Consider a heavenly body whose mass is 3x10^24 kg (half that of the earth) and radius is 3200km
(half that of the earth). What is the acceleration due to gravity of surface of the body?
(Take ge = 9.8 m/s?)
Answers
Let the radius and mass of Earth be R and M respectively.
We have,
Mass of Earth,
Radius of Earth,
Mass of the Heavenly body, ( Given in the question that the mass of the Heavenly body is half of that of Earth. )
Radius of the Heavenly body, ( Given in the question that the radius of the Heavenly body is half of that of Earth. )
We know that,
Where,
Gravitational constant = G,
Mass of the body = m,
Radius of the body = r
Ratio of Earth's g to that of Heavenly body's g is:
Hence, The acceleration due to gravity of the Heavenly body is two times to that of Earth's.
Therefore,
( Since the acceleration due to gravity of Earth is given as 9.8 m/s² )
Answer:
As we know g = Gm/r²
g on earth = 9.8 m/s²
Mass of heavenly body is half of earth
Radius of heavenly body is half of earth
Therfore: 1/2(m)(G)/1/2(r²) = 9.8
1/2{(m)(G)/(r²) = 9.8
m(G)/(r²) = 9.8 × 2
Since Gm/r2 = g
g = 19.6 m/s²
Therefore the Acceleration due to gravity on the heavenly body is 19.8 m/s²