Whenever two Apollo astronauts were on the surface of the Moon, a third astronaut orbited the Moon. Assume the orbit to be circular and 720 km above the surface of the Moon, where the acceleration due to gravity is 0.839 m/s2. The radius of the Moon is 1.70 106 m.
(a) Determine the astronaut's orbital speed.
m/s
(b) Determine the period of the orbit.
s
To determine the astronaut's orbital speed and period of the orbit, we can use the following formulas:
(a) Orbital speed (V) = √(GM/r)
(b) Period (T) = (2πr) / V
Where:
G is the gravitational constant (6.674 × 10^-11 m^3/kg/s^2),
M is the mass of the Moon (7.35 × 10^22 kg),
r is the orbit radius (720 km + 1.70 × 10^6 m).
Let's calculate the values step by step:
(a) Orbital speed (V):
Convert the orbit radius from km to m: 720 km = 720 × 1000 m = 720,000 m
Find the gravitational acceleration at the orbit radius, g = 0.839 m/s^2
Calculate the orbital speed using the formula:
V = √(GM/r)
V = √((6.674 × 10^-11 m^3/kg/s^2) × (7.35 × 10^22 kg) / (720,000 m + 1.70 × 10^6 m))
(b) Period (T):
Calculate the period using the formula:
T = (2πr) / V
Now, let's substitute the values into the formulas and calculate them:
(a) Orbital speed (V):
V = √((6.674 × 10^-11) × (7.35 × 10^22) / (720,000 + 1.70 × 10^6))
= √((5.423009 × 10^11) / (2.42 × 10^6))
= √(2.240039 × 10^5)
= 473.303 m/s (rounded to three significant figures)
Therefore, the astronaut's orbital speed is approximately 473.303 m/s.
(b) Period (T):
T = (2π × (720,000 + 1.70 × 10^6)) / 473.303
= (2π × 2.42 × 10^6) / 473.303
= 16154.418 s (rounded to three significant figures)
Therefore, the period of the orbit is approximately 16154.418 seconds.