Physic's qu on circular orbits
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Physic's qu on circular orbits

[From: ] [author: ] [Date: 11-11-01] [Hit: ]
dist = 2 pi r = speed x time = 39000m/s x 7.05yrs x 3.r = 1.therefore,M = v^2 r/G = 39000^2 x 1.38x10^12/6.......
Two newly discovered planets follow circular orbits around a star in a distant part of the galaxy. The orbital speeds of the planets are determined to be 39.0 km/s and 63.4 km/s. The slower planet's orbital period is 7.05 years. (a) What is the mass of the star? (b) What is the orbital period of the faster planet, in years?
Really not sure how to do this, thanks!

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we will equate centripetal and gravitational forces to find the mass of the star

m v^2/r = GMm/r^2

or M = v^2 r/G

m, M are the mass of the planet, star
r=distance from star

we know velocity and period, so can find r from

dist = 2 pi r = speed x time = 39000m/s x 7.05yrs x 3.15x10^7s/yr

r = 1.38x10^12 m

therefore,

M = v^2 r/G = 39000^2 x 1.38x10^12/6.67x10^-11 = 3.14x10^31 kg (about 16 solar masses)

knowing v for the second planet and the mass of the star, find its distance from the star from

v^2/r =GM/r^2 or

r = GM/v^2 = 6.67x10^-11x 3.14 x 10^31/(63400)^2

r = 5.21x10^11m

period = 2 pir/v = 5.17x10^7 s = 1.64yrs

(you could also get this result using the data in conjuction with Kepler's third law

we know from the problem that the ratios of velocities is 63.4/39=1.63; we know from equating forces that r from the star varies as 1/v^2:

GM/r^2 = v^2/r => r =GM/v^2

so if the velocities vary by this ratio, the distances vary by 1/1.63^2 or 0.378

kepler's third law tells us that P^2 = r^3 or P=r^(3/2)

therefore the periods have the ratio of (0..78)^(3/2) = 0.233

so if the slower planet has a period of 6.6yrs, the faster planet's period is 7.05yrs*0.233 = 1.64yrs)
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