IAO 2006, Практически тур, Група β — Задача 7. A Binary Radio Pulsar.
Practical round. Problems to solve · 10 ноември 2006 г.
Source annotations
The arc in 7.8 is DPB; an adjacent overwritten mark is not a further point. The Russian alternate explicitly annotates DPB. The original graph-paper orbit and the separate schematic are both preserved.
Archive workbook
The supplied workbook contains only the table and graph for 7.1. The complete chart is retained from a derived chart appendix because the original print layout splits it across two pages. Original workbook values and chart are unchanged.
Условие
A radio pulsar is a rapidly spinning neutron star emitting beams of radio waves from its magnetic poles. These are observed on Earth as series of pulses separated by the apparent pulse period . When the pulsar is a member of a binary system, this differs from the true pulse period () of the pulsar. The pulsar 0514-40 was studied with the GMRT, India in the year 2004. The true pulse period of this pulsar is ms.
Table gives data of as a function of time in units of orbital period, ( days). Assume the observer is located in the orbital plane of the pulsar.
7.1. Make a graph of this data set (scale the graph properly).
7.2. If the orbit had been circular, shape of the data curve would not be the same as the graph you have just drawn. Sketch the shape of the data curve for a circular orbit.
A graph of the elliptical orbit of the pulsar with eccentricity 0.866 is provided to you (for the pulsar 0514-40, the eccentricity of orbit is very close to this value). The arrow on the curve indicates the pulsar is moving anticlockwise in its orbit around the focus O. AP and BD are perpendicular to each other; AP is the major axis of the ellipse.

7.3. Taking at A as unit speed, calculate the transverse speeds at positions B, P and D (transverse speed of the pulsar at any point G is the magnitude of the component perpendicular to OG).
7.4. On the graph of orbit provided to you, a tangent to the curve at point B is shown on the graph. Using the graph of the orbit or otherwise, determine the total speed at B, P and D (, and respectively) in the same unit of speed.
7.5. From the shape of the versus data, infer whether the observer is located close to the major axis, or minor axis of the orbit and on which side.
7.6. Assume that the velocity component parallel to the major axis is maximum at the points B and D. Mark the points A, B, P and D on the graph of the data drawn by you.
7.7. Sketch roughly, on the same graph paper, the vs. curve if the observer was on the axis perpendicular to that in 7.5.
7.8. From the orbital graph, estimate the fraction of the total orbital period taken to travel the arc DPB. Estimate the same from the graph of the data you have drawn.
7.9. Draw a line on your graph of the data. Measure the areas under the curve you have drawn on the upper and lower sides of this line and use it to give an estimate of the major axis in light seconds.
Table of data
| (ms) | |
|---|---|
| 0.136661 | 4.990905 |
| 0.211722 | 4.991053 |
| 0.268267 | 4.991236 |
| 0.343782 | 4.991742 |
| 0.373530 | 4.992284 |
| 0.376936 | 4.992207 |
| 0.384599 | 4.990415 |
| 0.386994 | 4.989791 |
| 0.406418 | 4.989308 |
| 0.410888 | 4.989361 |
| 0.415199 | 4.989412 |
| 0.452397 | 4.989698 |
| 0.465116 | 4.989770 |
| 0.503379 | 4.989925 |
| 0.671385 | 4.990276 |
| 0.774999 | 4.990418 |
| 0.986004 | 4.990680 |

Note: Use the figure of the elliptical orbit drawn on the graph paper. Here it is drawn only schematically.
Решение
Внимание: Непълно решение
The archive workbook supplies the 17-row period table and its graph, corresponding to 7.1. It contains no answers or derivations for 7.2–7.9. Its two other sheets are empty.
| 0.136661 | 4.990905 |
| 0.211722 | 4.991053 |
| 0.268267 | 4.991236 |
| 0.343782 | 4.991742 |
| 0.373530 | 4.992284 |
| 0.376936 | 4.992207 |
| 0.384599 | 4.990415 |
| 0.386994 | 4.989791 |
| 0.406418 | 4.989308 |
| 0.410888 | 4.989361 |
| 0.415199 | 4.989412 |
| 0.452397 | 4.989698 |
| 0.465116 | 4.989770 |
| 0.503379 | 4.989925 |
| 0.671385 | 4.990276 |
| 0.774999 | 4.990418 |
| 0.986004 | 4.990680 |

Оригинал в Архива: betapract1.jpg · решение от архива: practsol.xls · допълнителни данни: betapract2.jpg · допълнителни данни: betapractdata1.jpg · допълнителни данни: betapractdata2.jpg