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Sagot :
Sure, let's tackle each part of this question step-by-step:
### (a) Calculate the force of attraction between the star and the Earth
Given data:
- Mass of the star [tex]\( M = 2.00 \times 10^{30} \, \text{kg} \)[/tex]
- Mass of the Earth [tex]\( m = 5.98 \times 10^{24} \, \text{kg} \)[/tex]
- Distance between the star and the Earth [tex]\( r = 1.50 \times 10^8 \, \text{km} \)[/tex]
- Gravitational constant [tex]\( G = 6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \)[/tex]
First, convert the distance from kilometers to meters:
[tex]\[ r = 1.50 \times 10^8 \, \text{km} = 1.50 \times 10^8 \times 10^3 \, \text{m} = 1.50 \times 10^{11} \, \text{m} \][/tex]
Newton's Law of Universal Gravitation states that the force [tex]\( F \)[/tex] between two masses is given by:
[tex]\[ F = G \frac{Mm}{r^2} \][/tex]
Substituting the given values:
[tex]\[ F = 6.67 \times 10^{-11} \, \frac{(2.00 \times 10^{30}) \times (5.98 \times 10^{24})}{(1.50 \times 10^{11})^2} \][/tex]
The computed force of attraction is:
[tex]\[ F \approx 3.545475555555556 \times 10^{22} \, \text{N} \][/tex]
### (b) Define Escape Velocity
Escape velocity is the minimum velocity an object must have to break free from the gravitational attraction of a massive body without further propulsion. In other words, it is the speed needed for an object to leave a planet or other body in space, overcoming its gravitational pull.
### (c) Differences between acceleration due to free fall ' g ' and Universal gravitation constant 'G'
1. Nature and Value:
- [tex]\( g \)[/tex]: It is the acceleration due to gravity experienced by an object close to the surface of a massive body, typically Earth. Its value is approximately [tex]\( 9.8 \, \text{m/s}^2 \)[/tex] on Earth (often rounded to [tex]\( 10 \, \text{m/s}^2 \)[/tex] for simplicity).
- [tex]\( G \)[/tex]: It is the universal gravitational constant, which is a proportionality factor in Newton's law of universal gravitation. Its value is approximately [tex]\( 6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \)[/tex].
2. Dependence:
- [tex]\( g \)[/tex]: The value of [tex]\( g \)[/tex] is dependent on the mass of the Earth (or other celestial bodies) and its radius. It varies slightly depending on altitude and geographical location.
- [tex]\( G \)[/tex]: The value of [tex]\( G \)[/tex] is a universal constant and does not change regardless of location or the masses involved. It is a fundamental constant of nature.
### (a) Calculate the force of attraction between the star and the Earth
Given data:
- Mass of the star [tex]\( M = 2.00 \times 10^{30} \, \text{kg} \)[/tex]
- Mass of the Earth [tex]\( m = 5.98 \times 10^{24} \, \text{kg} \)[/tex]
- Distance between the star and the Earth [tex]\( r = 1.50 \times 10^8 \, \text{km} \)[/tex]
- Gravitational constant [tex]\( G = 6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \)[/tex]
First, convert the distance from kilometers to meters:
[tex]\[ r = 1.50 \times 10^8 \, \text{km} = 1.50 \times 10^8 \times 10^3 \, \text{m} = 1.50 \times 10^{11} \, \text{m} \][/tex]
Newton's Law of Universal Gravitation states that the force [tex]\( F \)[/tex] between two masses is given by:
[tex]\[ F = G \frac{Mm}{r^2} \][/tex]
Substituting the given values:
[tex]\[ F = 6.67 \times 10^{-11} \, \frac{(2.00 \times 10^{30}) \times (5.98 \times 10^{24})}{(1.50 \times 10^{11})^2} \][/tex]
The computed force of attraction is:
[tex]\[ F \approx 3.545475555555556 \times 10^{22} \, \text{N} \][/tex]
### (b) Define Escape Velocity
Escape velocity is the minimum velocity an object must have to break free from the gravitational attraction of a massive body without further propulsion. In other words, it is the speed needed for an object to leave a planet or other body in space, overcoming its gravitational pull.
### (c) Differences between acceleration due to free fall ' g ' and Universal gravitation constant 'G'
1. Nature and Value:
- [tex]\( g \)[/tex]: It is the acceleration due to gravity experienced by an object close to the surface of a massive body, typically Earth. Its value is approximately [tex]\( 9.8 \, \text{m/s}^2 \)[/tex] on Earth (often rounded to [tex]\( 10 \, \text{m/s}^2 \)[/tex] for simplicity).
- [tex]\( G \)[/tex]: It is the universal gravitational constant, which is a proportionality factor in Newton's law of universal gravitation. Its value is approximately [tex]\( 6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \)[/tex].
2. Dependence:
- [tex]\( g \)[/tex]: The value of [tex]\( g \)[/tex] is dependent on the mass of the Earth (or other celestial bodies) and its radius. It varies slightly depending on altitude and geographical location.
- [tex]\( G \)[/tex]: The value of [tex]\( G \)[/tex] is a universal constant and does not change regardless of location or the masses involved. It is a fundamental constant of nature.
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