Physics Circle home Student Login

Home›Chapters›Gravitation PYQs›Escape Velocity›Question 9

MHT CET 2026 (16 Apr, Shift 1) · Escape Velocity

Escape Velocity question from MHT CET 2026 (16 Apr, Shift 1)

Q9MHT CET 2026 · 16 APR · SHIFT 1Difficult

Escape Velocity

The masses and radii of the earth and moon are M1M_1, R1R_1 and M2M_2, R2R_2 and respectively, Their centres are at a distance 'd' apart. The minimum speed with which body of mass 'm' should be projected from a distance 2d/32d/3 from the centre of M1M_1 so as to escape to infinity is

Show answer and solution

Answer: (c)

  1. Point is 2d3\dfrac{2d}{3} from M1M_1 and d3\dfrac{d}{3} from M2M_2.
  2. V=−3GM12d−3GM2d=−3G2d(M1+2M2)V = -\dfrac{3GM_1}{2d} - \dfrac{3GM_2}{d} = -\dfrac{3G}{2d}(M_1 + 2M_2)
  3. To escape: 12mv2=m∣V∣⇒v2=3Gd(M1+2M2)\tfrac12 mv^2 = m|V| \Rightarrow v^2 = \dfrac{3G}{d}(M_1+2M_2)
  4. So the answer is (c) 3Gd(M1+2M2)\sqrt{\dfrac{3G}{d}(M_1+2M_2)}.
Report a mistake

More questions on Escape Velocity