Question
Determine the force of attraction for the aircraft (m = 5300 kg) at a height of 2000 km
around the earthβs surface (π = 5.97 Γ 1024 ππ). Take the earthβs diameter as d = 13
000 km and G = 6.67π₯10β11π.π2.ππβ2.
around the earthβs surface (π = 5.97 Γ 1024 ππ). Take the earthβs diameter as d = 13
000 km and G = 6.67π₯10β11π.π2.ππβ2.
Answers
Answered by
GPT-4o mini
To determine the gravitational force of attraction between the Earth and an aircraft at a given height, we can use Newton's law of universal gravitation, which is given by the formula:
\[
F = \frac{G \cdot M \cdot m}{r^2}
\]
Where:
- \( F \) is the force of attraction,
- \( G \) is the gravitational constant, \( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \),
- \( M \) is the mass of the Earth, \( M = 5.97 \times 10^{24} \, \text{kg} \),
- \( m \) is the mass of the aircraft, \( m = 5300 \, \text{kg} \),
- \( r \) is the distance from the center of the Earth to the aircraft.
First, we need to calculate the distance \( r \) from the center of the Earth to the aircraft. The radius of the Earth can be obtained from its diameter:
\[
\text{Radius of the Earth} = \frac{d}{2} = \frac{13000 \, \text{km}}{2} = 6500 \, \text{km} = 6.5 \times 10^6 \, \text{m}
\]
The aircraft is at a height of \( 2000 \, \text{km} \), which we must convert to meters:
\[
\text{Height} = 2000 \, \text{km} = 2000 \times 10^3 \, \text{m} = 2.0 \times 10^6 \, \text{m}
\]
Now, the total distance \( r \) from the center of the Earth to the aircraft is:
\[
r = \text{Radius of the Earth} + \text{Height} = 6.5 \times 10^6 \, \text{m} + 2.0 \times 10^6 \, \text{m} = 8.5 \times 10^6 \, \text{m}
\]
Now, we can substitute all the values into the gravitational force equation:
\[
F = \frac{(6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2) \cdot (5.97 \times 10^{24} \, \text{kg}) \cdot (5300 \, \text{kg})}{(8.5 \times 10^6 \, \text{m})^2}
\]
Calculating \( (8.5 \times 10^6 \, \text{m})^2 \):
\[
(8.5 \times 10^6)^2 = 72.25 \times 10^{12} = 7.225 \times 10^{13} \, \text{m}^2
\]
Now substituting back into the force equation:
\[
F = \frac{(6.67 \times 10^{-11}) \cdot (5.97 \times 10^{24}) \cdot (5300)}{7.225 \times 10^{13}}
\]
Calculating the numerator:
\[
6.67 \times 10^{-11} \cdot 5.97 \times 10^{24} \cdot 5300 \approx 2.0972381 \times 10^{17} \, \text{N m}^2/\text{kg}
\]
Now performing the final calculation:
\[
F \approx \frac{2.0972381 \times 10^{17}}{7.225 \times 10^{13}} \approx 2.9 \, \text{N}
\]
Therefore, the force of attraction for the aircraft at a height of 2000 km around the Earth's surface is approximately:
\[
\boxed{2.9 \, \text{N}}
\]
\[
F = \frac{G \cdot M \cdot m}{r^2}
\]
Where:
- \( F \) is the force of attraction,
- \( G \) is the gravitational constant, \( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \),
- \( M \) is the mass of the Earth, \( M = 5.97 \times 10^{24} \, \text{kg} \),
- \( m \) is the mass of the aircraft, \( m = 5300 \, \text{kg} \),
- \( r \) is the distance from the center of the Earth to the aircraft.
First, we need to calculate the distance \( r \) from the center of the Earth to the aircraft. The radius of the Earth can be obtained from its diameter:
\[
\text{Radius of the Earth} = \frac{d}{2} = \frac{13000 \, \text{km}}{2} = 6500 \, \text{km} = 6.5 \times 10^6 \, \text{m}
\]
The aircraft is at a height of \( 2000 \, \text{km} \), which we must convert to meters:
\[
\text{Height} = 2000 \, \text{km} = 2000 \times 10^3 \, \text{m} = 2.0 \times 10^6 \, \text{m}
\]
Now, the total distance \( r \) from the center of the Earth to the aircraft is:
\[
r = \text{Radius of the Earth} + \text{Height} = 6.5 \times 10^6 \, \text{m} + 2.0 \times 10^6 \, \text{m} = 8.5 \times 10^6 \, \text{m}
\]
Now, we can substitute all the values into the gravitational force equation:
\[
F = \frac{(6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2) \cdot (5.97 \times 10^{24} \, \text{kg}) \cdot (5300 \, \text{kg})}{(8.5 \times 10^6 \, \text{m})^2}
\]
Calculating \( (8.5 \times 10^6 \, \text{m})^2 \):
\[
(8.5 \times 10^6)^2 = 72.25 \times 10^{12} = 7.225 \times 10^{13} \, \text{m}^2
\]
Now substituting back into the force equation:
\[
F = \frac{(6.67 \times 10^{-11}) \cdot (5.97 \times 10^{24}) \cdot (5300)}{7.225 \times 10^{13}}
\]
Calculating the numerator:
\[
6.67 \times 10^{-11} \cdot 5.97 \times 10^{24} \cdot 5300 \approx 2.0972381 \times 10^{17} \, \text{N m}^2/\text{kg}
\]
Now performing the final calculation:
\[
F \approx \frac{2.0972381 \times 10^{17}}{7.225 \times 10^{13}} \approx 2.9 \, \text{N}
\]
Therefore, the force of attraction for the aircraft at a height of 2000 km around the Earth's surface is approximately:
\[
\boxed{2.9 \, \text{N}}
\]