imported>Patrick: Created page with 'Io has an orbit of radius $r = 4.22 \times 10^8 m$ and the period $T = 1.77 days = 1.53 \times 10^5 s$ . From the period and the radius, we can find the spee…'

2011-05-27T02:52:13Z

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 <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 speed of Io:

<math>v = {2 \pi r \over T} = {2 \pi (4.22 \times 10^8) \over 1.53 \times 10^5} = 1.73 \times 10^4 m/s</math> 

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

<math>{G M_J m_I \over r^2} = m_I {v^2 \over r}</math>
 
Cancel out <math>r</math> and <math>m_I</math> and substitute known values into the equation.

<math>{(6.67 \times 10^{-11}) M_J \over 4.22 \times 10^8} = (1.73 \times 10^4)^2</math>

From this equation we compute <math>M_J = 1.9 \times 10^{27} kg</math>. 

'''Radius of Europa's orbit'''
 
The period of Europa is <math>T = 3.55 days = 3.07 \times 10^5 s</math>. We substitute the formula for <math>v = {2 \pi r \over T}</math> into the Newton's law

<math>{G M_J m_E \over r^2} = m_E {({2 \pi r \over T})^2 \over r}</math>

and then cancel out the mass of Europa (<math>m_E</math>) and substitute the known values of other constants to obtain:

<math>{(6.67 \times 10^{-11})(1.9 \times 10^{27}) \over r^3} = {4 \pi^2 \over (3.07 \times 10^5)^2}</math>
 
We solve to find <math>r = 6.71 \times 10^8 m</math>.

Newtons Laws EX 35 - Revision history

imported>Patrick: Created page with 'Io has an orbit of radius r = 4.22 \times 10^8 m and the period T = 1.77 days = 1.53 \times 10^5 s. From the period and the radius, we can find the spee…'

imported>Patrick: Created page with 'Io has an orbit of radius $r = 4.22 \times 10^8 m$ and the period $T = 1.77 days = 1.53 \times 10^5 s$ . From the period and the radius, we can find the spee…'