Whenever two Apollo astronauts were on the surface of the Moon, a third astronaut orbited the Moon. Assume the orbit to be circular and 400 km above the surface of the Moon. At this altitude, the free-fall acceleration is 1.11 m/s2. The radius of the Moon is 1.70 106 m.
(a) Determine the astronaut's orbital speed. (b) Determine the period of the orbit.
V^2/R = a = 1.11 m/s^2
For R, use the distance from the center of the Moon, 1.70*10^6 m + 4*10^5 m.
Solve for V
For the period P, solve
V*P = 2 pi R, the orbit circumference
To determine the astronaut's orbital speed and the period of the orbit, we can use the formulas related to circular motion and gravity.
(a) To find the astronaut's orbital speed, we can use the equation for centripetal acceleration:
ac = v^2 / r
Where ac is the centripetal acceleration, v is the orbital speed, and r is the radius of the orbit.
We are given that the free-fall acceleration on the lunar surface is 1.11 m/s^2, which is equivalent to the centripetal acceleration. The radius of the orbit is the sum of the radius of the Moon and the altitude of the orbit:
r = Radius of the Moon + Altitude of the orbit
= 1.70 * 10^6 m + 400 * 10^3 m
= 2.10 * 10^6 m
Substituting these values into the equation, we can solve for v:
1.11 m/s^2 = v^2 / (2.10 * 10^6 m)
Rearranging the equation, we have:
v^2 = 1.11 m/s^2 * 2.10 * 10^6 m
v^2 = 2.331 m^2/s^2
Taking the square root of both sides, we find:
v = √(2.331 m^2/s^2)
v ≈ 1.52 km/s
Therefore, the astronaut's orbital speed is approximately 1.52 km/s.
(b) To determine the period of the orbit, we can use the equation:
T = 2πr / v
Where T is the period of the orbit, r is the radius of the orbit, and v is the orbital speed.
Substituting the known values, we have:
T = 2π * (2.10 * 10^6 m) / (1.52 * 10^3 m/s)
T ≈ 8.77 * 10^5 s
Therefore, the period of the orbit is approximately 8.77 * 10^5 seconds.