NEET2026PhysicsChapterActual
A hypothetical planet has an average mass density 9 times that of Earth. If the escape velocity from the surface of this planet is identical to the escape velocity from the surface of Earth, what is the radius of this planet in terms of Earth's radius ( R_E )?
Options
- AR_E 9
- BR_E 3
- C3R_E
- D9R_E
Correct answer
B. R_E 3
Step-by-step solution
The escape velocity from the surface of a planet is given by v_e = 2GM R . The mass of the planet can be expressed in terms of its density and radius R as M = 4 3 R^3 . Substituting this into the escape velocity formula, we get: v_e = 2G R ( 4 3 R^3 ) = R 8 G 3 This implies that v_e R . Given that the escape velocity of the hypothetical planet is equal to that of Earth ( v_ ep = v_ eE ): R_p _p = R_E _E We are given _p = 9 _E . Substituting this value: R_p 9 _E = R_E _E 3R_p = R_E R_p = R_E 3 Answer: R_E 3