The radius of the earth is `6.37 xx 10^(6) m` and its mass is `5.975 xx 10^(24) kg`. Find the earth's average density to approopriate significant figures.

Explanation

To find the Earth's average density, we can follow these steps:

### Step 1: Understand the formula for density
Density (ρ) is defined as mass (m) divided by volume (V):
\[ \rho = \frac{m}{V} \]

### Step 2: Calculate the volume of the Earth
The Earth can be approximated as a sphere, and the volume (V) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
where \( r \) is the radius of the sphere.

### Step 3: Substitute the values into the volume formula
Given the radius of the Earth \( r = 6.37 \times 10^6 \, \text{m} \), we can calculate the volume:
\[ V = \frac{4}{3} \pi (6.37 \times 10^6)^3 \]

### Step 4: Calculate the mass of the Earth
The mass of the Earth is given as:
\[ m = 5.975 \times 10^{24} \, \text{kg} \]

### Step 5: Substitute the mass and volume into the density formula
Now we can substitute the mass and volume into the density formula:
\[ \rho = \frac{5.975 \times 10^{24}}{V} \]

### Step 6: Calculate the volume
First, we need to calculate \( (6.37 \times 10^6)^3 \):
\[ (6.37 \times 10^6)^3 = 2.573 \times 10^{20} \, \text{m}^3 \]
Now, substituting this into the volume formula:
\[ V = \frac{4}{3} \pi (2.573 \times 10^{20}) \approx 1.079 \times 10^{21} \, \text{m}^3 \]

### Step 7: Calculate the density
Now we can calculate the density:
\[ \rho = \frac{5.975 \times 10^{24}}{1.079 \times 10^{21}} \approx 5.54 \times 10^{3} \, \text{kg/m}^3 \]

### Step 8: Round to appropriate significant figures
The mass \( 5.975 \times 10^{24} \) has four significant figures, and the radius \( 6.37 \times 10^{6} \) has three significant figures. Therefore, we should express the density to three significant figures:
\[ \rho \approx 5.52 \times 10^{3} \, \text{kg/m}^3 \]

### Final Answer
The average density of the Earth is:
\[ \rho \approx 5.52 \times 10^{3} \, \text{kg/m}^3 \]