# Riddle #1

## An NEO with a peculiar orbit

The recently discovered asteroid 2020 HY5, firstly observed by Mt. Lem­ mon Survey on 23 April 2020, has an interesting particularity: it roughly passes half of its orbital period of about 387 days inside 1.3 au and the other half outside. Such distance is used for the definition of NEOs: the perihelion distance of an NEO must be below 1.3 au. 2020 HY5 actually spends 192.0 days below 1.3 au and 194.9 days above that distance.

And here is a riddle:

• Assuming an NEO that spends exactly 50% of its time inside 1.3 au and 50% of its time outside 1.3 au, what would be the maximum aphelion such an NEO could have?
• As a bonus, would you be able to find similar cases in our database? (Hint: you can use the advanced search functionality in our left menu)

## Solution by Daniel Estévez



In [1]:

%matplotlib inline
import numpy as np
import matplotlib.pyplot as plt
import pandas as pd



We assume that the NEO orbit is described by a Keplerian elliptical orbit. If we write $r_0 = 1.3\ \mathrm{au}$, the condition that the object spends exactly half of the time inside $r_0$ (and half the time outside $r_0$) can be rewritten by saying that when the object's mean anomaly $M$ equals $M_0 = \pi/2$, then the distance between the object and the Sun $r$ equals $r_0$. Indeed, since the mean anomaly grows uniformly in time, an object satisfying such condition would spend half of the orbital period having mean anomaly $M$ with $|M| < \pi/2$ and inside $r_0$, and half of the orbital period having mean anomaly $M$ with $|M| > \pi/2$ and outside $r_0$ (here we assume the mean anomaly to range between $-\pi$ and $\pi$).

By using Kepler's formula for the eccentric anomaly $E$, we see that when $M = M_0$, the eccentric anomaly $E_0$ is given as the solution of the equation $$M_0 = E_0 - \varepsilon \sin E_0,$$ where $\varepsilon$ is the eccentricity.

The formula for the heliocentric distance then yields the condition $$r_0 = a(1 - \varepsilon \cos E_0),$$ where $a$ is the semi-major axis. Now, the aphelion radius for such an orbit is $$a(1 + \varepsilon) = r_0 \frac{1+\varepsilon}{1 - \varepsilon \cos E_0}.$$

Since $r_0$ is fixed, we want to maximise $$f(\varepsilon, E_0) = \frac{1+\varepsilon}{1 - \varepsilon \cos E_0}$$ subject to the condition $$E_0 - \varepsilon \sin E_0 = M_0.$$

I haven't been able to find an analytic formula for the solution to this problem, but the solution can be approximated numerically, as we do below.

First note that $\varepsilon$ can be computed in terms of $E_0$ by $$\varepsilon = \frac{E_0 - M_0}{\sin E_0},$$ since $M_0 = \pi/2$ is fixed. As long as $\sin E_0 \neq 0$ this formula makes sense, but not all the values for $E_0$ are acceptable, since we need $0 \leq \varepsilon < 1$. Below we plot $\varepsilon$ in terms of $E_0$ for the acceptable range of parameters.



In [2]:

M0 = np.pi/2
E0 = np.linspace(1e-3, np.pi, 10000)
eps = (E0 - M0)/np.sin(E0)

plt.figure(figsize = (10,6), facecolor = 'w')
plt.plot(E0, eps)
plt.xlabel('$E_0$')
plt.ylabel('$\\varepsilon$')
plt.ylim((0,1))
plt.title('Eccentricity in terms of eccentric anomaly at $M = \\pi/2$');






Now that we have $\varepsilon$ in terms of $E_0$, we can compute and plot $f$ in terms of $E_0$. Below we plot $f$ and its maximum.



In [3]:

f = (1 + eps)/(1 - eps * np.cos(E0))
plt.figure(figsize = (10,6), facecolor = 'w')
plt.plot(E0, f)

idx = np.argmax(f)
plt.plot(E0[idx], f[idx], 'o', color = 'red')

plt.xlabel('$E_0$')
plt.ylabel('$f(\\varepsilon, E_0)$')
plt.title('Aphelion / critical radius ratio in terms of eccentric anomaly at $M = \\pi/2$');






Below we print out our results. We see that the maximum is attained at an eccentricity of approximately 0.55 (and this is indeed a valid value for $\varepsilon$) and that the aphelion is 1.23 times $r_0$.



In [4]:

print('Configuration where maximum is attained')
print('Eccentricity =', eps[idx])
print('E0 = ', E0[idx])
print('Aphelion / r_0 = ', f[idx])




Configuration where maximum is attained
Eccentricity = 0.5522225779735035
E0 =  2.0586080081674902
Aphelion / r_0 =  1.2330741623724644



The maximum aphelion that an object as described in this riddle can have is shown below.



In [5]:

f[idx] * 1.3




Out[5]:

1.6029964110842037



For the bonus part of the question there is some ambiguity about what "similar cases" means. If we understand this as objects that spend approximately half the time inside 1.3au and have an aphelion as large as possible given this condition, then a search in the database with the criteria that the eccentricity must be between 0.5 and 0.6 and the aphelion between 1.5 and 1.6 shows the following objects. This yields the objects we want, because the property that the object spends half the time inside 1.3au is completely determined by the relation between the eccentricity and the aphelion.

