A boy can jump 1m high on the earth's surface. What is the height he will be able to jump on a planet whose density is half of earth's density and radius is one-fourth's radius ?

Answer

To solve the problem of how high a boy can jump on a planet with half the density and one-fourth the radius of Earth, we can follow these steps:

### Step 1: Understand the relationship between jump height and gravitational acceleration

The height a boy can jump is determined by the gravitational acceleration on the planet. The formula for the maximum height \( h \) a person can jump is given by:

\[ h = \frac{v^2}{2g} \]

where \( v \) is the initial velocity and \( g \) is the acceleration due to gravity.

### Step 2: Calculate the gravitational acceleration on Earth

The gravitational acceleration on Earth is denoted as \( g \).

### Step 3: Calculate the gravitational acceleration on the new planet

The formula for gravitational acceleration \( g' \) on a planet is:

\[ g' = \frac{G \cdot M}{R^2} \]

where \( G \) is the universal gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius of the planet.

### Step 4: Express the mass of the new planet in terms of density

The mass \( M \) of the planet can be expressed in terms of its density \( \rho \) and volume \( V \):

\[ M = \rho \cdot V = \rho \cdot \frac{4}{3} \pi R^3 \]

### Step 5: Substitute the density and radius of the new planet

Given that the density of the new planet is half that of Earth, we have:

\[ \rho' = \frac{\rho_e}{2} \]

And since the radius is one-fourth that of Earth:

\[ R' = \frac{R_e}{4} \]

### Step 6: Substitute these values into the gravitational acceleration formula

Now, substituting these values into the gravitational acceleration formula:

\[ g' = \frac{G \cdot \left(\frac{\rho_e}{2} \cdot \frac{4}{3} \pi \left(\frac{R_e}{4}\right)^3\right)}{\left(\frac{R_e}{4}\right)^2} \]

### Step 7: Simplify the expression

After simplification, we find:

\[ g' = \frac{G \cdot \frac{\rho_e}{2} \cdot \frac{4}{3} \pi \cdot \frac{R_e^3}{64}}{\frac{R_e^2}{16}} = \frac{G \cdot \rho_e \cdot \frac{4}{3} \pi \cdot R_e^3}{2 \cdot \frac{R_e^2}{16}} = \frac{G \cdot \rho_e \cdot \frac{4}{3} \pi \cdot R_e^3 \cdot 16}{2 \cdot R_e^2} \]

This simplifies to:

\[ g' = \frac{8}{1} g = 8g \]

### Step 8: Relate the jump heights

Using the relationship established earlier, we can relate the jump heights on Earth and the new planet:

\[ \frac{h'}{h} = \frac{g}{g'} \]

Substituting \( h = 1 \, \text{m} \) (the height the boy can jump on Earth) and \( g' = 8g \):

\[ h' = 1 \cdot \frac{g}{8g} = \frac{1}{8} \, \text{m} \]

### Step 9: Calculate the new height

Since the boy can jump 1 meter on Earth, on the new planet, he will be able to jump:

\[ h' = 8 \, \text{m} \]

### Final Answer

The height the boy will be able to jump on the new planet is **8 meters**.

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