# Given: G=6.67259 x 10^-11 N m^2/kg^2 Mimas, a moon of Saturn, has an orbital radius of 1.88 x 10^8 m and an orbital period of about 23.48 h.Use Newton's version of Kepler's third law and these data...

Given: G=6.67259 x 10^-11 N m^2/kg^2

Mimas, a moon of Saturn, has an orbital radius of 1.88 x 10^8 m and an orbital period of about 23.48 h.

Use Newton's version of Kepler's third law and these data to find the mass of Saturn. Answer in units of kg.

Asked on by hrrjack

### 3 Answers |Add Yours

enotechris | College Teacher | (Level 2) Senior Educator

Posted on

Newton realized that two bodies orbit around a common center of mass, rather than one round the other. He therefore modified Kepler's 3rd Law to:

(m1 + m2) P^2 = (d1 + d2)^3 = R^3

where R is the total distance between the 2 centers of mass, and P is the orbital period, d is the distance between an object and the center of mass, and m is mass of an object.

Having the mass of Mimas along with the other data would be helpful.....

neela | High School Teacher | (Level 3) Valedictorian

Posted on

The Newton Kepler's Law of gravitation gives the relation of gravitational force and centrepetal force between planet and its satellite:

GMm/R^2 = mv^2/R^2 , where , M and m are the masses of planet and its satellite respectively. R is the distance between the centre of the planet and centre of its satellite and v is the velocity of the satellite.Replacing v by 2pR/T,( where T is the period of the satellite), we get M, the mass of the plannet:

M = (2P)^2* R^3/(G*T^2). ..........................(1)

By data, R = 1.88 x 10^8 m, distance between centres of Saturn and Mima .

G=6.67259 x 10^-11 N m^2 and

T = 23.48 hours = 23.48*60*60 seconds = 84528 seconds, is given to be the period of Mima, the moon of the Saturn.

Substituting the values in (1), we get:

M = 5.50221628*10^26 Kg is the estimated mass of the Saturn from this problem.

We’ve answered 317,697 questions. We can answer yours, too.