# Finding rate of change of the distance from an observer on earth to a satellite in orbit

This is a satellite motion related question. To give context, imagine a satellite orbiting Earth at a distance 'R' from the center of the Earth at constant velocity 'v'. An observer on the surface of the Earth will be 'X' distance from the center of the Earth. At position 'p' where the observer is standing, there will be a horizon which the satellite can be seen in. This horizon can be represented as a chord perpendicular to the line 'X'. As the satellite comes above the horizon, the angle of elevation 'θ' increases. Overhead would be 90 degrees. 'θ' would range from 0 to 180 degrees. Considering all of this, a line of sight from the observer at 'p' can be drawn, which is essentially the distance from the observer to the satellite. (Refer to diagram attached). Note that the satellite is considered to pass overhead (overhead orbit) .My question is can anyone derive a function of distance 'd' with respect to time 't'? I know that finding a relation for a function of distance 'd' with reference to angle of elevation 'θ' is easy, but I require the x-axis to be time

The time period can be calculated from 'v' and 'R'. If the observer is standing at the center of the Earth, the rate of change will be constant. But since the observer is 'x' distance from the center, how would the 'd' value change according with time?

I want an function that I can plot. y axis would be the distance 'd' and x axis would be time, with 'v', 'R' and 'X' being values I can substitute and change.

• Are you assuming that the observer is on the equator, and the satellite is in an equatorial orbit too ? Sep 5, 2022 at 11:03
• yes indeed. with these conditions, the satellite will pass overhead Sep 5, 2022 at 11:29

I attached a sketch. Try to have a look at it

• Thanks! this is exactly what I needed! Sep 5, 2022 at 12:45
• you're welcome! Sep 5, 2022 at 12:47

Let’s start by finding the angle between the observer and the satellite as seen from the centre of the Earth. Assuming the observer is on the equator then the observer revolves around the centre of the Earth once in $$24$$ hours. So this is an angular speed of $$15$$ degrees per hour. You can work out the angular speed of the satellite from its height (or, equivalently, from $$R$$). For equatorial low Earth orbits, the period is about $$1.5$$ hours, which is an angular speed of $$240$$ degrees per hour.

If you start a clock at $$t=0$$ when the satellite is overhead then $$t$$ hours later the angle at the centre of the Earth is

$$\phi = 240t - 15t = 225t \text{ degrees}$$

Note that this assumes a prograde orbit - for a retrograde orbit the equation is

$$\phi = 240t + 15t = 255t \text{ degrees}$$

Now you know $$R$$, $$X$$ and $$\phi$$, so you can find $$d$$ from the side-angle-side rule:

$$R^2+X^2-d^2 = 2RX \cos \phi$$

• Thanks for the solution! Sep 5, 2022 at 12:46