Final answer:
To find the depth where the value of acceleration due to gravity is 1/n times its surface value, we derive an equation considering Earth's gravity decreases linearly with depth within a uniformly dense Earth. The correct answer is R(n-1/n), which expresses the depth as a function of Earth's radius R and the factor n.The correct option is option b.
Step-by-step explanation:
The student question pertains to acceleration due to gravity and how its value decreases with depth. This question falls into the realm of classical Physics, namely gravitation. The formula for the acceleration due to gravity inside Earth at a depth d is derived from the law of universal gravitation and assumes a uniform Earth density.
Let R represent the radius of Earth and g is the acceleration due to gravity on the surface of Earth. At a depth d, the gravitational force, and thus the acceleration, is only due to the mass of the sphere with radius R - d (the mass above the depth d doesn't contribute to the gravity at that level).
Because gravity within Earth varies linearly with distance from the center (assuming uniform density), the acceleration due to gravity at depth d is g' = g * (1 - d/R). To find the depth where gravity is 1/n times its surface value, we set g' = g/n and solve for d:
g/n = g * (1 - d/R)
1/n = 1 - d/R
d/R = 1 - 1/n
Thus, d = R * (1 - 1/n) which corresponds to option B. R(n-1/n).