The depth $d$ at which the value of acceleration due to gravity becomes $\frac {1} { n}$ time the value at the surface is (R be the radius of the earth)
- R/n
- $R/n^2$
- $R\frac{(n-1)}{n}$
- $\frac{Rn}{(n-1)}$
The Correct Option is C
Solution and Explanation
The correct option is(C): \(R\frac{(n-1)}{n}\).
At depth d from earth's surface,
\(g'=g\left(1-\frac{d}{R} \right)\)
Given \(g'=\frac{g}{n}\)
So, \(\frac{g}{n}=g\left(1-\frac{d}{R}\right)\)
or \(\frac{1}{n}=1-\frac{d}{R}\) or \(\frac{d}{R}=1-\frac{1}{n}\)
\(=\frac{(n-1)R}{n}\)
\(\therefore\) \(d=\frac{(n-1)R}{n}\)
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Concepts Used:
Newtons Law of Gravitation
Gravitational Force
Gravitational force is a central force that depends only on the position of the test mass from the source mass and always acts along the line joining the centers of the two masses.
Newton’s Law of Gravitation:
According to Newton’s law of gravitation, “Every particle in the universe attracts every other particle with a force whose magnitude is,
- Directly proportional to the product of their masses i.e. F ∝ (M1M2) . . . . (1)
- Inversely proportional to the square of the distance between their center i.e. (F ∝ 1/r2) . . . . (2)
By combining equations (1) and (2) we get,
F ∝ M1M2/r2
F = G × [M1M2]/r2 . . . . (7)
Or, f(r) = GM1M2/r2 [f(r)is a variable, Non-contact, and conservative force]