To find the weight of an object at a certain distance from the center of the Earth, we can use the principle that gravitational force (and hence weight) is inversely proportional to the square of the distance from the center of the Earth. Here's a step-by-step solution:
1. Understand the given information:
- Weight on the surface of the Earth, [tex]\( W_{\text{surface}} \)[/tex] = 40 N.
- Distance from the center of the Earth where we need to find the weight, [tex]\( D \)[/tex] = 12800 km (which is [tex]\( 12800 \times 10^3 \)[/tex] meters).
- Radius of the Earth, [tex]\( R_{\text{earth}} = 6.4 \times 10^6 \)[/tex] meters.
2. Recall the relationship between weight and distance from the center of the Earth:
The gravitational force (weight) at a distance from the center of the Earth is given by:
[tex]\[
W \propto \frac{1}{D^2}
\][/tex]
This implies:
[tex]\[
\frac{W}{W_{\text{surface}}} = \left( \frac{R_{\text{earth}}}{D} \right)^2
\][/tex]
3. Substitute the given values into the equation:
[tex]\[
\frac{W}{40} = \left( \frac{6.4 \times 10^6}{12800 \times 10^3} \right)^2
\][/tex]
4. Simplify the ratio inside the parenthesis:
[tex]\[
\frac{6.4 \times 10^6}{12800 \times 10^3} = \frac{6.4 \times 10^6}{1.28 \times 10^7} = \frac{6.4}{12.8} \times \frac{10^6}{10^7} = 0.5 \times 10^{-1} = 0.05
\][/tex]
5. Calculate the square of the simplified ratio:
[tex]\[
\left( 0.05 \right)^2 = 0.0025
\][/tex]
6. Find the weight [tex]\( W \)[/tex] at distance [tex]\( D \)[/tex]:
[tex]\[
\frac{W}{40} = 0.0025
\][/tex]
[tex]\[
W = 40 \times 0.0025 = 0.1
\][/tex]
7. Final answer:
The weight of the object at a distance of 12800 km from the center of the Earth is 10 N.
