Question
Scientists are observing and recording data for two different planets, planet X and planet Y. Each planet orbits a different star. The planets have the same mass. The scientists record the following data.
Star X – Planet X Star Y – Planet Y
Distance (millions of kilometers) 100 100
Orbital period of the planet
(Earth days) 390 260
Which of the following statements is most likely true?
A.
The star close to planet Y has more mass.
B.
The star close to planet Y has less mass.
C.
The star close to planet Y is cooler.
D.
The star close to planet Y is hotter.
Star X – Planet X Star Y – Planet Y
Distance (millions of kilometers) 100 100
Orbital period of the planet
(Earth days) 390 260
Which of the following statements is most likely true?
A.
The star close to planet Y has more mass.
B.
The star close to planet Y has less mass.
C.
The star close to planet Y is cooler.
D.
The star close to planet Y is hotter.
Answers
GPT-4o mini
To determine which statement is most likely true, we can apply Kepler's Third Law of planetary motion, which states that the square of the orbital period (T) of a planet is directly proportional to the cube of the semi-major axis (distance) of its orbit (a). In a simplified form, for two planets orbiting different stars, we can express this as:
\[ T^2 \propto a^3 \]
Given that both planets (X and Y) are at the same distance from their respective stars and that a is constant at 100 million kilometers for both, we can say that their orbital periods should be related to the mass of their respective stars (M). Kepler's Third Law can be rewritten to express the relationship between orbital period and star mass:
\[ T^2 \propto \frac{a^3}{M} \]
Since they are both at the same distance (and we assume a spherical star system), we can conclude that:
\[ \frac{T_X^2}{T_Y^2} = \frac{M_Y}{M_X} \]
Given the orbital periods:
- Planet X has an orbital period of 390 days.
- Planet Y has an orbital period of 260 days.
Calculating the squares of these periods:
- \( T_X^2 = 390^2 = 152100 \)
- \( T_Y^2 = 260^2 = 67600 \)
Now we can set up a proportion based on the mass of the stars:
\[ \frac{152100}{67600} = \frac{M_Y}{M_X} \]
Now, we can calculate this ratio:
\[ \frac{152100}{67600} \approx 2.25 \]
This means that:
\[ M_Y \approx 2.25 \times M_X \]
Thus, Star Y has a greater mass than Star X, which means that Star Y is likely denser and more massive than Star X. The implications for temperature (since more massive stars tend to be hotter) lead us to conclude that:
- Star Y, being more massive, is likely hotter than Star X.
Thus, the most likely true statement is:
**D. The star close to planet Y is hotter.**
\[ T^2 \propto a^3 \]
Given that both planets (X and Y) are at the same distance from their respective stars and that a is constant at 100 million kilometers for both, we can say that their orbital periods should be related to the mass of their respective stars (M). Kepler's Third Law can be rewritten to express the relationship between orbital period and star mass:
\[ T^2 \propto \frac{a^3}{M} \]
Since they are both at the same distance (and we assume a spherical star system), we can conclude that:
\[ \frac{T_X^2}{T_Y^2} = \frac{M_Y}{M_X} \]
Given the orbital periods:
- Planet X has an orbital period of 390 days.
- Planet Y has an orbital period of 260 days.
Calculating the squares of these periods:
- \( T_X^2 = 390^2 = 152100 \)
- \( T_Y^2 = 260^2 = 67600 \)
Now we can set up a proportion based on the mass of the stars:
\[ \frac{152100}{67600} = \frac{M_Y}{M_X} \]
Now, we can calculate this ratio:
\[ \frac{152100}{67600} \approx 2.25 \]
This means that:
\[ M_Y \approx 2.25 \times M_X \]
Thus, Star Y has a greater mass than Star X, which means that Star Y is likely denser and more massive than Star X. The implications for temperature (since more massive stars tend to be hotter) lead us to conclude that:
- Star Y, being more massive, is likely hotter than Star X.
Thus, the most likely true statement is:
**D. The star close to planet Y is hotter.**