On August 12, 2005 the Mars Reconnaissance Orbiter (MRO) was launched from Earth and reached Mars on March 10, 2006. The orbiter is in roughly circular orbit about 390 km above the martian surface. Mars has a radius of 3396 km and a mass of 0.642 x 1024 kg (6.42E23 kg).
b) What is the radius of the MRO orbit?
m
c) What is the linear speed (v) of the orbiter above Mars?
m/s
d) What is the period of the orbiter in hours?
h
To calculate the radius of the MRO orbit, we can subtract the radius of Mars from the altitude of the orbiter above the Martian surface.
b) Radius of MRO orbit = Altitude of MRO above Martian surface - Radius of Mars
Radius of MRO orbit = 390 km - 3396 km
Radius of MRO orbit = -3006 km
Since the result is negative, we need to take the absolute value, which gives us:
Radius of MRO orbit = 3006 km (answer in kilometers)
c) To find the linear speed (v) of the orbiter above Mars, we can use the formula for circular motion:
v = √(GM/r)
where G is the gravitational constant (6.67430 x 10^-11 m^3 kg^-1 s^-2), M is the mass of Mars, and r is the radius of the MRO orbit.
To convert the radius of the MRO orbit from kilometers to meters:
Radius of MRO orbit = 3006 km * 1000 m/km = 3,006,000 meters
Plugging in the values into the formula:
v = √((6.67430 x 10^-11 m^3 kg^-1 s^-2) * (6.42E23 kg) / (3,006,000 m))
v ≈ 3241.7 m/s (rounded to 4 significant figures)
Therefore, the linear speed of the orbiter above Mars is approximately 3241.7 m/s.
d) To find the period of the orbiter in hours, we can use Kepler's Third Law of Planetary Motion:
T = 2π * √(r^3 / GM)
where T is the period of the orbiter, r is the radius of the MRO orbit, and G is the gravitational constant.
Plugging in the values into the formula:
T = 2π * √((3,006,000 m)^3 / ((6.67430 x 10^-11 m^3 kg^-1 s^-2) * (6.42E23 kg)))
T ≈ 28.54 hours (rounded to 2 decimal places)
Therefore, the period of the orbiter is approximately 28.54 hours.