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.
A orbiting satellite stays over a certain spot on the equator of (rotating) Earth. What is the altitude of the orbit (called a "synchronous orbit")? Write the equation for the period of a satellite in terms of the distance from
A orbiting satellite stays over a certain spot on the equator of (rotating) Venus. What is the altitude of the orbit (called a "synchronous orbit")? I found the radius and mass of Venus on the google, but I don't know what I
mercury rotates about its own axis once every 58.6 Earth days. a satellite orbits mercury once every murcurian day. how high is this satellite above mercury's surface? mass of mercury = 3.2x10^23 kg. the radius of mercury is
A geosynchronous equatorial orbiting satellite orbits 22,300 miles above the equator of Earth. It completes one full revolution each 24 hours. Assume Earth's radius is 3960 miles. a. How far will the GEO satellite travel in one
A satellite orbiting the earth in a circular path stays at a constant altitude of 100 kilometers throughout its orbit. Given that the radius of the earth is 6370 kilometers, find the distance that the satellite travels in
A satellite orbiting at an elevation of 3200 km does what a) Stays beyond the earth's gravity, which is why it stays up there. B) in an elliptical orbit, since its speed is less than the escape velocity. C) Falling toward earth