Exams › NEET › Physics

What is the proportion of the escape speed on Earth (ve) compared to that on a planet (vp) with double the radius and double the average density of Earth?

1. 1 : 2
2. 1 : 2√2
3. 1 : 4
4. 1 : 2

Correct answer: 1 : 2√2

Solution

Escape velocity is proportional to the square root of the product of the radius and density of the celestial body. Since the planet's radius and density are both twice that of Earth, the escape velocity ratio becomes √(2 × 2) = √4 = 2. Thus, ve : vp = 1 : 2√2.

Related NEET Physics questions

• Assertion Newton's law of gravitation resembles Coulomb's law of electrical forces. Reason Coulomb's law has the product of two charges in place of the product of the masses, and the electrostatic constant in place of the gravitational constant.
• How the gravitational constant will change if a brass plate is introduced between two bodies?
• Suppose that the acceleration of a free fall at the surface of a distant planet was found to be equal to that at the surface of the earth. If the diameter of the planet were twice the diameter of the earth, then the ratio of mean density of the planet to that of the earth would be:
• Suppose gravity of earth suddenly becomes zero, then in which direction will the moon begin to move if no other celestial body affects it?
• The distance between the sun and the earth is
• Gravitational constant is experimentally measured using a

⚔️ Practice NEET Physics free + battle 1v1 →