Answer :
Answer:
Time period for first satellites 24.46 days and for second satellites 37.67 days
Explanation:
Given :
Distance of first satellites [tex]r_{sat1} = 48000 \times 10^{3}[/tex] m
Distance of second satellites [tex]r _{sat2} = 64000 \times 10^{3}[/tex] m
Distance of charon [tex]r_{c} = 19600 \times 10^{3}[/tex] m
Time period of charon [tex]T_{c} = 6.39[/tex] days
From the kepler's third law,
Square of the time period is proportional to the cube of the semi major axis.
[tex]T^{2} = r^{3}[/tex]
[tex]\frac{T}{r^{\frac{3}{2} } } = constant[/tex]
For first satellites,
[tex]\frac{T_{c} }{r_{c} ^{\frac{3}{2} } } = \frac{T_{sat1} }{r_{sat1} ^{\frac{3}{2} } }[/tex]
[tex]{T_{sat1} } = 6.39 \times \frac{(48000 \times 10^{3} )^{\frac{3}{2} } }{(19600\times 10^{3} )^{\frac{3}{2} }}[/tex]
[tex]T_{sat1} = 24.46[/tex] days
For second satellites,
[tex]\frac{T_{c} }{r_{c} ^{\frac{3}{2} } } = \frac{T_{sat2} }{r_{sat2} ^{\frac{3}{2} } }[/tex]
[tex]{T_{sat2} } = 6.39 \times \frac{(64000 \times 10^{3} )^{\frac{3}{2} } }{(19600\times 10^{3} )^{\frac{3}{2} }}[/tex]
[tex]T_{sat2} = 37.67[/tex] days
Therefore, time period for first satellites = 24.46 days and for second satellites 37.67 days