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A distant star has a single planet circling it ina circular orbit of radius 6.65 × 10^11 m. Theperiod of the planet’s motion about the staris 680 days.What is the mass of the star? Thevalue of the universal gravitational constantis 6.67259 × 10^-11 N · m2/kg2.Answer in units of kg

1 Answer

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Answer:

mass of star = 5035.52 x 10^28 kg

Step-by-step explanation:

We would apply Kepler's third law. The equation is expressed as

T^2 = 4π^2r^3/Gm

By cross multiplying,

mT^2 = 4π^2r^3/G

Dividing both sides by T^2, it becomes

m = 4π^2r^3/GT^2

where

T is the time or period of the planet's motion

r is the radius of the orbiting planet

G is the universal gravitational constant

m is the mass of the star

From the information given,

T = 680 days

We would convert it to seconds. Recall,

1 day = 86400 seconds

680 days = 680 x 86400 = 58752000 seconds

G = 6.67259 × 10^-11 N · m2/kg2

r = 6.65 × 1011 m

π = 3.14

By substituting these values into the formula, we have

m = 4 x 3.14^2 x (6.65 × 10^11)^3/6.67259 × 10−11 x 58752000^2

m = 5035.52 x 10^28 kg

mass of star = 5035.52 x 10^28 kg

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