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.

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enotechris | College Teacher | (Level 2) Senior Educator

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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.....

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neela | High School Teacher | (Level 3) Valedictorian

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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.

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