Orbit of Comet 17P Holmes - equations

In summary, the conversation discusses the calculation of orbits, specifically for Comet 17P Holmes. The parameters used in the calculations include eccentricity, perihelion distance, ascending node, argument of perihelion, inclination, epoch, and orbital period. The semi-major and semi-minor axes, semi-latus rectum, and period in years are also calculated. The correct standard gravitational parameter is determined to be 1.32715 x 10^11 in km^3 / s^2. The formula for calculating orbital speed at any point in the orbit is provided, as well as the formula for finding the radial distance from the Sun at any given angle. There is a discrepancy in the value for r, which should
  • #1
CometDude
11
0
Im trying to learn how to calculate orbits.

Comet 17P Holmes
e: 0.432668
q: 2.053207
Ascending Node: 326.8586
w: 24.2856
Inclination: 19.1134
Epoch: 2454600.5 (2000)
Tp: 20070504.5695


I can calculate Semi Major Axis, Semi Minor Axis, semi-latus rectum, Period in years

[tex]a = q/(1-e)[/tex]
[tex]b = a\sqrt{1-e^2}[/tex]
[tex]l = a(1-e^2)[/tex]
[tex]T = \sqrt{a^3}[/tex]


semi-major axis a 3.61905727158
semi-minor axis b 3.2627731245014
semi-latus l 2.94156396627603518632608
Period T 6.8848294081932


How do I calculate
Perihelion date

at any date in the orbit
distance to the Sun, Earth in AU
orbital speed


thank you for you help.

Tony
 
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  • #2
Checking on wiki as I am trying to teach myself about orbits
[tex]\mu = GM[/tex] Standard gravitational parameter
On Wiki I also see that
[tex]\mu = 4\pi^2a^3/T^2[/tex] Standard gravitational parameter

but when I calc them they do not match
using the above orbital elements
I get [tex]\mu = 39.478417604357[/tex] when using [tex]\mu = 4\pi^2a^3/T^2[/tex]
and
[tex]\mu = 132712440018[/tex] [tex]\mu = GM[/tex] this number was taking right from the wiki page for the GM of the Sun.

lol I still don't have enough posts to post a link.

what is the correct Standard gravitational parameter? Am I missing a step or doing it out of order?
 
  • #3
CometDude said:
Checking on wiki as I am trying to teach myself about orbits
[tex]\mu = GM[/tex] Standard gravitational parameter
On Wiki I also see that
[tex]\mu = 4\pi^2a^3/T^2[/tex] Standard gravitational parameter

but when I calc them they do not match
using the above orbital elements
I get [tex]\mu = 39.478417604357[/tex] when using [tex]\mu = 4\pi^2a^3/T^2[/tex]
and
[tex]\mu = 132712440018[/tex] [tex]\mu = GM[/tex] this number was taking right from the wiki page for the GM of the Sun.

lol I still don't have enough posts to post a link.

what is the correct Standard gravitational parameter? Am I missing a step or doing it out of order?

I assume you are talking about this wiki page:

http://en.wikipedia.org/wiki/Standard_gravitational_parameter ?

From another source ( http://www.projectrho.com/rocket/Orbits.htm ) I get GM = 1.32715 x 10^11 in km^3 / s^2 which does not exactly match, but is pretty close to the wiki number you quote.

I have seen that in your first post you state that you have calculated the period T in years. but surely for the formula above you have to use the value in seconds. could you just repeat all the parameters you use as input for the formula together with their units (s, km, whatever) ?
 
Last edited by a moderator:
  • #4
Thanks for the reply

I have less then 15 posts so I can't post the link yet, but the wiki page I am talking about is the same one you posted.

This is right on the page. So I used period in years.
for elliptic orbits: [tex]4 \pi^2 a^3/T^2 = \mu [/tex] (with [tex]a[/tex] expressed in AU and [tex]T[/tex] in years, and with M the total mass relative to that of the Sun, we get [tex]a^3 / T^2 = M[/tex])

e: 0.432668
q: 2.053207 AU
Ascending Node: 326.8586 Deg
w: 24.2856 Deg
Inclination: 19.1134 Deg
Epoch: 2454600.5
Tp: 20070504.5695
Orbital period: 6.88 Years

semi-major axis a 3.61905727158 AU
semi-minor axis b 3.2627731245014 AU
semi-latus l 2.941563966276 AU
Period T 6.8848294081932 Years

