A orbiting satellite stays over a certain spot on the equator of (rotating) Mercury. What is the altitude of the orbit (called a "synchronous orbit")?
First you need to look up the period of rotation of Mercury. It is a rather slow rotator, locked to three rotations per two orbits about the sun. That make the sidereal rotation period 58.6 days. (That is NOT the length of a Mercury day).
Next you need the mass of Mercury to determine the geosynchronous orbital height. Set the gravitational pull equal to the centripetal force, and solve for R. The mass of the satellite will cancel out. The universal constant of gravity will appear in your formula.
To determine the altitude of a synchronous orbit above a rotating celestial body like Mercury, we need to consider the rotational characteristics of the body.
The concept of a synchronous orbit refers to the situation where a satellite orbits a planet at the same rate as the planet rotates, resulting in the satellite appearing to remain stationary relative to the planet's surface.
To calculate the altitude of a synchronous orbit, we need to account for the gravitational pull of the planet and its rotational period. Fortunately, the synchronous orbit altitude can be determined using the following formula:
Altitude = (GM T^2 / 4π^2)^(1/3) - R,
Where:
Altitude = altitude of the synchronous orbit above the surface of the planet
G = gravitational constant (approximately 6.67430 × 10^-11 m^3 kg^-1 s^-2)
M = mass of the planet
T = rotational period of the planet
R = radius of the planet
For Mercury:
Mass (M) = 3.3011 × 10^23 kg
Rotational Period (T) = approximately 58.6 days (or 5068800 seconds)
Radius (R) = 2440 km (or 2,440,000 meters)
Substituting these values into the formula, we can calculate the altitude:
Altitude = (6.67430 × 10^-11 * 3.3011 × 10^23 * (5068800^2) / (4π^2))^(1/3) - 2,440,000
After solving this equation, the altitude of the synchronous orbit above the equator of rotating Mercury would be obtained.