To determine the altitude of a synchronous orbit over the equator of rotating Pluto, we need to understand the concept of synchronous orbit.
A synchronous orbit, also known as a geostationary orbit, is an orbit at which a satellite's orbital period matches the rotational period of the celestial body it is orbiting. In this case, we assume that Pluto's rotation period is 6.39 Earth days or 153.36 hours.
Pluto's equatorial radius is approximately 1,185 kilometers. To find the altitude of the synchronous orbit, we can use Kepler's third law, which states that the square of the orbital period (T) is proportional to the cube of the semimajor axis (a).
First, we need to convert Pluto's rotation period to seconds:
153.36 hours * 60 minutes * 60 seconds = 553,296 seconds
Next, we use the equation for the period of a circular orbit:
T^2 = (4 * ฯ^2 * a^3) / (G * M)
Where:
T = Orbital period (in seconds)
ฯ = Pi (approximately 3.14159)
a = Semimajor axis (orbital radius + planet radius)
G = Gravitational constant (approximately 6.67430 x 10^-11 m^3 kg^-1 s^-2)
M = Mass of Pluto (approximately 1.303 x 10^22 kg)
Rearranging the equation, we get:
a = ((T^2 * G * M) / (4 * ฯ^2))^(1/3)
Plugging in the values, we have:
a = ((553,296^2 * 6.67430 x 10^-11 * 1.303 x 10^22) / (4 * ฯ^2))^(1/3)
Calculating the value of a using a scientific calculator or software, we find a โ 1,176,150 kilometers.
Finally, we subtract Pluto's equatorial radius (1,185 kilometers) from the semimajor axis to get the altitude of the synchronous orbit:
Altitude = a - Pluto's equatorial radius
Altitude = 1,176,150 kilometers - 1,185 kilometers
Altitude โ 1,175,965 kilometers
Therefore, for a satellite to stay over a certain spot on the equator of rotating Pluto, it would need to be at an altitude of approximately 1,175,965 kilometers.