NEET Exam > NEET Questions > The ratio of escape velocity at earth (ve) to...
Start Learning for Free
The ratio of escape velocity at earth (ve) to the escape velocity at a planet (vp) whose radius and mean density are twice as that of earth is :
- a)1:√2
- b)1:2
- c)1:2√2
- d)1:4
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
The ratio of escape velocity at earth (ve) to the escape velocity at a...
Calculation of escape velocity for Earth and the planet
Escape velocity is the minimum velocity required by an object to escape from the gravitational pull of a celestial body. It is given by the formula:
VE = (2GM/R)^(1/2)
Where VE is the escape velocity, G is the gravitational constant, M is the mass of the celestial body, and R is the radius of the celestial body.
For Earth, the values of G, M, and R are 6.67 × 10^-11 Nm^2/kg^2, 5.97 × 10^24 kg, and 6.38 × 10^6 m, respectively. Substituting these values in the above formula, we get:
VE = (2 × 6.67 × 10^-11 × 5.97 × 10^24 / 6.38 × 10^6)^(1/2) = 11.2 km/s
For the planet whose radius and mean density are twice that of Earth, the values of G, M, and R are the same as that of Earth, but the mass is twice and the radius is twice. Substituting these values in the above formula, we get:
VP = (2 × 6.67 × 10^-11 × 2 × 5.97 × 10^24 / 2 × 2 × 6.38 × 10^6)^(1/2) = 7.9 km/s
Calculation of the ratio of escape velocities
The ratio of escape velocities is given by:
VE/VP = 11.2/7.9 = 1.41 ≈ 1.4
Therefore, the correct option is C) 1:2.
Escape velocity is the minimum velocity required by an object to escape from the gravitational pull of a celestial body. It is given by the formula:
VE = (2GM/R)^(1/2)
Where VE is the escape velocity, G is the gravitational constant, M is the mass of the celestial body, and R is the radius of the celestial body.
For Earth, the values of G, M, and R are 6.67 × 10^-11 Nm^2/kg^2, 5.97 × 10^24 kg, and 6.38 × 10^6 m, respectively. Substituting these values in the above formula, we get:
VE = (2 × 6.67 × 10^-11 × 5.97 × 10^24 / 6.38 × 10^6)^(1/2) = 11.2 km/s
For the planet whose radius and mean density are twice that of Earth, the values of G, M, and R are the same as that of Earth, but the mass is twice and the radius is twice. Substituting these values in the above formula, we get:
VP = (2 × 6.67 × 10^-11 × 2 × 5.97 × 10^24 / 2 × 2 × 6.38 × 10^6)^(1/2) = 7.9 km/s
Calculation of the ratio of escape velocities
The ratio of escape velocities is given by:
VE/VP = 11.2/7.9 = 1.41 ≈ 1.4
Therefore, the correct option is C) 1:2.
Free Test
FREE
| Start Free Test |
Community Answer
The ratio of escape velocity at earth (ve) to the escape velocity at a...
|
Explore Courses for NEET exam
|
|
Similar NEET Doubts
Top Courses for NEETView all
The ratio of escape velocity at earth (ve) to the escape velocity at a planet (vp) whose radius and meandensity are twice as that of earth is :a)1:√2b)1:2c)1:2√2d)1:4Correct answer is option 'C'. Can you explain this answer?
Question Description
The ratio of escape velocity at earth (ve) to the escape velocity at a planet (vp) whose radius and meandensity are twice as that of earth is :a)1:√2b)1:2c)1:2√2d)1:4Correct answer is option 'C'. Can you explain this answer? for NEET 2025 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The ratio of escape velocity at earth (ve) to the escape velocity at a planet (vp) whose radius and meandensity are twice as that of earth is :a)1:√2b)1:2c)1:2√2d)1:4Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The ratio of escape velocity at earth (ve) to the escape velocity at a planet (vp) whose radius and meandensity are twice as that of earth is :a)1:√2b)1:2c)1:2√2d)1:4Correct answer is option 'C'. Can you explain this answer?.
The ratio of escape velocity at earth (ve) to the escape velocity at a planet (vp) whose radius and meandensity are twice as that of earth is :a)1:√2b)1:2c)1:2√2d)1:4Correct answer is option 'C'. Can you explain this answer? for NEET 2025 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The ratio of escape velocity at earth (ve) to the escape velocity at a planet (vp) whose radius and meandensity are twice as that of earth is :a)1:√2b)1:2c)1:2√2d)1:4Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The ratio of escape velocity at earth (ve) to the escape velocity at a planet (vp) whose radius and meandensity are twice as that of earth is :a)1:√2b)1:2c)1:2√2d)1:4Correct answer is option 'C'. Can you explain this answer?.
Solutions for The ratio of escape velocity at earth (ve) to the escape velocity at a planet (vp) whose radius and meandensity are twice as that of earth is :a)1:√2b)1:2c)1:2√2d)1:4Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for NEET.
Download more important topics, notes, lectures and mock test series for NEET Exam by signing up for free.
Here you can find the meaning of The ratio of escape velocity at earth (ve) to the escape velocity at a planet (vp) whose radius and meandensity are twice as that of earth is :a)1:√2b)1:2c)1:2√2d)1:4Correct answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
The ratio of escape velocity at earth (ve) to the escape velocity at a planet (vp) whose radius and meandensity are twice as that of earth is :a)1:√2b)1:2c)1:2√2d)1:4Correct answer is option 'C'. Can you explain this answer?, a detailed solution for The ratio of escape velocity at earth (ve) to the escape velocity at a planet (vp) whose radius and meandensity are twice as that of earth is :a)1:√2b)1:2c)1:2√2d)1:4Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of The ratio of escape velocity at earth (ve) to the escape velocity at a planet (vp) whose radius and meandensity are twice as that of earth is :a)1:√2b)1:2c)1:2√2d)1:4Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice The ratio of escape velocity at earth (ve) to the escape velocity at a planet (vp) whose radius and meandensity are twice as that of earth is :a)1:√2b)1:2c)1:2√2d)1:4Correct answer is option 'C'. Can you explain this answer? tests, examples and also practice NEET tests.
|
|
Explore Courses for NEET exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup