44% i) Sun Now 12 10:08 AM Halliday, Fundamentals of Physics, 10e PHYS PRI OF NA

Question

44% i) Sun Now 12 10:08 AM Halliday, Fundamentals of Physics, 10e PHYS PRI OF NANO SCIENCE 1 (NENG1 Assignment Gradebook ORION Downloadable eTextbook nt MESSAGE MY INSTRUCTOR FULL SCREEN PRINTER VERSION BACK Chapter 13, Problem 048 The mean distance of an asteroid from the Sun is 3.06 times that of Earth from the Sun. From Kepler's law of periods, calculate the number of years required for the asteroid to make one revolution around the Sun. Number Units the tolerance is +/-2% LINK TO TEXT LINK TO SAMPLE PROBLEM MATH HELP By accessing this Question Assistance, you wil learn while you earn points based on the Point Potential Policy set by your instructor Question Attempts: 0 of 5 used SAVE FOR LATER SUBMIT ANSWER Earn Maximum Points available only if you answer this question correctly in three attempts or less.

Explanation / Answer

The distance from the earth to the sun is 149.6 * 10^9 m

the radius of the satellite is r = 3.06 * 149.6 * 10^9 m

the mass of the sun is M = 1.989 * 10^30 kg

The time period of asteroid is

T^2 = (4 * pi^2 * r^3) / (G * M)

T^2 = (4 * pi^2 * (3.06 * 149.6 * 10^9)^3) / (6.67 * 10^-11 * 1.989 * 10^30)

T = 168958128.3155 s

we know 1 s = 1/60 min

1min = 1/60 hr

1hr = 1/24 days

1day = 1/365 years

in years T = 168958128.3155 / (60 * 60 * 24 * 365)

T = 5.357 years

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