Question

Consider an imaginary planet whose mass is 8 times that of the earth and whose radius is 2 times that of the earth. What is the value of acceleration due to gravity on the surface of that planet? If a man of mass of 60 kg stands on the surface of that planet, What will be his weight? Given that the acceleration due to gravity on the surface of the earth is 9.8 ms^-2.

Answer 1

Given:

The mass of an imaginary planet (M') is 8 times that of the earth. i.e., M' = 8M, where M is the mass of the Earth.

The radius of the planet is 2 times that of the earth. i.e., R' = 2R, where R is the radius of the Earth.

The mass (m) of a man on the given planet = 60 kg.

Required to find:

The acceleration due to gravity (g') on the surface of the planet.

Solution:

The acceleration due to the gravity (g) on the surface of the earth is given by $g=\frac{GM}{R^2}$.

By substituting the given values in the above equation, we get $g\text{'}=\frac{GM\text{'}}{R{\text{'}}^2}=\frac{Gx8M}{{\left(2R\right)}^2}=\frac{8GM}{4R^2}=\frac{2GM}{R^2}=2g$

Therefore, the value of the acceleration due to gravity (g') on the surface of the given planet is double the value of the ‘g’ on the surface of the earth.

Hence, g' = 2g = 2 x 9.8 = 19.6 ms^-2.

The weight (W) of an object is given by W = mg', where m is the mass of the given object.

By substituting the value of the mass (m) of the man in the above equation, we get W = mg' = 60 x 19.6 = 1,176 N.

Hence, the mass of the man on the given planet is 1,176 N.