# Orbit of Comet 17P Holmes - equations

• CometDude
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
CometDude
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

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

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

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

but when I calc them they do not match
using the above orbital elements
I get $$\mu = 39.478417604357$$ when using $$\mu = 4\pi^2a^3/T^2$$
and
$$\mu = 132712440018$$ $$\mu = GM$$ 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?

CometDude said:
Checking on wiki as I am trying to teach myself about orbits
$$\mu = GM$$ Standard gravitational parameter
On Wiki I also see that
$$\mu = 4\pi^2a^3/T^2$$ Standard gravitational parameter

but when I calc them they do not match
using the above orbital elements
I get $$\mu = 39.478417604357$$ when using $$\mu = 4\pi^2a^3/T^2$$
and
$$\mu = 132712440018$$ $$\mu = GM$$ 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?

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:

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: $$4 \pi^2 a^3/T^2 = \mu$$ (with $$a$$ expressed in AU and $$T$$ in years, and with M the total mass relative to that of the Sun, we get $$a^3 / T^2 = M$$)

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 $$v_\mathrm{per} = \sqrt{ \frac{(1+e)\mu}{(1-e)a} } \,$$ at minimum (periapsis) distance $$r_\mathrm{per}=(1-e)a\!\,$$

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

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

CometDude said:

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: $$4 \pi^2 a^3/T^2 = \mu$$ (with $$a$$ expressed in AU and $$T$$ in years, and with M the total mass relative to that of the Sun, we get $$a^3 / T^2 = M$$)

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 $\mu$ you arrived at above, your answer will be in AU/yr.
(The reason you got a different value for $\mu$ than wiki was that you were using different base units.)
Periapsis: maximum speed $$v_\mathrm{per} = \sqrt{ \frac{(1+e)\mu}{(1-e)a} } \,$$ at minimum (periapsis) distance $$r_\mathrm{per}=(1-e)a\!\,$$

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

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

You can use:

$$v=\sqrt{\mu \left ( \frac{2}{r}- \frac{1}{a} \right )}$$

where $r$ is the radial distance from the Sun at that part of the orbit.

You can find $r$ for any angle $\theta$ as measured from perihelion by

$$r = a \left ( \frac{1-e^2}{1+ e \cos{\theta}} \right )$$

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 $\mu$ you arrived at above, your answer will be in AU/yr.
(The reason you got a different value for $\mu$ than wiki was that you were using different base units.)

Janus said:
$$v=\sqrt{\mu \left ( \frac{2}{r}- \frac{1}{a} \right )}$$

where $r$ is the radial distance from the Sun at that part of the orbit.

You can find $r$ for any angle $\theta$ as measured from perihelion by

$$r = a \left ( \frac{1-e^2}{1+ e \cos{\theta}} \right )$$
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 $$r_\mathrm{per}=(1-e)a\!\,$$
5.1849075431599 AU $$r_\mathrm{ap}=(1+e)a\!\,$$

