The following two questions (12 and 13) pertain to you, the science officer on a

Question

The following two questions (12 and 13) pertain to you, 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 times 10^7m. You have previously determined that the planet orbits 2.9 times 10^11m from its star with a period of 402 earth days. Once on the surface you find that the acceleration due to gravity is 19.5 m/s^2. What is the mass of the star? 1.2 times 10^28kg 1.2 times 10^31kg 2.4 times 10^28kg 2.4 times 10^31kg 0.6 times 10^28kg What is the mass of the planet? 0.6 times 10^25kg 1.2 times 10^22kg 1.2 times 10^25kg 2.4 times 10^25kg 2.4 times 10^22kg

Explanation / Answer

Given data
diameter = 1.8*10^7 m
period = 402 earth days
planet orbits = 2.9*10^11 m
g = 19.5 m/s^2

we can find the mass of the planet from g=GM/R^2 where G is newtonian grav cst, M the mass of the planet, R the radius and g thelocal accel due to grav

What is the mass of the planet?
What is the mass of the star?

solving for M:

M=gR^2/G=(19.5*(9*10^6)^2)/(6.67*10^-11)
M=2.4*10^25 kg--------------Mass of the Planet

find the mass of the star from Kepler's 3rd law

equate grav force to centripetal force

GMm/r^2=mv^2/r
M=mass of star
m=mass planet
r=size of orbit
v=orbital speed

this becomes M=v^2r/G
v=circumference of orbit/time = 2 pi r/time
v=2 pi * 2.9*10^11/402days * 86,400sec/day
v = 52434.58 m/s
M= (( 52434.58)^2*(2.9*10^11))/(6.67*10^-11)
M=1.2*10^31 kg -------------Mass of the star

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