A hydrogen balloon released on the moon

Options

A. climbs up with an acceleration of `6xx9.8ms^(-2)`
B. falls down with an acceleration of `9.8xx6ms^(-2)`
C. falls with acceleration of `9.8 ms^(-2)`
D. falls with an acceleration of `(9.8)/6 ms^(-2)`

Answer

To solve the problem of a hydrogen balloon released on the moon, we need to analyze the forces acting on the balloon and the acceleration due to gravity on the moon. Here’s a step-by-step solution:

### Step 1: Understand the context
A hydrogen balloon is filled with a gas that is lighter than the surrounding air. On Earth, such a balloon would rise due to the buoyant force acting on it. However, the situation is different on the moon.

### Step 2: Determine the acceleration due to gravity on the moon
The acceleration due to gravity on Earth (g) is approximately 9.8 m/s². The acceleration due to gravity on the moon (g_moon) is about 1/6th of that on Earth. Therefore, we can calculate:
\[ g_{moon} = \frac{g}{6} = \frac{9.8 \, \text{m/s}^2}{6} \approx 1.63 \, \text{m/s}^2 \]

### Step 3: Analyze the forces on the balloon
When the hydrogen balloon is released on the moon, two forces act on it:
1. The buoyant force (upward)
2. The weight of the balloon (downward)

On the moon, the weight of the balloon can be calculated using:
\[ \text{Weight} = m \cdot g_{moon} \]
where \( m \) is the mass of the balloon.

### Step 4: Determine the net force and acceleration
Since the balloon is filled with hydrogen, it is less dense than the surrounding lunar atmosphere (which is negligible). Therefore, the buoyant force will be less than the weight of the balloon, causing it to fall.

The net force acting on the balloon can be expressed as:
\[ F_{net} = \text{Weight} - \text{Buoyant Force} \]

However, since the buoyant force is much smaller than the weight, we can approximate that the balloon will fall with an acceleration equal to the acceleration due to gravity on the moon.

### Step 5: Conclusion
The balloon will fall with an acceleration of approximately \( g_{moon} \), which is:
\[ \text{Acceleration} = g_{moon} \approx 1.63 \, \text{m/s}^2 \]

### Final Answer
The balloon falls with an acceleration of \( \frac{9.8}{6} \, \text{m/s}^2 \).