also shouldn't $$r_\mathrm{ap}-r_\mathrm{per}= a$$semi-major axis a 3.61905727158 AU $$a = q/(1-e)$$
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$$r = a \left ( \frac{1-e^2}{1+ e \cos{\theta}} \right )$$
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
19 2.1715141859572
20 2.1849772516919
21 2.1992823989505
22 2.2144550834088
23 2.2305227314112
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
34 2.4766151353939
35 2.5065029310671
36 2.5379034238784
37 2.570893698276
38 2.6055570823145
39 2.6419837480088
40 2.6802713843081
41 2.7205259530347
42 2.7628625398569
43 2.8074063144199
44 2.8542936162206
45 2.9036731857594
46 2.9557075640553
47 3.0105746879096
48 3.0684697135179
49 3.129607107396
50 3.1942230513766
51 3.2625782180279
52 3.3349609847099
53 3.4116911692337
54 3.4931243885281
55 3.5796571648968
56 3.6717329337628
57 3.7698491441095
58 3.8745656906265
59 3.9865149782539
60 4.106414
61 4.2350789139802
62 4.373442744468
63 4.5225770168713
64 4.6837183857538
65 4.8583016539398
66 5.0480010467544
67 5.2547822538187
68 5.480968664426
69 5.7293265276795
70 6.0031756610456
71 6.3065351198844
72 6.6443174238784
73 7.022591334367
74 7.4489431735921
75 7.93298267207
76 8.4870656081401
77 9.1273499739255
78 9.8753802247691
79 10.760535665251
80 11.823947867384
81 13.125029701556
82 14.752901141357
83 16.847608616174
84 19.642563717239
85 23.557908352002
86 29.433927631311
87 39.231288532668
88 58.831994555751
89 117.64606831604
90 3.3532522407569E+16
91 -117.64606831604
92 -58.831994555752
93 -39.231288532668
94 -29.433927631311
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135 -2.9036731857594
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147 -2.4481686593938
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149 -2.3953398575963
150 -2.370839228304
151 -2.347542575636
152 -2.3254007560651
153 -2.3043680873945
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155 -2.2654632668592
156 -2.2475148694826
157 -2.2305227314112
158 -2.2144550834088
159 -2.1992823989505
160 -2.184977251692
161 -2.1715141859572
162 -2.1588695990416
163 -2.1470216342617
164 -2.1359500838071
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166 -2.1160631180766
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169 -2.0916362490989
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173 -2.0686262262009
174 -2.0645166381578
175 -2.0610499171829
176 -2.0582207222335
177 -2.0560247104094
178 -2.0544585206212
179 -2.053519761006
180 -2.053207
181 -2.053519761006
182 -2.0544585206212
183 -2.0560247104094
184 -2.0582207222335
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210 -2.370839228304
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213 -2.4481686593938
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225 -2.9036731857594
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227 -3.0105746879096
228 -3.0684697135179
229 -3.129607107396
230 -3.1942230513766
231 -3.2625782180279
232 -3.3349609847099
233 -3.4116911692337
234 -3.4931243885281
235 -3.5796571648968
236 -3.6717329337628
237 -3.7698491441095
238 -3.8745656906265
239 -3.9865149782539
240 -4.106414
241 -4.2350789139802
242 -4.373442744468
243 -4.5225770168713
244 -4.6837183857538
245 -4.8583016539398
246 -5.0480010467544
247 -5.2547822538187
248 -5.480968664426
249 -5.7293265276795
250 -6.0031756610456
251 -6.3065351198844
252 -6.6443174238784
253 -7.022591334367
254 -7.4489431735921
255 -7.93298267207
256 -8.4870656081401
257 -9.1273499739255
258 -9.875380224769
259 -10.760535665251
260 -11.823947867384
261 -13.125029701556
262 -14.752901141357
263 -16.847608616174
264 -19.642563717239
265 -23.557908352002
266 -29.43392763131
267 -39.231288532668
268 -58.831994555752
269 -117.64606831604
270 -1.117750746919E+16
271 117.64606831605
272 58.831994555752
273 39.231288532668
274 29.433927631311
275 23.557908352002
276 19.642563717239
277 16.847608616174
278 14.752901141357
279 13.125029701556
280 11.823947867384
281 10.760535665251
282 9.8753802247691
283 9.1273499739255
284 8.4870656081401
285 7.93298267207
286 7.4489431735921
287 7.022591334367
288 6.6443174238784
289 6.3065351198844
290 6.0031756610456
291 5.7293265276795
292 5.480968664426
293 5.2547822538187
294 5.0480010467544
295 4.8583016539398
296 4.6837183857538
297 4.5225770168713
298 4.373442744468
299 4.2350789139802
300 4.106414
301 3.9865149782539
302 3.8745656906265
303 3.7698491441095
304 3.6717329337628
305 3.5796571648968
306 3.4931243885281
307 3.4116911692337
308 3.3349609847099
309 3.2625782180279
310 3.1942230513766
311 3.129607107396
312 3.0684697135179
313 3.0105746879096
314 2.9557075640553
315 2.9036731857594
316 2.8542936162206
317 2.8074063144199
318 2.7628625398569
319 2.7205259530347
320 2.6802713843081
321 2.6419837480088
322 2.6055570823145
323 2.570893698276
324 2.5379034238784
325 2.5065029310671
326 2.4766151353939
327 2.4481686593938
328 2.4210973520319
329 2.3953398575963
330 2.370839228304
331 2.347542575636
332 2.3254007560651
333 2.3043680873945
334 2.2844020923972
335 2.2654632668592
336 2.2475148694826
337 2.2305227314112
338 2.2144550834088
339 2.1992823989505
340 2.184977251692
341 2.1715141859572
342 2.1588695990416
343 2.1470216342617
344 2.1359500838071
345 2.1256363005512
346 2.1160631180766
347 2.1072147782501
348 2.099076865761
349 2.0916362490989
350 2.0848810275101
351 2.0788004835238
352 2.0733850406882
353 2.0686262262009
354 2.0645166381578
355 2.0610499171829
356 2.0582207222335
357 2.0560247104094
358 2.0544585206212
359 2.053519761006
360 2.053207

Last edited:
You made a mistake somewhere with the equation. You should never get an negative answer. The limits for cos$\theta$ are -1 and 1. Thus the limits of your answer will come out between:

$$r = a \left ( \frac{1-e^2}{1+ e }} \right )$$

and

$$r = a \left ( \frac{1-e^2}{1- e }} \right )$$

Last edited:
Okay, here's your problem. Instead of adding e times cos $\theta$ to 1 like you should, you are adding 1 to e and multiplying this sum by cos $\theta$.

ohhhhhhhhhhhhhhh its a good thing my webcam is off I have a big red face.

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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