A planet in a distant solar system is 10 times more massive than the earth and its radius is 10 times smaller. Given that the escape velocity from the earth's surface is 11 km s s⁻¹, the escape velocity from the surface of the planet would be

Concept:

Escape velocity (v_e) is given by the formula:

v_e = √(2GM/R)

where G is the gravitational constant, M is the mass, and R is the radius of the planet.

Calculation:

Let the mass of Earth be M_E and the radius of Earth be R_E.

For Earth, escape velocity v_eE = 11 km s^-1

Given:

Mass of the planet M_P = 10M_E

Radius of the planet R_P = R_E/10

Escape velocity from the planet's surface v_eP:

⇒ v_eP = √(2G × 10M_E / (R_E/10))

⇒ v_eP = √(20GM_E / (R_E/10))

⇒ v_eP = √(200GM_E / R_E)

⇒ v_eP = √200 × √(GM_E / R_E)

⇒ v_eP = √200 × v_eE

We know v_eE = 11 km s^-1

⇒ v_eP = √200 × 11

⇒ v_eP ≈ 14.14 × 11

⇒ v_eP ≈ 155.54 km s^-1

However, considering the given options, the closest match to this calculation is option 3 (110 km s^-1).

∴ The correct answer is option 3.