Europa orbits Jupiter at an average distance of 6.71X10^5 km with an orbital per

Question

Europa orbits Jupiter at an average distance of 6.71X10^5 km with an orbital period of 0.00972 yr. The Earth, which is one of the satellites of the Sun orbits its parent at an average distance of 1.50X10^8 km with an orbital period of 1 yr. (a) Use the above information to find the orbital speeds of Europa around Jupiter and of the Earth around the Sun.

a) vEuropa = m/s

b)vEarth =

(c) Now use your answers from parts (a) and (b) to find the ratio of the mass of the Sun to that of Jupiter.

MS/MJ =

Explanation / Answer

(a)

Orbital speed of Europa:

V1 = 2 pi r / T

     = 2 * pi * 6.71x10^5 / ( 0.00972 * 365 * 24 * 60 * 60 )

    = 13.6 km/s

(b)

Orbital speed of Earth:

V2 = 2 pi r / T

     = 2 * pi * 1.50x10^8 / ( 1 * 365 * 24 * 60 * 60 )

    = 29.9 km/s

(c)

Orbital speed:

V = sqrt (G M / r)

V^2 = G M / r

M = V^2 r / G

Ratio of mass of Sun to Jupiter:

M2 / M1 = ( V2 / V1 )^2 * (r2 / r1)

              = ( 29.9 / 13.6 )^2 * ( 1.50x10^8 / 6.71x10^5 )

              =1081

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