MHT CET202613 April 2026Evening ShiftPhysicsGravitationActual
If earth has a mass nine times and radius twice to the planet P. Then v_e 3 x ms ⁻¹ will be the minimum velocity required by a rocket to pull out of gravitational force of P, where v_e is escape velocity on earth. The value of x is
Options
- A2
- B3
- C18
- D1
Correct answer
A. 2
Step-by-step solution
The escape velocity on Earth is given by v_e = 2GM_e R_e The escape velocity on planet P is given by v_p = 2GM_p R_p Given that M_e = 9M_p M_p = M_e 9 and R_e = 2R_p R_p = R_e 2 Substituting these values into the expression for v_p : v_p = 2G ( M_e 9 ) R_e 2 v_p = 2GM_e R_e 2 9 v_p = 2GM_e R_e 2 3 v_p = v_e 3 2 Comparing this with the given expression v_e 3 x , we get x = 2 . Answer: 2