Acceleration Due to Gravity and its variation: Practice Problem & Solution
For moon, its mass is $\frac{1}{81}$ of earth mass and its diameter is $\frac{1}{3.7}$ of earth diameter. If acceleration due to gravity at earth surface is $9.8\text{ m/s}^2$ then at moon its value is: (1999)
Solution Explained:
To solve this problem, we apply the core principles of Acceleration Due to Gravity and its variation. Understanding the underlying formula is key to arriving at the correct answer below:
Using $g \propto \frac{M}{R^2}$, we have $g_m = g_e \times (\frac{M_m}{M_e}) \times (\frac{R_e}{R_m})^2$. Substituting the values: $g_m = 9.8 \times \frac{1}{81} \times (3.7)^2 \approx 1.65\text{ m/s}^2$.
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