The density of the earth is given by `rho=Kg^(x)r^(y)G^(z)`. Obtain the values of x,y and z

Explanation

To solve the problem of finding the values of \( x \), \( y \), and \( z \) in the equation for the density of the Earth given by \( \rho = k g^x r^y G^z \), we will follow these steps:

### Step 1: Write down the dimensions of each variable
The density \( \rho \) has the dimension of mass per unit volume, which is given by:
\[ [\rho] = M L^{-3} \]
where \( M \) is mass and \( L \) is length.

The dimensions of the other variables are:
- Gravitational acceleration \( g \):
\[ [g] = L T^{-2} \]
- Radius \( r \):
\[ [r] = L \]
- Gravitational constant \( G \):
\[ [G] = M^{-1} L^3 T^{-2} \]

### Step 2: Substitute the dimensions into the equation
We can substitute the dimensions into the equation:
\[ \rho = k g^x r^y G^z \]
This gives us:
\[ M L^{-3} = k (L T^{-2})^x (L)^y (M^{-1} L^3 T^{-2})^z \]

### Step 3: Expand the right-hand side
Expanding the right-hand side:
\[ M L^{-3} = k \cdot L^x T^{-2x} \cdot L^y \cdot M^{-z} L^{3z} T^{-2z} \]
Combining the terms:
\[ M L^{-3} = k \cdot M^{-z} \cdot L^{x+y+3z} \cdot T^{-2x-2z} \]

### Step 4: Compare dimensions on both sides
Now we will compare the dimensions of both sides:
1. For mass \( M \):
\[ 1 = -z \quad \Rightarrow \quad z = -1 \]
2. For length \( L \):
\[ -3 = x + y + 3z \]
Substituting \( z = -1 \):
\[ -3 = x + y + 3(-1) \quad \Rightarrow \quad -3 = x + y - 3 \quad \Rightarrow \quad x + y = 0 \]
3. For time \( T \):
\[ 0 = -2x - 2z \]
Substituting \( z = -1 \):
\[ 0 = -2x - 2(-1) \quad \Rightarrow \quad 0 = -2x + 2 \quad \Rightarrow \quad 2x = 2 \quad \Rightarrow \quad x = 1 \]

### Step 5: Solve for \( y \)
Now substituting \( x = 1 \) into the equation \( x + y = 0 \):
\[ 1 + y = 0 \quad \Rightarrow \quad y = -1 \]

### Final Values
Thus, we have:
\[ x = 1, \quad y = -1, \quad z = -1 \]

### Summary
The values of \( x \), \( y \), and \( z \) are:
- \( x = 1 \)
- \( y = -1 \)
- \( z = -1 \)