Answer:
a) They are equal.
b) 1 hour and 24 minutes.
Step-by-step explanation:
We are going to solve this exercise applying the third law of Kepler.
Third law of Kepler : ''The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit''
Let be ''T'' the orbital period and ''a'' the semi-major axis ⇒
[tex]\frac{T^{2}}{a^{3}}=K[/tex]
Where K is a constant.
a)
Both planets have the same period because they have equal semi-major axis ⇒ T1 = T2
Being T1 the period of the first planet and T2 the period of the second planet.
b)
Let be ''r'' the Earth's radius
For the Moon :
[tex]\frac{T^{2}}{a^{3}}=K[/tex]
[tex]\frac{(27.3days)^{2}}{(60r)^{3}}=K[/tex]
For the satellite :
[tex]\frac{T^{2}}{r^{3}}=K[/tex]
[tex]K=K[/tex] ⇒
[tex]\frac{(27.3days)^{2}}{(60r)^{3}}=\frac{T^{2}}{r^{3}}[/tex]
[tex]\frac{745.29(days^{2})}{216000(r^{3})}=\frac{T^{2}}{r^{3}}[/tex]
[tex]T^{2}=(3.45041666)(10)^{-3}(days)^{2}[/tex]
[tex]T=0.05874024days[/tex]
[tex]1 day =24h[/tex] ⇒
[tex]0.05874024days=x=1.40976576h[/tex]
[tex]1.40976576h-1h=0.40976576h[/tex]
[tex]1h=60min[/tex]
[tex]0.40976576h=x=24.5859min[/tex]
T ≅ 1 hour and 24 minutes
