A planet of mass m = 4.75 × 1024 kg orbits a star of mass M = 2.85 × 1029 kg in a circular path. The radius of the orbit is R = 7.35 x 10’ km. What is the orbital period Tplanet of the planet in Earth days? days Tplanet
A planet of mass m = 4.75 × 1024 kg orbits a star of mass M = 2.85 × 1029 kg in a circular path. The radius of the orbit is R = 7.35 x 10’ km. What is the orbital period Tplanet of the planet in Earth days? days Tplanet
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![### Orbital Period Calculation
#### Problem Statement:
A planet of mass \( m = 4.75 \times 10^{24} \) kg orbits a star of mass \( M = 2.85 \times 10^{29} \) kg in a circular path. The radius of the orbit is \( R = 7.35 \times 10^{7} \) km.
#### Question:
What is the orbital period \( T_{\text{planet}} \) of the planet in Earth days?
\[ T_{\text{planet}} = \ \_\_\_\_\_\_\_\_\_\_ \]
[Text Box - T_{\text{planet}} in days]
#### Explanation:
To determine the orbital period of the planet, the following formula derived from Kepler's third law of planetary motion can be used:
\[ T_{\text{planet}} = 2\pi \sqrt{\frac{R^3}{G(M + m)}} \]
- \( T_{\text{planet}} \) is the orbital period.
- \( R \) is the radius of the orbit.
- \( G \) is the gravitational constant (\( G \approx 6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \)).
- \( M \) is the mass of the star.
- \( m \) is the mass of the planet.
However, since often \( M \gg m \), we can approximate the formula as:
\[ T_{\text{planet}} \approx 2\pi \sqrt{\frac{R^3}{GM}} \]
Make sure to convert all units consistently (e.g., converting \( R \) from kilometers to meters).
After calculation, the orbital period \( T_{\text{planet}} \) should be converted from seconds to Earth days by:
\[ 1 \, \text{day} = 24 \times 3600 \, \text{seconds} \]
This will yield the result in Earth days.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5039f9f1-9031-4cf0-b840-7355f91f2cd2%2F59a4f971-275b-4262-b141-10176cd5233a%2Fyqrxnhh.png&w=3840&q=75)
Transcribed Image Text:### Orbital Period Calculation
#### Problem Statement:
A planet of mass \( m = 4.75 \times 10^{24} \) kg orbits a star of mass \( M = 2.85 \times 10^{29} \) kg in a circular path. The radius of the orbit is \( R = 7.35 \times 10^{7} \) km.
#### Question:
What is the orbital period \( T_{\text{planet}} \) of the planet in Earth days?
\[ T_{\text{planet}} = \ \_\_\_\_\_\_\_\_\_\_ \]
[Text Box - T_{\text{planet}} in days]
#### Explanation:
To determine the orbital period of the planet, the following formula derived from Kepler's third law of planetary motion can be used:
\[ T_{\text{planet}} = 2\pi \sqrt{\frac{R^3}{G(M + m)}} \]
- \( T_{\text{planet}} \) is the orbital period.
- \( R \) is the radius of the orbit.
- \( G \) is the gravitational constant (\( G \approx 6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \)).
- \( M \) is the mass of the star.
- \( m \) is the mass of the planet.
However, since often \( M \gg m \), we can approximate the formula as:
\[ T_{\text{planet}} \approx 2\pi \sqrt{\frac{R^3}{GM}} \]
Make sure to convert all units consistently (e.g., converting \( R \) from kilometers to meters).
After calculation, the orbital period \( T_{\text{planet}} \) should be converted from seconds to Earth days by:
\[ 1 \, \text{day} = 24 \times 3600 \, \text{seconds} \]
This will yield the result in Earth days.
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