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Velocity of a point on the equator of a rotating spherical planet is v. The angular velocity of the planet is such that the apparent value of acceleration due to gravity ‘g’ at the equator is half of that at the poles. The escape velocity of a particle from the surface of the planet is
- a)v
- b)2v
- c)3v
- d)4v
Correct answer is option 'B'. Can you explain this answer?
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Velocity of a point on the equator of a rotating spherical planet is v...
Apparent value of g at pole is equal to
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Velocity of a point on the equator of a rotating spherical planet is v...
Analysis:
To solve this problem, we need to consider the concept of centripetal acceleration, as well as the relationship between gravitational acceleration and escape velocity.
Centripetal Acceleration:
When an object moves in a circular path, it experiences centripetal acceleration directed towards the center of the circle. This acceleration can be calculated using the formula:
a = ω²r
Where a is the centripetal acceleration, ω is the angular velocity, and r is the radius of the circular path.
Gravitational Acceleration:
The acceleration due to gravity at a certain point on a planet's surface depends on the mass of the planet and the distance from its center. It can be calculated using the formula:
g = GM/r²
Where g is the gravitational acceleration, G is the gravitational constant, M is the mass of the planet, and r is the distance from the center of the planet.
Apparent Value of g at the Equator and the Poles:
The apparent value of acceleration due to gravity at the equator is less than at the poles due to the effect of rotation. At the equator, the centrifugal force due to the rotation of the planet reduces the effective gravitational force. As a result, the apparent value of acceleration due to gravity is less than at the poles.
Derivation:
Let's assume the radius of the planet is R and the angular velocity is ω. The velocity of a point on the equator can be calculated using the formula:
v = ωR
We are given that the apparent value of acceleration due to gravity at the equator is half of that at the poles. Let's denote the acceleration due to gravity at the equator as g_e and at the poles as g_p. Therefore, we have:
g_e = (1/2)g_p
We know that gravitational acceleration can be calculated using the formula:
g = GM/R²
Using this formula, we can write the equations for g_e and g_p as:
g_e = GM/R²
g_p = GM/2R²
Escape Velocity:
The escape velocity is the minimum velocity required for an object to escape the gravitational field of a planet. It can be calculated using the formula:
v_e = √(2gR)
Substituting the values of g_e and g_p in the above equation, we can write:
v_e = √(2g_eR)
v_e = √(2GM/R)
Solution:
To find the escape velocity in terms of v, we need to compare the above equation with the equation for v:
v = ωR
Comparing the two equations, we can write:
√(2GM/R) = v
2GM/R = v²
GM/R² = v²/2
To solve this problem, we need to consider the concept of centripetal acceleration, as well as the relationship between gravitational acceleration and escape velocity.
Centripetal Acceleration:
When an object moves in a circular path, it experiences centripetal acceleration directed towards the center of the circle. This acceleration can be calculated using the formula:
Where a is the centripetal acceleration, ω is the angular velocity, and r is the radius of the circular path.
Gravitational Acceleration:
The acceleration due to gravity at a certain point on a planet's surface depends on the mass of the planet and the distance from its center. It can be calculated using the formula:
Where g is the gravitational acceleration, G is the gravitational constant, M is the mass of the planet, and r is the distance from the center of the planet.
Apparent Value of g at the Equator and the Poles:
The apparent value of acceleration due to gravity at the equator is less than at the poles due to the effect of rotation. At the equator, the centrifugal force due to the rotation of the planet reduces the effective gravitational force. As a result, the apparent value of acceleration due to gravity is less than at the poles.
Derivation:
Let's assume the radius of the planet is R and the angular velocity is ω. The velocity of a point on the equator can be calculated using the formula:
We are given that the apparent value of acceleration due to gravity at the equator is half of that at the poles. Let's denote the acceleration due to gravity at the equator as g_e and at the poles as g_p. Therefore, we have:
We know that gravitational acceleration can be calculated using the formula:
Using this formula, we can write the equations for g_e and g_p as:
Escape Velocity:
The escape velocity is the minimum velocity required for an object to escape the gravitational field of a planet. It can be calculated using the formula:
Substituting the values of g_e and g_p in the above equation, we can write:
Solution:
To find the escape velocity in terms of v, we need to compare the above equation with the equation for v:
Comparing the two equations, we can write:
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Velocity of a point on the equator of a rotating spherical planet is v. The angular velocity of the planet is such that the apparent value of acceleration due to gravity g at the equator is half of that at the poles. The escape velocity of a particle from the surface of the planet is a)v b)2vc)3vd)4v Correct answer is option 'B'. Can you explain this answer?
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Velocity of a point on the equator of a rotating spherical planet is v. The angular velocity of the planet is such that the apparent value of acceleration due to gravity g at the equator is half of that at the poles. The escape velocity of a particle from the surface of the planet is a)v b)2vc)3vd)4v Correct 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 Velocity of a point on the equator of a rotating spherical planet is v. The angular velocity of the planet is such that the apparent value of acceleration due to gravity g at the equator is half of that at the poles. The escape velocity of a particle from the surface of the planet is a)v b)2vc)3vd)4v Correct 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 Velocity of a point on the equator of a rotating spherical planet is v. The angular velocity of the planet is such that the apparent value of acceleration due to gravity g at the equator is half of that at the poles. The escape velocity of a particle from the surface of the planet is a)v b)2vc)3vd)4v Correct answer is option 'B'. Can you explain this answer?.
Velocity of a point on the equator of a rotating spherical planet is v. The angular velocity of the planet is such that the apparent value of acceleration due to gravity g at the equator is half of that at the poles. The escape velocity of a particle from the surface of the planet is a)v b)2vc)3vd)4v Correct 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 Velocity of a point on the equator of a rotating spherical planet is v. The angular velocity of the planet is such that the apparent value of acceleration due to gravity g at the equator is half of that at the poles. The escape velocity of a particle from the surface of the planet is a)v b)2vc)3vd)4v Correct 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 Velocity of a point on the equator of a rotating spherical planet is v. The angular velocity of the planet is such that the apparent value of acceleration due to gravity g at the equator is half of that at the poles. The escape velocity of a particle from the surface of the planet is a)v b)2vc)3vd)4v Correct answer is option 'B'. Can you explain this answer?.
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