Question

The mass of Planet X is one-tenth that of the Earth, and its diameter is one-half that of the Earth. The
acceleration due to gravity at the surface of Planet X is most nearly

Answer 1

Answer:-

Let the mass of the earth be M and let it's diameter be d.

Radius of earth (r) = d/2

Given:-

Diameter of the planet X is 1/4 of the earth's.

→ Diameter of the planet X = d/4

→ Radius of the planet X = (d/2) / 2 = (d/4)

mass of the planet X = 1/10 * M = M/10

The value of g on earth's surface is 10 m/s² (approx.).

We know that,

g = GM/r²

where,

• G is gravitational constant
• M is mass of the body
• R is its radius/distance.

Hence,

→ GM/(d/2)² = 10

→ GM * 4/d² = 10

→ GM/d² = 10/4 -- equation (1)

Now,

value of on planet X = G*(M/10) / (d/4)²

→ value of g on planet X = GM/10 * (16 / d²)

→ value of g on planet X = GM/d² * 16/10

Putting the value from equation (1) we get,

→ value of g on planet X = 10/4 * 16/10

→ Value of g on planet X = 4 m/s²

Therefore, the acceleration due to gravity on planet X will be 4 m/s².