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Problem: There is an satellite that orbits the Earth. Its orbit is circular and its orbital plane is perpendicular to Earth's Equatorial Plane. It orbits is such that a person who is on equator see this satellite on Zenith every 12Hours.

a. Find Satellite's Period.

b. What is distance of satellite to person when it Sets on person's horizon?

My problem with this question is:

For the first part: somewhere said period is 12hours. Somewhere else said the person on equator also moves (Earth Rotation) so 12 hours is P/2 and P= 24hours. What is the right answer?

For the second part: Which of the following figures show the position of satellite when it sets? (Figure 1 or 2?)

figure 1

figure 2

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    $\begingroup$ You probably got voted down cause this sounds a lot like a homework question. It's a fun question though. Zenith is directly overhead. It's hot hard to work out if you think about where the person on the equator is every 12 hours. $\endgroup$ – userLTK Mar 28 '17 at 8:42
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    $\begingroup$ Note, though, that the rotation period of the Earth is not 24 hours, but shorter, suggesting the question is rubbish. $\endgroup$ – Walter Apr 2 '17 at 11:51
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I'm treating this as a homework question. You have made a clear attempt, so it is not off topic, but you should clarify the source of the problem.

You will first need to establish if the rotation of the Earth is to be considered (it makes a big difference, but in a homework, sometimes the Earth stops spinning!) You have then correctly found the two possible answers, either 24 or 12 hours.

You can also consider the motion of the Earth around the sun (as the Earth only takes 23hr, 56min to spin once, relative to the stars). What this means is that, after 12 hours, the Earth has turned more than 180 degrees, and there is no orbital plane that passes over the person, and through the origin, and is perpendicular to the equator. However the error is only about 1 degree, so probably can be neglected for the purposes of homework.

For the second part, the first picture is certainly wrong, as the line of sight (in green) passes through the Earth. The second shows the line of sight (now in red) being tangent to the Earth's surface. You will need to use orbital mechanics to find the distance from the Earth's centre to the satellite, and find out the distance from the Earth's centre to its surface (green lines), giving two sides of a right-angled triangle. And the rest is maths.

There are a few details that may affect the answer: you should consider if the motion of the person will affect the answer to part (b) (and if not, why not). You should know that the Earth is not spherical. You should consider whether the optical refraction of light by the Earth's atmosphere should be considered. However, in a homework problem, such issues can often be ignored, as their effects are small.

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  • $\begingroup$ Such satellite orbit is impossible. The Earth spin period is not 24 hours. If the person was moving (by about 360/365.25 degrees per day, and the Earth's orbit was circular) and the satellites orbit had the same period as the Earth's spin period, something similar is possible. $\endgroup$ – Walter Apr 2 '17 at 11:57
  • $\begingroup$ I guess you are noting that the sidereal day is about 4 min shorter than the solar day. That's true, and could go into the "few details that may affect the answer". As noted, in homework sometimes the earth stops spinning and sometimes it it stops orbiting the sun. $\endgroup$ – James K Apr 2 '17 at 12:13
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Part (1). Due to the earth's rotation, an observer on the equator moves 180 deg every 12 hours. Your satellite is moving perpendicular to the equatorial plane and it meets the observer on the other side of its orbit after 12 hours. Therefore it too has travelled 180 degrees. To revolve a further 180 degrees (and complete a full 360 degree revolution), in time to meet the observer again, the satellite would take another 12 hours. Therefore the time period of the satellite would have to be 12h + 12h = 24h.

(2) A satellite with a time period of 24h is called a geosynchronous satellite. The radius $r$ of the orbit of this type of satellite is approximately 36,000 km, can be determined by working further into this formula...

$$ \frac{v^2}{r} = \frac{GM}{r^2}$$

So, in your second diagram, we can see that the radius of the satellite is 36,000 km, and we know the radius of the earth is 6400km, so by Pythagoras rule, the distance $d$ to the satellite when it sets can be obtained by,

$$d = \sqrt{36000^2 - 6400^2}$$

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The question is wrong: such a satellite orbit does not exist. Due to angular-momentum conservation, the orbital plane of the satellite is fixed in space, but the position of the observer rotates by about $\left( 360+\frac{360}{365.25} \right)^{\circ}$ every $24$ hours. This is simply because the Earth rotates more than once in an average day, i.e. the time between consecutive noons (Sun in zenith on the equator).

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  • $\begingroup$ @TheDarkSide According to this answer in the SE MathJax tutorial, "The degree symbol for angles is not ^\circ. Although many people use this notation, the result looks quite different from the canonical degree symbol shipped with the font" I just copy ° from Wikipedia or google. $\endgroup$ – uhoh Jul 2 '17 at 6:42
  • $\begingroup$ @uhoh So what is the correct LaTeX way for generating the degree symbol? $\endgroup$ – Walter Jul 2 '17 at 8:53
  • $\begingroup$ a good question. Different computer OS may offer different keyboard shortcuts for just typing it directly rather than copy/paste like I do, or resorting to commanding LaTeX or MathJax to "generate" it. After I read this excellent and elaborate answer to my question about a possesive apostrophe, I decided that I'd get by in life without understanding characters. Actually I decided that when ctrl-G stopped ringing the bell i.stack.imgur.com/EKxX9.jpg 10 print "^G", 20 GOTO 10 $\endgroup$ – uhoh Jul 2 '17 at 10:04
  • $\begingroup$ However, you might want to read about sun-syncrhonous oribits before declaring the question "wrong". $\endgroup$ – uhoh Jul 2 '17 at 10:09
  • $\begingroup$ @uhoh but such an orbit cannot be (and remain) in a plane perpendicular to the orbital plane of the Earth, can it? $\endgroup$ – Walter Jul 2 '17 at 10:24

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