Designator Eccentricity Aphelion (au)
2001XX4 0.556838 1.5669
2006EK53 0.517586 1.5559
2007DA 0.573595 1.5894
2007US 0.577134 1.5121
2008LC2 0.515329 1.5285
2009UM1 0.59611 1.5344
2011CK22 0.565387 1.5757
2012VE37 0.536756 1.5985
2013AH76 0.546486 1.5228
2016GG216 0.515344 1.528
2016VU2 0.547716 1.5271
2017SK21 0.506605 1.5429
2018CC1 0.500576 1.5917
2018BB5 0.592965 1.5233
2019QZ4 0.560951 1.5978
2019SM2 0.5503 1.5186
2019EO 0.531979 1.523
2019GD20 0.536367 1.5772
2019SC8 0.530063 1.5955
2019YR6 0.516992 1.5625
3753 Cruithne 0.514844 1.5114
88213 2001AF2 0.595312 1.5219
141531 2002GB 0.528936 1.5168
250577 2005AC 0.517868 1.5936
439437 2013NK4 0.549901 1.5856
481025 2004VA1 0.51237 1.59
511684 2015BN509 0.568469 1.5787
512234 2015VO66 0.599967 1.5234
512244 2015YE18 0.540916 1.5985
539856 2017EV2 0.55758 1.5735

If we understand "similar objects" just as objects that spend half the time inside 1.3au, then the search is more complex. We need to pull out from the database the eccentricity and aphelion for all the objects and select those whose eccentricity and aphelion divided by 1.3au lies close to the curve depicted below.

This curve shows the relation between the eccentricity $\varepsilon$ and the quotient between the aphelion and $r_0$ for those objects that spend half the time inside $r_0$.



In [6]:

plt.figure(figsize = (10,6), facecolor = 'w')
plt.plot(eps, f)
plt.xlim((0,1))
plt.ylim((1,1.25))
plt.xlabel('$\\varepsilon$')
plt.ylabel('$f(\\varepsilon, E_0)$')
plt.title('Aphelion / critical radius ratio in terms of eccentricity');






By selecting all objects in the database with eccentricity between 0 and 1 and aphelion between 0 and 10000, we can get a CSV file listing the eccentricity and aphelion of all the objects. Here we load this file.



In [7]:

r0 = 1.3
objects = pd.read_csv('export_1591718497855.csv', names = ['Name', 'Eccentricity', 'Aphelion'])
objects




Out[7]:

.dataframe tbody tr th:only-of-type {
vertical-align: middle;
}

.dataframe tbody tr th {
vertical-align: top;
}

text-align: right;
}

Name
Eccentricity
Aphelion

0
433 Eros
0.222951
1.7831

1
719 Albert
0.546558
4.0808

2
887 Alinda
0.570332
3.8846

3
1036 Ganymed
0.533046
4.0851

4
1221 Amor
0.435285
2.7550

...
...
...
...

22901
539063 2016MK1
0.715769
2.5753

22902
539694 2016TE93
0.175686
1.2173

22903
539856 2017EV2
0.557580
1.5735

22904
539940 2017HW1
0.458870
3.1183

22905
543319 2013YE38
0.624424
3.3568

22906 rows × 3 columns



Now we select only those objects which are close enough to the eccentricity-aphelion curve.



In [8]:

tolerance = 0.01
good_objects = objects[np.abs(np.interp(objects['Eccentricity'], eps, f) - objects['Aphelion']/r0) < tolerance]
good_objects




Out[8]:

.dataframe tbody tr th:only-of-type {
vertical-align: middle;
}

.dataframe tbody tr th {
vertical-align: top;
}

text-align: right;
}

Name
Eccentricity
Aphelion

48
1993KA
0.197658
1.5037

235
1999RJ33
0.189328
1.4924

377
2000PN8
0.217068
1.5241

391
2000WN148
0.173545
1.4853

496
2001XP31
0.385333
1.5881

...
...
...
...

22544
506590 2005XB1
0.418766
1.6050

22600
512244 2015YE18
0.540916
1.5985

22664
518507 2006EE1
0.276396
1.5350

22840
528284 2008SW11
0.408322
1.5972

22892
537395 2015LG2
0.181574
1.4881

259 rows × 3 columns



Below we plot the eccentricity and aphelion of those satellites in relation to the curve. We see that most objects have eccentricities around 0.2-0.3, away from the maximum at 0.55.



In [9]:

plt.figure(figsize = (10,6), facecolor = 'w')
plt.plot(eps, f)
plt.xlim((0,1))
plt.ylim((1,1.25))
plt.xlabel('$\\varepsilon$')
plt.ylabel('$f(\\varepsilon, E_0)$');
plt.plot(good_objects['Eccentricity'], good_objects['Aphelion']/r0, '.')
plt.title('Objects that spend approximately half the time inside 1.3au');