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The mass and diameter of a planet have twice the value of the corresponding parameters of earth. Acceleration due to gravity on the surface of the planet is
- a)9.8 m/sec2
- b)4.9 m/sec2
- c)980 m/sec2
- d)19.6 m/sec2
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
The mass and diameter of a planet have twice the value of the correspo...
Given: The mass and diameter of the planet are twice that of the Earth.
To find: The acceleration due to gravity on the surface of the planet.
Solution:
Let M be the mass of the planet and R be the radius of the planet.
Given, the mass of the planet is twice that of the Earth, i.e., M = 2M⊕, where M⊕ is the mass of the Earth.
Also, the diameter of the planet is twice that of the Earth, i.e., 2R⊕ = R.
Using the formula for acceleration due to gravity, we have:
g = GM/R²
where G is the universal gravitational constant.
Substituting M = 2M⊕ and R = 2R⊕/2 in the above equation, we get:
g = G(2M⊕)/(2R⊕/2)²
g = G(2M⊕)/(R⊕)²
g = 2GM⊕/(R⊕)²
But we know that the acceleration due to gravity on the surface of the Earth is g⊕ = GM⊕/(R⊕)².
Substituting this value in the above equation, we get:
g = 2g⊕
Therefore, the acceleration due to gravity on the surface of the planet is twice that of the Earth, i.e., g = 4.9 m/s².
Hence, option (b) is the correct answer.
To find: The acceleration due to gravity on the surface of the planet.
Solution:
Let M be the mass of the planet and R be the radius of the planet.
Given, the mass of the planet is twice that of the Earth, i.e., M = 2M⊕, where M⊕ is the mass of the Earth.
Also, the diameter of the planet is twice that of the Earth, i.e., 2R⊕ = R.
Using the formula for acceleration due to gravity, we have:
g = GM/R²
where G is the universal gravitational constant.
Substituting M = 2M⊕ and R = 2R⊕/2 in the above equation, we get:
g = G(2M⊕)/(2R⊕/2)²
g = G(2M⊕)/(R⊕)²
g = 2GM⊕/(R⊕)²
But we know that the acceleration due to gravity on the surface of the Earth is g⊕ = GM⊕/(R⊕)².
Substituting this value in the above equation, we get:
g = 2g⊕
Therefore, the acceleration due to gravity on the surface of the planet is twice that of the Earth, i.e., g = 4.9 m/s².
Hence, option (b) is the correct answer.
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Community Answer
The mass and diameter of a planet have twice the value of the correspo...
Given that the mass and radius (similar to diameter) gets doubled. The acceleration due to gravity is directly proportional to the ratio of mass to the square of radius.
So, the case of this planet the mass to radius square ratio will be 2/4 = 1/2. This means the acceleration due to gravity on that planet is half the value on Earth and is equal to 4.9 m/s^2
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The mass and diameter of a planet have twice the value of the corresponding parameters of earth. Acceleration due to gravity on the surface of the planet isa)9.8 m/sec2b)4.9 m/sec2c)980 m/sec2d)19.6 m/sec2Correct answer is option 'B'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The mass and diameter of a planet have twice the value of the corresponding parameters of earth. Acceleration due to gravity on the surface of the planet isa)9.8 m/sec2b)4.9 m/sec2c)980 m/sec2d)19.6 m/sec2Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The mass and diameter of a planet have twice the value of the corresponding parameters of earth. Acceleration due to gravity on the surface of the planet isa)9.8 m/sec2b)4.9 m/sec2c)980 m/sec2d)19.6 m/sec2Correct answer is option 'B'. Can you explain this answer?.
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