Answer :
To determine which of the given options best compares satellites X and Y, we need to analyze their orbital periods and tangential speeds based on their distances from Earth.
### Step 1: Analyzing Orbital Periods
The orbital period of a satellite (the time it takes to complete one orbit around Earth) is related to its distance from Earth by Kepler's third law. Kepler's third law states that the square of the orbital period \(T\) is proportional to the cube of the average orbital radius \(r\):
[tex]\[ T^2 \propto r^3 \][/tex]
This means that if we have two satellites at different distances from Earth, the satellite farther from Earth will have a longer orbital period. Let's denote the distances of satellites X and Y from Earth as \( r_X = 1.2 \times 10^5 \, \text{m} \) and \( r_Y = 1.9 \times 10^5 \, \text{m} \) respectively.
Since \( r_X < r_Y \),
[tex]\[ (1.2 \times 10^5)^3 < (1.9 \times 10^5)^3 \][/tex]
This implies that:
[tex]\[ T_X < T_Y \][/tex]
So, satellite X has a shorter orbital period than satellite Y.
### Step 2: Analyzing Tangential Speeds
The tangential speed \(v\) of a satellite in orbit is derived from the balance of gravitational force and centripetal force. It is given by the formula:
[tex]\[ v = \sqrt{\frac{GM}{r}} \][/tex]
where \(G\) is the gravitational constant, \(M\) is the mass of the Earth, and \(r\) is the distance from the Earth.
Given \(v \propto \frac{1}{\sqrt{r}}\), the satellite closer to Earth (with a smaller \(r\)) will have a higher tangential speed. Since the distances are \( r_X = 1.2 \times 10^5 \, \text{m} \) and \( r_Y = 1.9 \times 10^5 \, \text{m} \),
[tex]\[ \frac{1}{\sqrt{1.2 \times 10^5}} > \frac{1}{\sqrt{1.9 \times 10^5}} \][/tex]
Thus,
[tex]\[ v_X > v_Y \][/tex]
So, satellite X has a faster tangential speed than satellite Y.
### Conclusion
Based on the analyses above:
- Satellite \(X\) has a shorter period than Satellite \(Y\).
- Satellite \(X\) has a faster tangential speed than Satellite \(Y\).
The best comparison is:
Satellite [tex]$X$[/tex] has a shorter period and a faster tangential speed than Satellite [tex]$Y$[/tex].
### Step 1: Analyzing Orbital Periods
The orbital period of a satellite (the time it takes to complete one orbit around Earth) is related to its distance from Earth by Kepler's third law. Kepler's third law states that the square of the orbital period \(T\) is proportional to the cube of the average orbital radius \(r\):
[tex]\[ T^2 \propto r^3 \][/tex]
This means that if we have two satellites at different distances from Earth, the satellite farther from Earth will have a longer orbital period. Let's denote the distances of satellites X and Y from Earth as \( r_X = 1.2 \times 10^5 \, \text{m} \) and \( r_Y = 1.9 \times 10^5 \, \text{m} \) respectively.
Since \( r_X < r_Y \),
[tex]\[ (1.2 \times 10^5)^3 < (1.9 \times 10^5)^3 \][/tex]
This implies that:
[tex]\[ T_X < T_Y \][/tex]
So, satellite X has a shorter orbital period than satellite Y.
### Step 2: Analyzing Tangential Speeds
The tangential speed \(v\) of a satellite in orbit is derived from the balance of gravitational force and centripetal force. It is given by the formula:
[tex]\[ v = \sqrt{\frac{GM}{r}} \][/tex]
where \(G\) is the gravitational constant, \(M\) is the mass of the Earth, and \(r\) is the distance from the Earth.
Given \(v \propto \frac{1}{\sqrt{r}}\), the satellite closer to Earth (with a smaller \(r\)) will have a higher tangential speed. Since the distances are \( r_X = 1.2 \times 10^5 \, \text{m} \) and \( r_Y = 1.9 \times 10^5 \, \text{m} \),
[tex]\[ \frac{1}{\sqrt{1.2 \times 10^5}} > \frac{1}{\sqrt{1.9 \times 10^5}} \][/tex]
Thus,
[tex]\[ v_X > v_Y \][/tex]
So, satellite X has a faster tangential speed than satellite Y.
### Conclusion
Based on the analyses above:
- Satellite \(X\) has a shorter period than Satellite \(Y\).
- Satellite \(X\) has a faster tangential speed than Satellite \(Y\).
The best comparison is:
Satellite [tex]$X$[/tex] has a shorter period and a faster tangential speed than Satellite [tex]$Y$[/tex].
