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

in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.

the answer is a 3.8cm

explanation: