Consider a satellite at the altitude of geostationary...

Consider a satellite at the altitude of geostationary satellites but whose orbital plane is inclined to the equatorial plane by an angle $\phi$. To a stationary user on the earth's surface at north latitude $\phi$, does this satellite appear motionless in the sky? If not, describe its motion.

Key Concepts

Geosynchronous vs Geostationary Orbits

A geosynchronous orbit is one in which a satellite’s orbital period matches the rotation period of the Earth. However, a satellite in a geosynchronous orbit is only considered geostationary if it also lies in the Earth’s equatorial plane. When the satellite’s orbit is inclined relative to the equator, although its period remains synchronized with Earth’s rotation, its position in the sky shifts periodically, meaning it does not remain fixed relative to any point on the Earth’s surface.

Orbital Inclination

Orbital inclination is the angle between the plane of the satellite’s orbit and the Earth’s equatorial plane. This tilt causes the satellite’s ground track to vary in latitude, leading to an apparent oscillatory motion in the sky as seen by an observer on Earth. The higher the inclination, the greater the extent of the north–south movement observed from the ground.

Apparent Motion of Satellites

The apparent motion of a satellite in an inclined geosynchronous orbit is a result of the projection of its orbital motion onto the sky as observed from a given location on Earth. Unlike a geostationary satellite that appears fixed at one position, an inclined geosynchronous satellite will trace a periodic path—often resembling a north–south oscillation or a figure?8 pattern (analemma) depending on the observer’s position—with the satellite periodically reaching higher and lower elevations.

Observer's Latitude and Viewing Geometry

The latitude of an observer plays a significant role in how the satellite’s motion is perceived. When an observer’s latitude matches the orbital inclination of the satellite, the geometry aligns such that the satellite’s apparent motion is more directly vertical, potentially crossing close to the zenith. This specific alignment does not result in a stationary appearance; instead, the satellite oscillates above and below the observer’s zenith in a regular cycle.

Transcript

00:01 In this problem, we're going to determine the radius and hence the altitude of what's called a geosynchronous orbit.

00:11 In a geosynchronous orbit, basically your satellite is circling the earth with a period of one day.

00:23 So maybe the first thing to do is take one day and convert it into a number of seconds so that we may be.

00:32 Use our mks system or system international set of units.

00:40 So we'll go ahead and do that.

00:42 There are 24 hours a day and 3 ,600 seconds per hour.

00:50 So yeah, lots of seconds in a day.

01:02 And the best way to solve this is to assume that the satellite is in a circular orbit.

01:09 And what's holding it in that circular orbit is the force of gravity from the earth.

01:18 So we're going to set the force of gravity acting on that satellite equal to its mass times its centripetal acceleration.

01:31 And we'll use newton's law of gravity for the gravity side.

01:41 And we'll show r here is the distance between the satellites.

01:45 Light and the center of the earth.

01:48 It's the radius of its orbit.

01:54 And centripetal acceleration is the same thing as angular velocity squared times r.

02:04 And here, just to note that the angular velocity is related to the period in a very simple way.

02:14 So we can put all that together.

02:18 And what we want to do is solve for the radius.

02:22 Which appears on both sides of the equation.

02:27 That should be an r squared in gravity.

02:30 Let's get the force of gravity correct.

02:39 But we'll have to do a little bit of algebra, and it's not too bad...

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