Astronomy and Astrophysics (Other (Not Listed) Sample)

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Astronomy and Astrophysics
Question #1
Given data
Semi major axis of the mars, a=1.5 AU
At this point, we can apply Kepler’s Third Law, which states that the ratio of the squares of the orbital period for two planets is equal to the ratio of the cubes of their mean orbit radius
T2=a3
Where T is the Time period in Earth Years and 'a' is in AU
Substituting the variables
T2=1.53
T2=3.375
T=3.375
T=1.837 Earth years
Question #2
The information given from the question
The mass of Mars, m=6.39×1023kg
The semi-major axis of phobos around Mars, r=23,000 km=23000×103m=23×106m
The gravitational constant G=6.67×10-11m3kg-1s-1
The formula for orbital period of phobos around Mars is , T=2πr3GM
Hence, T=2π(23×106m)3(6.67×10-11m3kg-1s-1)×(6.39×1023kg)
T=2π(23×106)3(6.67×10-11)×(6.39×1023)×m3m3kg-1s-1kg
T=2π12167×109(6.67×6.39)×(10-11×1023)×m3m3kg-1s-1kg
T=2π12167×10942.6213×1012×m3m3kg-1s-1kg
T=2π×16895.78630431496×m3m3kg-1s-1kg
T=106159.35626051784 s
T=106159.35626051784×186400 days
T=1.2286962530152527595 days
T=1.2 days
Question #3
Information provided from the question
Distance between mars and earth = 0.5 AU= 0.5×1.496 ×1011 meters
The speed of light is given, 3×108m/s
Time in seconds =distance in metersspeed in m/s
Time in seconds =0.5×1.496 ×1011m3×108m/s
Time in seconds =0.53×1.496 ×1011-8mm/s
Time in seconds =0.24933 ×103s
Time in seconds ≈250 s
Question #4
Given information
A mirror diameter of the Keck Telescope = 10 m 
Visible wavelengths, λ = 700 nm = 7×10-7m
Find the resolving power
The formula is given by, R.P=D1.22λ
R.P=10 m1.22×7×10-7m
R.P=11.22×7×101--7
R.P=0.11709601874×108
R.P=1.1709601874×107
R.P=1.2×107
At the point that visible