Consider a NASA spacecraft in a circular orbit 440 km above the surface of Mars. What is the spacecraft's orbital period in hours?
To determine the spacecraft's orbital period in hours, we need to use Kepler's third law of planetary motion, which states that the square of the orbital period of a planet or spacecraft is directly proportional to the cube of the semi-major axis of its orbit.
1. First, we need to find the semi-major axis of the spacecraft's orbit. The altitude above the surface of Mars is given as 440 km, so we add it to the radius of Mars to get the total distance from the center of Mars to the spacecraft's orbit. The radius of Mars is approximately 3,389.5 km (or 3,389,500 meters).
Semi-Major Axis = Altitude + Radius of Mars
= 440 km + 3,389.5 km
= 3,829.5 km
= 3,829,500 meters
2. Next, we can use Kepler's third law to find the orbital period. The equation we can use is:
T^2 = (4π^2 / GM) * a^3
Where T is the orbital period, G is the gravitational constant (6.67430 x 10^-11 m^3 / (kg * s^2)), M is the mass of Mars (6.39 x 10^23 kg), and a is the semi-major axis.
Rearranging the equation to solve for T:
T = √((4π^2 / GM) * a^3)
Plugging in the values:
T = √((4 * π^2) / (6.67430 x 10^-11) * (6.39 x 10^23) * (3,829,500)^3)
Calculating this equation will give us the spacecraft's orbital period in seconds.
3. Finally, we can convert the orbital period from seconds to hours. There are 3600 seconds in one hour, so we divide the orbital period in seconds by 3600.
By following these steps, you can calculate the orbital period in hours for a NASA spacecraft in a circular orbit 440 km above the surface of Mars.