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Main Menu Contents Gla ??Course Contents » O Timer ?Notes d Evaluate Feedback-Print (m1nfo » Asteroid orbits the Sun Due in 3 hours, 10 minu An asteroid, whose mass is 4.30x10 times the mass of Earth, revolves in a circular orbit around the Sun at a distance that is 4 times the Earth's distance from the Sun. Calculate the period of revolution of the asteroid Submit Answer You have entered thatSubmission not graded. Use Tries 0/10 Previous answer before more digits. What is the ratio of the kinetic energy of the asteroid to the kinetic energy of Earth? 15x1044 Submit Answer Inco Post Discussion rrect. Tries 1/10 Previous Tries Send Feedback O Type here to search os W F3 F4 · F5 Fe

Explanation / Answer

1] Period of orbit

The orbit variables can be determined from Newtons laws,

GM_sun*m/R^2 = m*v^2/R

GM_sun/R = v^2 = (2*pi*R/T)^2

T^2 = (2*pi)^2*R^3/GM_sun

T = K*R^(3/2)

Tasteriod/Tearth = Rasteroid/Rearth

Tasteroid = 4^(3/2)*1 = 8 years

2] KEasteriod/KEearth = 0.5*Ma*Va^2 / (0.5MeVe^2)

= Ma*(wa*Ra)^2/ [Me*(we*Re)^2]

= 4.30^10^-4*Me*(2*pi/Ta)^2*(4*Re)^2 /[Me*(2*pi/Te)^2*Re^2]

= 4.30*10^-4*16*(1/8)^2

= 1.075*10^-4

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