We have to find the "g" of that planet as we know
g = GM/R^2 where g = 9.8 m/s^2
=> Here in above formula M amd R are Mass of Earth and Radius of Earth respectively.
But in the given data we have
mass of planet = M/7
radius of planet = R/2
Putting these value in the equation we have
g' = (G × M/7) ÷ (R/2)^2
= GM/5 ÷ R^2 / 4
= GM/5 × 4/R^2
= 4GM/5R^2
we can write it as :-
= 4/5 (GM/R^2)
= 4/5 × 9.8 ( GM/R^2 = 9.8m/s^2)
= 7.84 m/s^2
Here we got the acceleration due to gravity of that planet g' = 7.84 m/s^2.
Then Weight of body =
g' × mass = 7.84 m/s^2 × 70 = 548.8N
Hence the weight of the body is 548.8 Newton.
》Method - 02
Weight is nothing but the force acquired by Gravitational force of Earth or any other body. We have formula as follows:-
F = G × m × M / R^2
where g = 6.67 × 10^(-11)
Where m = mass of body
M = mass of earth
R = Radius of Earth
We know Radius of Earth is 6.37× 10^6m
also the mass of Earth is 5.97× 10^24 kg
But the Radius of that Planet is 1/2 of earth so its radius is
6.37 × 10^6m ÷ 2 = 3.2 × 10^6 m
And the Mass of that body is 1/7 of earth so its mass is
5.97 ×10^24 ÷ 7 = 0.86 × 10^24 kg
Gravitational Constant would remain the same.
Substituting all the calculated value in the Gravitational Force Formula we get
F = G × m × M / R^2
= [6.67 × 10^(-11) × 70 × 0.86 × 10^24] ÷ (3.2 × 10^6 )^2
= [6.67 × 70 × 0.86 × 10^13]/ [3.2 × 3.2 × 10^12]
= (401.534 × 10^13) ÷ (10.24 × 10^12)
= (401.53 ÷ 10.24) × 10
= 54.88 × 10
= 548.8 Newton
Hence the Weight of the body is 548.8 Newton. (ANS)