part a and b Class Management I Help HW: Kepler\'s Laws Begin Date: 3/212018 12:

Question

part a and b

Class Management I Help HW: Kepler's Laws Begin Date: 3/212018 12:01:00 AM Due Date: 43 2018 11:59:00 PM End Date: 4 10/2018 11:59:00 PM (13%) Problem 5: Astronomical observations of our Milky Way galaxy indicate that it has a mass of about 80-1011 solar masses. A star orbiting on the galaxy's periphery is about 6.0 . 10 light years from its center Assignment Status Click here for detailed view 50% Part (a) what should the orbital perud of that star be in years, Grade Summary 5% 95 Problem Status potential 1 Completed LD cos tan) 78 9 O Partial 3 Completed 4 Completed atan) acota) sinho tanh cotanh Attermpts remaining 4 (5% per attempt) detailed view cosh( Partial Degrees Radians 10. BACKSPACE -DELI CLEAR 7 Partial Hmt Hints: 0-deduction perhint Hutiretaining Feedback: 2 deduction per fedbuck Sg 50% Part (b) If its period is 60+ 10 existence of "dark matter" in the universe years instead, what is the nass of the galaxy in ? Such calculations are used to imply the and have indicated, for example, the existence of very massive black holes at the centers of some type here to search 612 PM /31/2018

Explanation / Answer

a) T = sqrt ( 4*pi2*r3 / GM) ---------- 1

T = 2*pi *sqrt ( r3 / GM)

T = 2*pi*sqrt ( (6e4*3e8*365.25*24*3600)3 / 6.67e-11*8e11*1.99e30)

Now in the above equation, I have changes the orbital distance from light years to m and mass from solar masses to mass.

T = 2*pi*sqrt ((5.6803e20)3 / 1.0619e32)

T = 2*pi*1.31375e15

T = 8.25E15 seconds

or, T = 8.3 * 1015 seconds

b) Just use the formula as above ( marked as 1)

M = sqrt ( 4*pi2*r3 / GT2)

squaring both sides,

M = sqrt ( 4*pi2*(5.6803e20)3 / 6.67e-11*(6e7*365.25*25*3600)2

M = sqrt ( 7.2356e63 / 2.39e20

M = 3.025e49

However, we need to divide it by 1 solar mass ,

so m = M / 1.99E30

m = 3.025e49 / 1.99e30

m = 1.52*1019 solar masses

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