Question:
When a ball is dropped from a height it takes \(t\) sec to reach the ground. If the same experiment is done on a different planet having mass \(100\) times the earth's mass and radius \(10\) times the earth's radius, then the time it will take to cover the same height in the new planet is:
When a ball is dropped from a height it takes \(t\) sec to reach the ground. If the same experiment is done on a different planet having mass \(100\) times the earth's mass and radius \(10\) times the earth's radius, then the time it will take to cover the same height in the new planet is:
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Acceleration due to gravity depends on
\[
g=\frac{GM}{R^2}.
\]
If both mass and radius change proportionally so that \(\frac{M}{R^2}\) remains constant, then \(g\) remains unchanged.
Updated On: Jun 24, 2026
- \(t\ \text{s}\)
- \(100t\ \text{s}\)
- \(\dfrac{t}{100}\ \text{s}\)
- \(\dfrac{t}{10}\ \text{s}\)
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The Correct Option is A
Solution and Explanation
Step 1: Use the formula for acceleration due to gravity.
Acceleration due to gravity is \[ g=\frac{GM}{R^2} \] For the new planet: \[ M'=100M \] and \[ R'=10R \] Thus, \[ g'=\frac{G(100M)}{(10R)^2} \] \[ g'=\frac{100GM}{100R^2} \] \[ g'=\frac{GM}{R^2} \] \[ g'=g \]
Step 2: Use the equation of motion.
For a body dropped from rest, \[ h=\frac{1}{2}gt^2 \] Since the same height is covered and \[ g'=g, \] the time remains unchanged.
Therefore, \[ t'=t \]
Step 3: Final conclusion.
Hence, the required time is \[ \boxed{t\ \text{s}} \]
Acceleration due to gravity is \[ g=\frac{GM}{R^2} \] For the new planet: \[ M'=100M \] and \[ R'=10R \] Thus, \[ g'=\frac{G(100M)}{(10R)^2} \] \[ g'=\frac{100GM}{100R^2} \] \[ g'=\frac{GM}{R^2} \] \[ g'=g \]
Step 2: Use the equation of motion.
For a body dropped from rest, \[ h=\frac{1}{2}gt^2 \] Since the same height is covered and \[ g'=g, \] the time remains unchanged.
Therefore, \[ t'=t \]
Step 3: Final conclusion.
Hence, the required time is \[ \boxed{t\ \text{s}} \]
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