A hypothetical planet has a density 4 times that of the earth and a radius 2 times that of the earth. If the escape velocity from the surface of the earth is 11.2 ~km/s , the escape velocity from the surface of the hypothetical planet will be :

Step-by-step solution

The escape velocity from the surface of a planet is given by v_ e = 2GM R . Since mass M = 4 3 R³ , we can write: v_ e = 2G R ( 4 3 R³ ) = 8 G 3 R² Thus, v_ e R . For the hypothetical planet and the earth, the ratio of their escape velocities is: v_ p v_ e = ( R_ p R_ e ) _ p _ e Given that R_ p = 2R_ e and _ p = 4 _ e , we have: v_ p v_ e = (2) 4 = 4 Therefore, the escape velocity from the hypothetical planet is: v_ p = 4 11.2 ~km/s = 44.8 ~km/s . Answer: 44.8 ~km/s