Gravitational Constant And Universality
The universal gravitational constant G is described as a fundamental constant of nature in the study of gravitation. Which statement about G is correct?
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Solution
Its value is the same everywhere in the universe and independent of the medium between masses
NCERT Class 11, Chapter 8 (Gravitation) explains that on a rotating Earth a portion of the true gravitational pull must be diverted to provide the centripetal force for an observer who is carried in a daily circle of latitude. The apparent gravity is therefore reduced to g′=g−ω2Rcos2λ, where λ is the latitude, ω the Earth's angular speed, and R its radius. The reduction term is largest when cos2λ is maximum, that is at the equator where λ=0∘ and the observer moves in the widest circle at the greatest linear speed. At the poles, λ=90∘, so cosλ=0 and rotation has no effect on apparent gravity at all. At 45∘ latitude the effect is only intermediate, not the greatest. It is therefore certainly not uniform everywhere on the surface. As a plausibility check, points on the equator trace the largest daily circle and so require the most centripetal force, making the rotational reduction in measured gravity greatest there, exactly matching the cos2λ dependence in the formula and explaining why measured g is slightly smaller at the equator than at the poles. In addition, this rotational effect adds to the geometric flattening of the Earth, both of which combine to make effective gravity weakest at the equator and strongest at the poles, a result confirmed by precise gravimeter measurements across latitudes.
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About This Question
- Subject
- physics
- Chapter
- gravitation
- Topic
- gravitational constant and universality
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
Its value is the same everywhere in the universe and independent of the medium between masses
NCERT Class 11, Chapter 8 (Gravitation) explains that on a rotating Earth a portion of the true gravitational pull must be diverted to provide the centripetal force for an observer who is carried in a daily circle of latitude. The apparent gravity is therefore reduced to g′=g−ω2Rcos2λ, where λ is the latitude, ω the Earth's angular speed, and R its radius. The reduction term is largest when cos2λ is maximum, that is at the equator where λ=0∘ and the observer moves in the widest circle at the greatest linear speed. At the poles, λ=90∘, so cosλ=0 and rotation has no effect on apparent gravity at all. At 45∘ latitude the effect is only intermediate, not the greatest. It is therefore certainly not uniform everywhere on the surface. As a plausibility check, points on the equator trace the largest daily circle and so require the most centripetal force, making the rotational reduction in measured gravity greatest there, exactly matching the cos2λ dependence in the formula and explaining why measured g is slightly smaller at the equator than at the poles. In addition, this rotational effect adds to the geometric flattening of the Earth, both of which combine to make effective gravity weakest at the equator and strongest at the poles, a result confirmed by precise gravimeter measurements across latitudes.
This easy difficulty physics question is from the chapter gravitation, covering the topic of gravitational constant and universality. It appeared in the 2025 exam.
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