Question

At a height 'R' above the earth's surface the gravitational acceleration is (R = radius of earth, g = acceleration due to gravity on earth's surface)

• A. $g$
• B. $\frac{g}{8}$
• C. $\frac{g}{4}$
• D. $\frac{g}{2}$

Hint:

Always remember that gravity follows an inverse-square law relative to the distance from the center of the planet. Doubling your distance from the center ($R \rightarrow 2R$) means the gravitational pull drops by a factor of $2^2 = 4$, yielding $\frac{g}{4}$ instantly!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We need to find the acceleration due to gravity ($g'$) at an altitude height $h = R$ above the surface of the Earth, expressed in terms of the standard surface gravitational acceleration $g$.

Step 2: Key Formula or Approach:
The acceleration due to gravity at any distance $r$ measured from the center of the Earth is given by the inverse-square law: $$g' = \frac{GM}{r^2}$$ The total distance from the center is $r = R + h$, where $R$ is the Earth's radius and $h$ is the altitude. At the surface ($h = 0$), $g = \frac{GM}{R^2}$. We can establish a direct proportional relationship: $$g' = g \left( \frac{R}{R + h} \right)^2$$

Step 3: Detailed Explanation:
Substitute the given altitude $h = R$ into our proportional equation: $$g' = g \left( \frac{R}{R + R} \right)^2$$ Simplify the expression inside the parentheses: $$g' = g \left( \frac{R}{2R} \right)^2$$ The radius variable $R$ cancels out completely: $$g' = g \left( \frac{1}{2} \right)^2 = \frac{g}{4}$$ This matches option (C).

Step 4: Final Answer:
The acceleration due to gravity at a height $R$ is $\frac{g}{4}$, which corresponds to option (C).