AP EAMCET202018 Sep 2020Morning ShiftPhysicsGravitationActual
The escape velocity for a planet whose radius is (1.7 10^6 ~m ) and acceleration due to gravity is (1.7 ~ms ⁻² ) is
Options
- A(1.7 kms ⁻¹ )
- B(2.89 kms ⁻¹ )
- C(1.7 2 kms ⁻¹ )
- D(3.4 kms ⁻¹ )
Correct answer
C. (1.7 2 kms ⁻¹ )
Step-by-step solution
Radius of planet, (R=1.7 10^6 ~m ) Acceleration due to gravity, (g=1.7 ~ms ⁻² ) ( ) Escape velocity on the surface of planet is given as ( aligned v_e & = 2 g R = 2 1.7 1.7 10^6 & =1.7 2 10^3 ~ms ⁻¹=1.7 2 ~km ~s ⁻¹ aligned )