whenever two Apollo astronauts were on the surface of the moon, a third astronaut orbited the moon, assume the orbit to be circular and 100 km above the surface of the moon, where the acceleration due to gravity is 1.52 m/s^2. the radius of the moon is 1.70* 10^6 m.(a) determine the astronaut's orbital speed, and (b) the period of the orbit
Orbital speed derives from
Vc = sqrt(µ/r) where µ = the Moon's gravitational constant, 4.9033x10^12m^3/sec.^2 and r = the Moon's radius, 1738km, and the orbital altitude = 1838km = 1,838,000m or
Vc = sqrt(4.9033x10^12/1,828,000m) = 1638m/sec.
The orbital period derives from
T = 2Pisqrt(r^3/µ) = 6.28sqrt[(1,838,000^3/4.9033x10^12]/60 = 118min.
To determine the astronaut's orbital speed and the period of the orbit, we can use the principles of circular motion and the laws of gravity.
(a) To find the astronaut's orbital speed, we can use the formula for the centripetal force:
F = ma
Where F is the gravitational force between the moon and the astronaut, m is the mass of the astronaut, and a is the centripetal acceleration. In this case, the centripetal acceleration is the same as the gravitational acceleration on the surface of the moon, which is given as 1.52 m/s^2.
The gravitational force can be calculated using Newton's law of universal gravitation:
F = G * ((m1 * m2) / r^2)
Where G is the gravitational constant, m1 and m2 are the masses of the moon and the astronaut respectively, and r is the distance between the centers of the moon and the orbiting astronaut.
Since the orbit is assumed to be circular, the centripetal force is provided entirely by the gravitational force. Thus, the formulas can be equated:
F = ma = G * ((m1 * m2) / r^2)
Plugging in the known values:
m * a = G * ((M * m) / r^2)
Here, M is the mass of the moon and m is the mass of the astronaut. Since we are not given the mass of the astronaut, we can cancel it out from both sides:
a = (G * M) / r^2
Substituting the given values:
a = (6.67 * 10^-11 N * m^2 / kg^2 * 7.35 * 10^22 kg) / (1.70 * 10^6 m + 100000 m)^2
Now we can solve for the centripetal acceleration, a:
a = 1.52 m/s^2 (given)
To find the astronaut's orbital speed, we can use the formula for centripetal acceleration in circular motion:
a = (v^2) / r
Now we can solve for v, the orbital speed:
v = sqrt(a * r)
Plugging in the given values:
v = sqrt(1.52 m/s^2 * (1700000 m + 100000 m))
Therefore, the astronaut's orbital speed is the square root of (1.52 m/s^2 multiplied by the sum of the radius of the moon and the altitude of the orbit around the moon).
(b) To find the period of the orbit, we can use the formula for the period of a circular orbit:
T = (2πr) / v
Plugging in the known values:
T = (2π * (1700000 m + 100000 m)) / v
Therefore, the period of the orbit is equal to 2π multiplied by the sum of the radius of the moon and the altitude of the orbit, divided by the astronaut's orbital speed.