### 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.

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

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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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.

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