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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