The Apollo 8 mission in 1968 included a circular orbit at an altitude of 111 km above the Moon’s surface. What was the period of this orbit? The Moon’s mass is Mm=7.35E22 kg and its radius is Rm= 1.74E6 m
You need the orbit radius, and that is
R = 1.74*10^6 m + 0.111*10^6 m
= 1.85*10^6 m
Gravity pull = centripetal force
G*Mm*m/R^2 = m*V^2/R
Satellite mass m cancels out.
V = sqrt(G*Mm/R)
Period = 2 pi R/V
Substitute the previous equation for R.
To find the period of the orbit of the Apollo 8 mission around the Moon, we can use Kepler's third law of planetary motion. This law states that the square of the period of an orbit (T) is equal to the cube of the semi-major axis of the orbit (a) divided by the sum of the masses of the objects involved (M + m), where M is the mass of the central body (in this case, the Moon) and m is the mass of the orbiting body (the Apollo 8 spacecraft).
The semi-major axis of a circular orbit is equal to the radius of the orbit. In this case, the radius of the orbit is the sum of the radius of the Moon (Rm) and the altitude of the orbit above the Moon's surface. So, the semi-major axis (a) can be calculated as:
a = Rm + altitude
Given:
Rm = 1.74E6 m (radius of the Moon)
altitude = 111 km = 111000 m
a = 1.74E6 m + 111000 m
a = 1854000 m
Now, let's calculate the period (T):
T^2 = (a^3) / (M + m)
M = 7.35E22 kg (mass of the Moon)
m = mass of the Apollo 8 spacecraft (which we'll assume to be negligible compared to the Moon's mass)
T^2 = (1854000^3) / (7.35E22 + m)
To simplify the equation, we can assume that m is negligible compared to the Moon's mass, so we can neglect it in the denominator.
T^2 ≈ (1854000^3) / (7.35E22)
Now, let's calculate T by taking the square root of both sides of the equation:
T ≈ sqrt((1854000^3) / (7.35E22))
Using a calculator, the approximate value for T is:
T ≈ 4808 seconds
Therefore, the period of the Apollo 8 mission's orbit around the Moon was approximately 4808 seconds.