Question 9A newly discovered planet, Eagal, has two known moons, Dorainn and Bron. The table belowgives some of the known information about Dorainn and Bron.DorainnBronMass1.427 x 1015 kg1.427 x 1015 kgRadius6.142 kmOrbital Period1.091 x 105 sOrbital Radius23,640 km27,650 kma. Determine the gravitational field strength at the surface of Dorainn.b. Using only the information provided for each moon and proportionality, determineBron's orbital period. Clearly show or explain all workings.c. Use the data provided for Dorainn to determine the mass of Eagal.d. Determine the orbital speed of Bron.

Given data: Mass of each moon: m=1.427×1015kg Radius of Dorainn, rD=6.142 km=6.142×103m Orbital…

Answered: Question 9 A newly discovered planet,…

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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Expert Solution

Step 1

Given data:

Mass of each moon: $m = 1.427 \times 10^{15} kg$

Radius of Dorainn, $r_{D} = 6.142 km = 6.142 \times 10^{3} m$

Orbital period of Dorainn, $T_{D} = 1.091 \times 10^{5} s$

Orbital radius: $R_{D} = 23640 \times 10^{3} m; R_{B} = 27650 \times 10^{3} m$

Step 2

To determine:

Gravitational field strength at the surface of Dorainn.
Bron's orbital period.
Mass of Eagal.
Orbital speed of Bron.

Step 3

Explanation:

a. Gravitational field strength can be calculated as follows:

$g_{D} = \frac{G m}{r_{D}^{2}} g_{D} = \frac{(6.67 \times 10^{- 11} N . m^{2} / {kg}^{2}) (1.427 \times 10^{15} kg)}{{(6.142 \times 10^{3} m)}^{2}} g_{D} = 2.52 \times 10^{- 3} m / s^{2}$

b. Using Kepler's third law, the orbital speed of Bron can be calculated as follows:

$T^{2} \propto R^{3} {(\frac{T_{D}}{T_{B}})}^{2} = {(\frac{R_{D}}{R_{B}})}^{3} T_{D}^{2} R_{B}^{3} = T_{B}^{2} R_{D}^{3} T_{B} = \sqrt{\frac{T_{D}^{2} R_{B}^{3}}{R_{D}^{3}}} T_{B} = \sqrt{\frac{{(1.091 \times 10^{5} s)}^{2} {(27650 \times 10^{3} m)}^{3}}{{(27640 \times 10^{3} m)}^{3}}} T_{B} = 1.092 \times 10^{5} s$