Newtons Laws EX 35

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Revision as of 02:52, 27 May 2011 by imported>Patrick (Created page with 'Io has an orbit of radius <math>r = 4.22 \times 10^8 m</math> and the period <math>T = 1.77 days = 1.53 \times 10^5 s</math>. From the period and the radius, we can find the spee…')
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Io has an orbit of radius r=4.22×108m and the period T=1.77days=1.53×105s. From the period and the radius, we can find the speed of Io:

v=2πrT=2π(4.22×108)1.53×105=1.73×104m/s

Mass of Jupiter

Then we can use the Newton's second law to find the mass of Jupiter:

GMJmIr2=mIv2r

Cancel out r and mI and substitute known values into the equation.

(6.67×1011)MJ4.22×108=(1.73×104)2

From this equation we compute MJ=1.9×1027kg.

Radius of Europa's orbit

The period of Europa is T=3.55days=3.07×105s. We substitute the formula for v=2πrT into the Newton's law

GMJmEr2=mE(2πrT)2r

and then cancel out the mass of Europa (mE) and substitute the known values of other constants to obtain:

(6.67×1011)(1.9×1027)r3=4π2(3.07×105)2

We solve to find r=6.71×108m.