Question:
Acceleration due to gravity on the surface of earth is \( g \). If the diameter of earth is reduced to one third of its original value and mass remains unchanged, then the acceleration due to gravity on the surface of the earth is ________________ g.
Acceleration due to gravity on the surface of earth is \( g \). If the diameter of earth is reduced to one third of its original value and mass remains unchanged, then the acceleration due to gravity on the surface of the earth is ________________ g.
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Remember that the acceleration due to gravity depends inversely on the square of the radius of the Earth. When the radius decreases by a factor, the acceleration due to gravity increases by the square of that factor.
Updated On: Jan 14, 2026
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Approach Solution - 1
The acceleration due to gravity \( g \) on the surface of the Earth is given by the formula: \[ g = \frac{GM}{R^2}, \] where:
- \( G \) is the gravitational constant,
- \( M \) is the mass of the Earth,
- \( R \) is the radius of the Earth. If the diameter is reduced to one third of its original value, the new radius \( R' \) becomes: \[ R' = \frac{R}{3}. \] Since mass \( M \) remains unchanged, the new acceleration due to gravity \( g' \) is: \[ g' = \frac{GM}{(R/3)^2} = \frac{GM}{R^2} \times 9 = 9g. \] Thus, the acceleration due to gravity increases by a factor of 9.
Final Answer: \( 9g \).
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Approach Solution -2
Step 1: Understand the formula for acceleration due to gravity.
The acceleration due to gravity \( g \) on the surface of the Earth is given by the formula: \[ g = \frac{GM}{R^2}, \] where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.
Step 2: Analyze the changes in the problem.
- The diameter of the Earth is reduced to one third of its original value. Since the diameter is reduced to one third, the radius \( R \) becomes one third of its original value. \[ R' = \frac{R}{3}. \] - The mass \( M \) of the Earth remains unchanged.
Step 3: Apply the formula for the new acceleration due to gravity.
The new acceleration due to gravity \( g' \) is: \[ g' = \frac{GM}{R'^2}. \] Substituting \( R' = \frac{R}{3} \): \[ g' = \frac{GM}{\left(\frac{R}{3}\right)^2} = \frac{GM}{\frac{R^2}{9}} = 9 \times \frac{GM}{R^2} = 9g. \] Thus, the new acceleration due to gravity is 9 times the original value of \( g \).
Final Answer:
\[ \boxed{9g}. \]
The acceleration due to gravity \( g \) on the surface of the Earth is given by the formula: \[ g = \frac{GM}{R^2}, \] where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.
Step 2: Analyze the changes in the problem.
- The diameter of the Earth is reduced to one third of its original value. Since the diameter is reduced to one third, the radius \( R \) becomes one third of its original value. \[ R' = \frac{R}{3}. \] - The mass \( M \) of the Earth remains unchanged.
Step 3: Apply the formula for the new acceleration due to gravity.
The new acceleration due to gravity \( g' \) is: \[ g' = \frac{GM}{R'^2}. \] Substituting \( R' = \frac{R}{3} \): \[ g' = \frac{GM}{\left(\frac{R}{3}\right)^2} = \frac{GM}{\frac{R^2}{9}} = 9 \times \frac{GM}{R^2} = 9g. \] Thus, the new acceleration due to gravity is 9 times the original value of \( g \).
Final Answer:
\[ \boxed{9g}. \]
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