### 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)...

See similar textbooks

Similar questions

Planet X orbits the star Omega with a "year" that is 450 earth days long. Planet Y circles Omega at four times the distance of Planet X. How long is a year on Planet Y? Express your answer in earth days.
B8
What is the radius of geosynchronous orbit around the earth? (where the period of orbit is the same as the earths rotational period) Using the mass of the earth as me = 5.98 x 1024 kg and the sidereal day as 23 hours 56 minutes and 4.0905 s. Number Units
The International Space Station is in a 210-mile-high orbit. What is the station's orbital speed? The radius of Earth is 6.37×10^6 m, its mass is 5.98×10^24kg. Express your answer with the appropriate units. What is the station's orbital period? Express your answer in minutes.
Please Aspa
An intergalactic spaceship arrives at a distant planet that rotates on its axis with a period of T = 31 hours. The mass of the planet is M = 2.1 • 1025 kg. The spaceship enters a circular orbit with an orbital period that is equal to the planet's period for the rotation about its axis, T. a. Enter an expression for the radius of the spaceship’s orbit. b. Calculate the orbital radius in meters.
Assume that planet A has radius RA and mass MA planet B has radius Rg and mass Mg. Let gA and g; are the accelerations due to gravity at their surfaces, respectively. If MA = Mg, and RA = 2Rg The ratio (ga/gB) is: and !! %3D А. 1 В. 2 C. 4 D. 0.5 E. 0.25
A landing craft with mass M is in a circular orbit a distance d above the surface of a planet. The period ofthe orbit is T. The astronauts in the landing craft measure the diameter of the planet to be D. The landing craft sets down at the north pole of the planet. a)What is the weight of a person of mass m as they step out onto the plant’s surface? b)Suppose days on this planet last t seconds (i.e. the planet rotates about its axis once every t seconds).Write an expression for the astronaut’s perceived weight at the equator in terms of their weight at the north pole. (Hint: think about centripetal force)
The gas-giant planet Rom (mass 5.7⨯1026 kg) goes around the star Galla (mass 2.0⨯1030 kg) in a circular orbit. If the orbital radius of Rom is 2.5×1011 m … Note: G = 6.67⨯10-11 N·m2/kg2 i) What is the gravitational force of Rom on Galla?
An undiscovered planet, many lightyears from Earth, has one moon in a periodic orbit. This moon takes 20.0 days on average to complete one nearly circular revolution around the unnamed planet. If the distance from the center of the moon to the surface of the planet is 2.420 × 10° km and the planet has a radius of 4.100 × 10³ km, calculate the moon's radial acceleration a. a = m/s? & TOOLS x10
The moon's period of revolution around the Earth is 27.3 d. How far away is the moon from the surface of the Earth? O A) 1.33× 107 m B) 3.83 × 108 m C) 3.77 × 108 m D) 4.88 × 10⁹ m
1