I wish that I could post links but on the wiki page wiki/Periapsis_distance

its does not give the units of speed. would it be km/s ?
Periapsis: maximum speed [tex]v_\mathrm{per} = \sqrt{ \frac{(1+e)\mu}{(1-e)a} } \,[/tex] at minimum (periapsis) distance [tex]r_\mathrm{per}=(1-e)a\!\,[/tex]

Apoapsis: minimum speed [tex]v_\mathrm{ap} = \sqrt{ \frac{(1-e)\mu}{(1+e)a} } \,[/tex] at maximum (apoapsis) distance [tex]r_\mathrm{ap}=(1+e)a\!\,[/tex]


How do I calculate the orbital speed at times other than Perihelion and Aphelion
 
  • #5
CometDude said:
Thanks for the reply

I have less then 15 posts so I can't post the link yet, but the wiki page I am talking about is the same one you posted.

This is right on the page. So I used period in years.
for elliptic orbits: [tex]4 \pi^2 a^3/T^2 = \mu [/tex] (with [tex]a[/tex] expressed in AU and [tex]T[/tex] in years, and with M the total mass relative to that of the Sun, we get [tex]a^3 / T^2 = M[/tex])

e: 0.432668
q: 2.053207 AU
Ascending Node: 326.8586 Deg
w: 24.2856 Deg
Inclination: 19.1134 Deg
Epoch: 2454600.5
Tp: 20070504.5695
Orbital period: 6.88 Years

semi-major axis a 3.61905727158 AU
semi-minor axis b 3.2627731245014 AU
semi-latus l 2.941563966276 AU
Period T 6.8848294081932 Years

I wish that I could post links but on the wiki page wiki/Periapsis_distance

its does not give the units of speed. would it be km/s ?
The units of speed will depend on the units used. If you use meters, kilograms and seconds, your answer will be in meters/sec. If you use AUs, years and the [itex]\mu[/itex] you arrived at above, your answer will be in AU/yr.
(The reason you got a different value for [itex]\mu[/itex] than wiki was that you were using different base units.)
Periapsis: maximum speed [tex]v_\mathrm{per} = \sqrt{ \frac{(1+e)\mu}{(1-e)a} } \,[/tex] at minimum (periapsis) distance [tex]r_\mathrm{per}=(1-e)a\!\,[/tex]

Apoapsis: minimum speed [tex]v_\mathrm{ap} = \sqrt{ \frac{(1-e)\mu}{(1+e)a} } \,[/tex] at maximum (apoapsis) distance [tex]r_\mathrm{ap}=(1+e)a\!\,[/tex]


How do I calculate the orbital speed at times other than Perihelion and Aphelion

You can use:

[tex]v=\sqrt{\mu \left ( \frac{2}{r}- \frac{1}{a} \right )}[/tex]

where [itex]r[/itex] is the radial distance from the Sun at that part of the orbit.

You can find [itex]r[/itex] for any angle [itex]\theta[/itex] as measured from perihelion by

[tex]r = a \left ( \frac{1-e^2}{1+ e \cos{\theta}} \right )[/tex]
 
  • #6
Janus said:
The units of speed will depend on the units used. If you use meters, kilograms and seconds, your answer will be in meters/sec. If you use AUs, years and the [itex]\mu[/itex] you arrived at above, your answer will be in AU/yr.
(The reason you got a different value for [itex]\mu[/itex] than wiki was that you were using different base units.)

Thank you, your answer made it very clear to me, now that I know about AU/yr vs km/s.
 
  • #7
Janus said:
[tex]v=\sqrt{\mu \left ( \frac{2}{r}- \frac{1}{a} \right )}[/tex]

where [itex]r[/itex] is the radial distance from the Sun at that part of the orbit.

You can find [itex]r[/itex] for any angle [itex]\theta[/itex] as measured from perihelion by

[tex]r = a \left ( \frac{1-e^2}{1+ e \cos{\theta}} \right )[/tex]
I'm having a problem with the value for r. r should not be smaller than 2.053207 or larger than 5.1849075431599 right?

