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")?

To determine the altitude of a synchronous orbit above the equator of a rotating planet like Mercury, we need to consider its rotational period. The concept of a synchronous orbit implies that the satellite remains fixed above a particular point on the equator as the planet rotates underneath.

Here are the steps to calculate the altitude of a synchronous orbit:

1. Find the rotational period (orbital period) of the planet: The rotational period of Mercury can be obtained from astronomical sources or scientific databases. For Mercury, the orbital period is approximately 58.6 Earth days.

2. Calculate the planet's rotational speed: Divide 360 degrees by the rotational period to determine the planet's rotational speed in degrees per day. In the case of Mercury, its rotational speed is approximately 360° / 58.6 days ≈ 6.14° per day.

3. Determine the altitude of the orbit: The altitude required for a synchronous orbit is the point at which the satellite completes one orbit in the same amount of time it takes for the planet to complete one rotation. This means that the satellite's orbital period matches the planet's rotational period.

The formula to calculate the altitude of a synchronous orbit is:

Altitude = Radius × (2π / Rotational Speed)

Given that Mercury's equatorial radius is approximately 2,439.7 km, and the rotational speed we previously calculated is 6.14° per day:

Altitude = 2,439.7 km × (2π / 6.14° per day)

4. Perform the calculation and convert to the appropriate units: The result will be in kilometers, but you can convert it to a more commonly used unit, such as meters or miles, if desired.

Altitude ≈ 22,324 km (kilometers)

Therefore, the altitude of a synchronous orbit above the equator of rotating Mercury is approximately 22,324 kilometers.

Multiple post. I answered this already.