You are the science officer on a visit to a distant solar system. Prior to landing on a planet you measure its diameter to be 1.8×(10^7) m and its rotation period to be 22.3 hours. You have previously determined that the planet orbits 2.2×(10^11) m from its star with a period of 432 earth days. Once on the surface you find that the free-fall acceleration is 12.2 m/s^2.What is the mass of (a) the planet and (b) the star?

Mp= 1.48×10^23 Kg and M = 4.47×10^30 Kg

Explanation:

Given that

Diameter of planet D = 1.8×10^6m

Radius of planet Rp = 0.9×10^6m

Period of rotation of planet = 22.3 hrs = 80280s

Radius of orbit r = 2.2 × 10^11 m

Period of revolution around star T =432days = 432×24×60×60 = 37324800s

Acceleration of gravity on the surface of planet gp = 12.2m/s^2

gp = GMp/(Rp)^2

Mp = gp×(Rp)^2 /G

= 12.2 * (0.9*10^6)^2 ÷ 6.67×10^-11

9.882×10^12 ÷ 6.67×10^-11

Mp= 1.4×10^23 Kg

To determine the mass of the star, we consider the revolution of the planet around the star with period T

T^2 = (4π^2/GM)r^3

M = 4π^2r^3 ÷ GT^2

M = 4π^2* (2.2×10^11)^3 ÷ 6.67×10^-11 × ( 37324800)^2

M= 4.47×10^30 Kg

50 inches should be the resultant vector

terd muffin is the correct answer