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:

• A. \(t\ \text{s}\)
• B. \(100t\ \text{s}\)
• C. \(\dfrac{t}{100}\ \text{s}\)
• D. \(\dfrac{t}{10}\ \text{s}\)

Hint:

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.

Answer 1

The Correct Option is A

Step 1: Recall the formula for gravitational acceleration.
At the surface of a planet, the acceleration due to gravity is:
\[ g = \frac{GM}{R^2} \] where $G$ is the gravitational constant, $M$ is the planet's mass, and $R$ is its radius.

Step 2: Write the ratio of gravitational accelerations.
For the new planet: $M' = 100M$ and $R' = 10R$.
\[ \frac{g'}{g} = \frac{M'/R'^2}{M/R^2} = \frac{100M / (10R)^2}{M/R^2} = \frac{100M}{100R^2} \cdot \frac{R^2}{M} = 1 \]

Step 3: Interpret the result.
Since $g' = g$, the gravitational pull at the surface of the new planet is exactly the same as on Earth.

Step 4: Use kinematics for the fall.
For a ball dropped from rest through height $h$:
\[ h = \frac{1}{2}g t^2 \]

Step 5: Compare times on both planets.
Since $h$ is the same and $g' = g$, the time to fall must also be the same:
\[ t' = t \]

Step 6: State the answer.
The ball takes the same time $t$ seconds on the new planet.
\[ \boxed{t \text{ s}} \]