Find the acceleration due to gravity of the moon at a point 1000 km above the moon's surface. The mass of the moon is `7.4 x10^22 ' kg and its radius is 1740 km.

Explanation

To find the acceleration due to gravity (g') on the moon at a point 1000 km above its surface, we can use the formula for gravitational acceleration at a distance from the center of a celestial body:

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

Where:
- \( G \) is the universal gravitational constant, approximately \( 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)
- \( M \) is the mass of the moon, which is \( 7.4 \times 10^{22} \, \text{kg} \)
- \( r \) is the distance from the center of the moon to the point where we want to find the acceleration due to gravity.

### Step 1: Determine the total distance from the center of the moon
The radius of the moon is given as \( 1740 \, \text{km} \) and the height above the moon's surface is \( 1000 \, \text{km} \).

To find the total distance \( r \) from the center of the moon to the point 1000 km above its surface, we add the radius of the moon and the height:

\[ r = \text{Radius of the moon} + \text{Height above the surface} \]
\[ r = 1740 \, \text{km} + 1000 \, \text{km} = 2740 \, \text{km} \]

Convert this distance into meters:

\[ r = 2740 \, \text{km} \times 1000 \, \text{m/km} = 2.74 \times 10^6 \, \text{m} \]

### Step 2: Substitute values into the formula for g'
Now that we have \( r \), we can substitute the values into the formula for gravitational acceleration:

\[ g' = \frac{G \cdot M}{r^2} \]
\[ g' = \frac{(6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2) \cdot (7.4 \times 10^{22} \, \text{kg})}{(2.74 \times 10^6 \, \text{m})^2} \]

### Step 3: Calculate \( r^2 \)
Calculate \( r^2 \):

\[ r^2 = (2.74 \times 10^6 \, \text{m})^2 = 7.5076 \times 10^{12} \, \text{m}^2 \]

### Step 4: Calculate \( g' \)
Now substitute \( r^2 \) back into the equation for \( g' \):

\[ g' = \frac{(6.67 \times 10^{-11}) \cdot (7.4 \times 10^{22})}{7.5076 \times 10^{12}} \]

Calculating the numerator:

\[ 6.67 \times 10^{-11} \cdot 7.4 \times 10^{22} = 4.9388 \times 10^{12} \]

Now, divide by \( r^2 \):

\[ g' = \frac{4.9388 \times 10^{12}}{7.5076 \times 10^{12}} \approx 0.657 \, \text{m/s}^2 \]

### Final Answer
Thus, the acceleration due to gravity on the moon at a point 1000 km above its surface is approximately:

\[ g' \approx 0.657 \, \text{m/s}^2 \]