# Astronomy and Astrophysics (Other (Not Listed) Sample)

Astronomy and Astrophysics

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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

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