In gravitational physics, if the value of acceleration due to gravity at the surface of the earth is g, what is its value at a point located at a height of 2R above the surface (that is, at a distance 3R from the centre of the earth), where R is the radius of the earth?

Introduction / Context:
The acceleration due to gravity g at the surface of the earth has a standard approximate value of 9.8 metres per second squared. However, this value changes as we move away from the surface. The question asks you to determine how g changes at a specific height above the surface, expressed in terms of the earth's radius R. This is a typical conceptual numerical question in gravitation, testing your understanding of the inverse square dependence of gravitational field on distance from the centre of a spherical mass.

Given Data / Assumptions:

• Acceleration due to gravity at the surface of the earth is g.
• Radius of the earth is R.
• We consider a point at a height of 2R above the surface, so its distance from the centre is 3R.
• The earth is treated as a spherically symmetric mass distribution.

Concept / Approach:
Outside a spherically symmetric body like the earth, the gravitational field behaves as if all mass were concentrated at the centre. The acceleration due to gravity at a distance r from the centre is given by g(r) = G * M / r^2, where G is the universal gravitational constant and M is the mass of the earth. At the surface, r = R and g = G * M / R^2. At a distance r = 3R, the new acceleration g(3R) becomes G * M / (3R)^2. By taking the ratio of g(3R) to g, we can find the new value in terms of g without knowing numerical values of G or M.

Step-by-Step Solution:
Step 1: Write the expression for acceleration due to gravity at distance r: g(r) = G * M / r^2.Step 2: At the surface, r = R, so g = G * M / R^2.Step 3: At height 2R above the surface, the distance from the centre is r = R + 2R = 3R.Step 4: The new acceleration is g(3R) = G * M / (3R)^2 = G * M / (9R^2).Step 5: Divide g(3R) by g to find the ratio: g(3R) / g = (G * M / 9R^2) / (G * M / R^2) = 1/9, so g(3R) = g / 9.

Verification / Alternative check:
You can check the trend qualitatively. As we move farther from the earth, the acceleration due to gravity must decrease. The distance has become three times the radius, so the inverse square law suggests the acceleration becomes one ninth of its surface value. This matches the derived result g/9. Since we expect a significant decrease but not negative or zero value at this moderate height, the answer g/9 is reasonable and consistent with physical intuition and the mathematical formula.

Why Other Options Are Wrong:
Option A, g/3, corresponds to an inverse first power dependence on distance, which is not correct for gravitational field; gravity follows a 1/r^2 law, not 1/r. Option B, g/4, would correspond to doubling the distance, not tripling it. Option D, g/2, again does not fit the inverse square relationship and would predict too weak a decrease. Only option C, g/9, is consistent with the factor of three increase in distance and the inverse square law, since 3^2 is 9.

Common Pitfalls:
Students often confuse height above the surface with distance from the centre, forgetting to add the radius of the earth. Another common mistake is applying a linear 1/r dependence rather than 1/r^2, leading to wrong factors such as g/3. To avoid these errors, always express distances from the centre of the earth and substitute carefully into the formula g(r) = G * M / r^2. Doing a quick sanity check on whether gravity should decrease strongly or slowly as distance changes also helps confirm the result.

Final Answer:
g/9