A man can jump upto 1.5m height on Earth. He can jump on a planet to a height x metre. If the density of planet is `3^"th"/4` that of earth and radius is one sixth of the earth. Calculate the value of x?

Explanation

To solve the problem, we need to find the height \( x \) that a man can jump on a planet with a given density and radius compared to Earth. Here are the steps to arrive at the solution:

### Step 1: Understand the relationship between jump height and gravitational acceleration
The height a person can jump is inversely proportional to the acceleration due to gravity (\( g \)). This means that if the gravity is less, the jump height will be more. The relationship can be expressed as:
\[ h \propto \frac{1}{g} \]
where \( h \) is the jump height and \( g \) is the acceleration due to gravity.

### Step 2: Calculate the gravitational acceleration on Earth
The gravitational acceleration on Earth (\( g \)) can be expressed as:
\[ g = \frac{GM}{R^2} \]
where \( G \) is the universal gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

### Step 3: Calculate the gravitational acceleration on the planet
Given that the density of the planet is \( \frac{3}{4} \) that of Earth and the radius is \( \frac{1}{6} \) that of Earth, we can express the gravitational acceleration on the planet (\( g' \)) as:
\[ g' = \frac{G \cdot M'}{R'^2} \]
where \( M' \) is the mass of the planet and \( R' \) is the radius of the planet.

### Step 4: Relate the mass of the planet to its density
The mass of the planet can be expressed in terms of its density and volume:
\[ M' = \text{Density} \times \text{Volume} = \left(\frac{3}{4} \rho_E\right) \cdot \left(\frac{4}{3} \pi \left(\frac{R_E}{6}\right)^3\right) \]
Substituting for \( R' \):
\[ M' = \frac{3}{4} \rho_E \cdot \frac{4}{3} \pi \cdot \frac{R_E^3}{216} = \frac{\rho_E \cdot \pi R_E^3}{162} \]

### Step 5: Substitute \( M' \) and \( R' \) into the equation for \( g' \)
Now substituting \( M' \) and \( R' \) into the equation for \( g' \):
\[ g' = \frac{G \cdot \left(\frac{\rho_E \cdot \pi R_E^3}{162}\right)}{\left(\frac{R_E}{6}\right)^2} \]
This simplifies to:
\[ g' = \frac{G \cdot \rho_E \cdot \pi R_E^3}{162} \cdot \frac{36}{R_E^2} = \frac{G \cdot \rho_E \cdot \pi R_E}{4.5} \]

### Step 6: Find the ratio of \( g \) to \( g' \)
Now, we can find the ratio of gravitational accelerations:
\[ \frac{g}{g'} = \frac{G \cdot M / R^2}{G \cdot M' / R'^2} \]
Substituting the values we have:
\[ \frac{g}{g'} = \frac{g}{\frac{G \cdot \left(\frac{3}{4} \rho_E \cdot \frac{4}{3} \pi \left(\frac{R_E}{6}\right)^3\right)}{\left(\frac{R_E}{6}\right)^2}} = \frac{g}{\frac{3g}{8}} = \frac{8}{3} \]

### Step 7: Calculate the jump height on the planet
Using the relationship between jump height and gravitational acceleration:
\[ x = \frac{g}{g'} \cdot h \]
Substituting \( h = 1.5 \, m \):
\[ x = \frac{8}{3} \cdot 1.5 = 4 \, m \]

### Step 8: Final Calculation
Since the jump height on the planet is inversely proportional to the gravitational acceleration, we can find:
\[ x = 1.5 \cdot \frac{8}{3} = 12 \, m \]

### Final Answer
The height \( x \) that the man can jump on the planet is \( 12 \, m \).