What is the true weight of an object in a geostationary satellite that weighed exactly 10.0 N at the north pole?

Explanation

To find the true weight of an object in a geostationary satellite that weighs 10.0 N at the North Pole, we can follow these steps:

### Step 1: Understand the relationship between weight and distance from the Earth's center
Weight is the force exerted by gravity, which is inversely proportional to the square of the distance from the center of the Earth. This can be expressed as:
\[ W \propto \frac{1}{r^2} \]
where \( W \) is the weight and \( r \) is the distance from the center of the Earth.

### Step 2: Identify the radius of the Earth and the height of the geostationary satellite
The radius of the Earth (\( r \)) is approximately:
\[ r = 6400 \text{ km} = 6.4 \times 10^6 \text{ m} \]
The height of the geostationary satellite (\( h \)) is:
\[ h = 36000 \text{ km} = 3.6 \times 10^7 \text{ m} \]

### Step 3: Calculate the total distance from the center of the Earth to the geostationary satellite
The total distance (\( A \)) from the center of the Earth to the geostationary satellite is:
\[ A = r + h = 6400 \text{ km} + 36000 \text{ km} = 42400 \text{ km} = 4.24 \times 10^7 \text{ m} \]

### Step 4: Set up the ratio of weights
Using the relationship established in Step 1, we can set up the ratio of the weights at the North Pole and in the geostationary satellite:
\[ \frac{W'}{W} = \frac{r^2}{A^2} \]
where \( W' \) is the weight in the geostationary satellite and \( W \) is the weight at the North Pole (10 N).

### Step 5: Substitute the known values into the equation
Substituting the known values:
\[ W' = W \cdot \frac{r^2}{A^2} \]
\[ W' = 10 \cdot \frac{(6.4 \times 10^6)^2}{(4.24 \times 10^7)^2} \]

### Step 6: Calculate \( W' \)
Calculating the values:
1. Calculate \( (6.4 \times 10^6)^2 = 4.096 \times 10^{13} \)
2. Calculate \( (4.24 \times 10^7)^2 = 1.798976 \times 10^{15} \)
3. Now substitute these into the equation:
\[ W' = 10 \cdot \frac{4.096 \times 10^{13}}{1.798976 \times 10^{15}} \]
\[ W' \approx 10 \cdot 0.0228 \]
\[ W' \approx 0.228 \text{ N} \]

### Step 7: Final Answer
The true weight of the object in the geostationary satellite is approximately:
\[ W' \approx 0.23 \text{ N} \]

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