Planet K2-116b has an Average orbital radius of 7.18x10^9 m around the star K2-116. It has a mass of about 0.257 times the mass of the earth and an orbital period of 2.7 days. What is the orbital speed of the planet?
Determine the mass of the star.

a) v = 1.94 × 10^5 m/s

b) Ms = 2.09 × 10^24 kg

Explanation:

Given:

m = 0.257M (M = mass of earth = 5.972×10^24 kg)

= 1.535×10^24 kg

r = 7.18×10^9 m

T = 2.7 days × (24 hr/1 day) × (3600 s/1 hr)

= 2.3328×10^5 s

a) To find the orbital speed of the planet, we need to find the circumference of the planet's orbit first:

C = 2×(pi)×r

= 2(3.14)(7.18×10^9m)

= 4.51×10^10 m

The orbital speed v is then given by

v = C/T

= (4.51×10^10 m)/(2.33×10^5 s)

= 1.94 × 10^5 m/s

b) We know that centripetal force Fc is given by

Fc = mv^2/r

where v = orbital speed

r = average orbital radius

m = mass of planet

We also know that the gravitational force FG between the star K2-116 and the planet is given by

FG = GmMs/r^2

where m = mass of planet

Ms = mass of star K2-116

r. = average orbital radius

G = universal gravitational constant

= 6.67 × 10^-11 m^3/kg-s^2

Equating Fc and FG together, we get

Fc = FG

mv^2/r = GmMs/r^2

Note that m and one of the r's get cancelled out so we are left with

v^2 = GMs/r

Solving for the mass of the star Ms, we get

Ms = rv^2/G

=(7.18 × 10^9 m)(1.94 × 10^5 m/s)^2/(6.67 × 10^-11 m^3/kg-^2)

= 2.09 × 10^24 kg

force tends to be low and long diustance masses are high

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