I calculate
2.053207 AU [tex] r_\mathrm{per}=(1-e)a\!\,[/tex]
5.1849075431599 AU [tex] r_\mathrm{ap}=(1+e)a\!\,[/tex]

also shouldn't [tex] r_\mathrm{ap}-r_\mathrm{per}= a[/tex]semi-major axis a 3.61905727158 AU [tex]a = q/(1-e)[/tex]
e: 0.432668
q: 2.053207 AU
Ascending Node: 326.8586 Deg
w: 24.2856 Deg
Inclination: 19.1134 Deg
Epoch: 2454600.5
Tp: 20070504.5695
Orbital period: 6.88 Years[tex]r = a \left ( \frac{1-e^2}{1+ e \cos{\theta}} \right )[/tex]
Code:
deg  r given in AU
0 2.053207
1 2.053519761006
2 2.0544585206212
3 2.0560247104094
4 2.0582207222335
5 2.0610499171829
6 2.0645166381578
7 2.0686262262009
8 2.0733850406882
9 2.0788004835238
10 2.0848810275101
11 2.0916362490989
12 2.099076865761
13 2.1072147782501
14 2.1160631180766
15 2.1256363005512
16 2.1359500838071
17 2.1470216342617
18 2.1588695990416
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21 2.1992823989505
22 2.2144550834088
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24 2.2475148694826
25 2.2654632668592
26 2.2844020923972
27 2.3043680873945
28 2.3254007560651
29 2.347542575636
30 2.370839228304
31 2.3953398575963
32 2.4210973520319
33 2.4481686593938
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35 2.5065029310671
36 2.5379034238784
37 2.570893698276
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39 2.6419837480088
40 2.6802713843081
41 2.7205259530347
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87 39.231288532668
88 58.831994555751
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273 39.231288532668
274 29.433927631311
275 23.557908352002
276 19.642563717239
277 16.847608616174
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320 2.6802713843081
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324 2.5379034238784
325 2.5065029310671
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329 2.3953398575963
330 2.370839228304
331 2.347542575636
332 2.3254007560651
333 2.3043680873945
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344 2.1359500838071
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354 2.0645166381578
355 2.0610499171829
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358 2.0544585206212
359 2.053519761006
360 2.053207
 
Last edited:
  • #8
You made a mistake somewhere with the equation. You should never get an negative answer. The limits for cos[itex] \theta[/itex] are -1 and 1. Thus the limits of your answer will come out between:

[tex]r = a \left ( \frac{1-e^2}{1+ e }} \right )[/tex]

and

[tex]r = a \left ( \frac{1-e^2}{1- e }} \right )[/tex]
 
Last edited:
  • #9
Okay, here's your problem. Instead of adding e times cos [itex]\theta[/itex] to 1 like you should, you are adding 1 to e and multiplying this sum by cos [itex]\theta[/itex].
 
  • #10
ohhhhhhhhhhhhhhh its a good thing my webcam is off I have a big red face.

yes works great now I also had to deg2rad the angle.

thank you for your help
 
  • #11
Ok I have been away for a while trying to get past the next step, but I'm not having much luck in finding out how to do it.

I would like to be able to calculate velocity and distance and position at a given time. if you can point me in the right direction a website or a book/video that would be huge help.

I am thinking of ordering Orbital Mechanics by John Prussing and Astronomical Algorithms by Jean Meeus. Does any recommend these books or some other book?
 
Last edited:

1. What is the equation for calculating the orbit of Comet 17P Holmes?

The equation for calculating the orbit of Comet 17P Holmes is known as Kepler's Third Law, which states that the square of the orbital period (P) is proportional to the cube of the semi-major axis (a) of the orbit. The equation is represented as P^2 = 4π^2a^3/GM, where G is the gravitational constant and M is the mass of the central body.

2. How is the semi-major axis of Comet 17P Holmes' orbit determined?

The semi-major axis of Comet 17P Holmes' orbit is determined by measuring the distance between the comet and the Sun at two different points in its orbit. This distance is then divided by 2 to get the semi-major axis, which is the longest radius of the elliptical orbit.

3. What factors can affect the orbit of Comet 17P Holmes?

The orbit of Comet 17P Holmes can be affected by several factors, including the gravitational pull of other celestial bodies, such as planets and stars, as well as the comet's own mass and velocity. Changes in these factors can cause the comet's orbit to become more elliptical or to shift in its orientation.

4. What is the significance of understanding the orbit of Comet 17P Holmes?

Understanding the orbit of Comet 17P Holmes can provide valuable insights into the composition and behavior of comets. By studying its orbit, scientists can learn more about the formation and evolution of our solar system, as well as the conditions of the early universe. Additionally, knowing the orbit can help predict the comet's future movements and potential hazards to Earth.

5. Can the orbit of Comet 17P Holmes change over time?

Yes, the orbit of Comet 17P Holmes can change over time due to various factors such as gravitational interactions with other objects, outgassing, and collisions. These changes can cause the comet's orbit to become more elongated or to shift in its orientation, leading to different orbital periods and paths. However, these changes are relatively small and can be predicted and accounted for by scientists studying the comet's orbit.

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