If earth has a mass nine times and radius twice to that of a planet P. Then vₑ/3 √x ms⁻¹ will be the minimum velocity required by a rocket to pull out of gravitational force of P, where vₑ is escape velocity on earth. The value of x is

Solution

Physics sir · 3 steps

Escape speed goes as √(M/R), so plug in the planet's mass and radius as fractions of Earth's.

1. Escape speed formula. For any planet, v = √(2GM/R). Only the ratio M/R changes from one planet to another.
2. Planet in Earth terms. Earth has 9 times the mass and 2 times the radius, so M_P = M_E/9 and R_P = R_E/2.

   v_P = √((2G (M_E/9))/(R_E/2)) = √(2/9) √(2GM_E/R_E) = √2/3 vₑ

3. So. v_P = vₑ/3√2, which matches vₑ/3√x with x = 2. The answer is (c).

The idea to keep. Escape speed scales as √(M/R). Write the new planet's mass and radius as multiples of Earth's, and the ratio of escape speeds follows in one line.

Answer from the NTA final answer key. Solution by PrepWiser.