Question

Halley’s comet last passed perihelion on February 9, 1986. The famous comet’s orbit has semi-major axis a = 17.8 AU and eccentricity e = 0.967.

Solve Kepler’s equation and calculate the quantities E (radians), v = teta (degrees), and the length of the radius vector r (from the sun, in AU) for July 20,2000. For simplicity, take the time of day for all dates to be exactly noon, local time.

How does your value of r compare with the actual current position of the comet (no need to compare the current values of E and teta)? What might be a reason for any difference?

Answer 1

For halley's comet
semi major axis is a = 17.8 AU
eccentrity is e = 0.967

a. E is eccentric anomaly
time of perihilion passage, T = Feb 9 , 1986
so,
r = rp(1 - e*cos(E))/(1 - e)
where rp = a(1 - e)

also, E - e*sinE = 2*pi(t - T)/T'
where T'is time priod of the orbit of halleys comet
now, T'= 2*pi(sqrt(a^3/GM))
now, G = 6.67*10^-11, M = mass of sun = 1.989*10^30 kg
a = 17.8 AU = 2.6628*10^12 m
hence,
T'= 2370322190.175975549043 s = 27434.2846085 days = 75.1109777 years
so mean anomaly, M = (2*pi/T')(t - T)
todays date is 20 july 2020
number of years passed , t - T = 12580 days = 34.44216290212183 years
hence,
M = 2.881156636762800858

hence, eccentric anomaly, E
E - e*sin(E) = M = 2.8804695564837636245344919915346
E - 0.967*sin(E) = 2.88046955648376362453449199153
solving this, we get E = 3.008 radians = 172.34570477535162 deg

now, v is true anomaly
now,
tan(v/2) = sqrt((1+e)/(1-e))*tan(E/2)
v = 2*arctan(115.4106937361624) = 179.0071224 deg

and r = rp(1 - e*cos(E))/(1 - e) = a(1 - e*cos(E))
r = 17.8(1 - 0.969*cos(172.34570477535162 ))
r = 34.89451448522130 AU
so length of r vector on 20 July 2020 will be 34.894514485221309658 AU
current distacne from web, for halleys comet is 35.93161 AU ( for 20 July)
% diff = (35.93161 - 34.894514485221309658)*100/35.93161 = 2.88630%
which is less than 3% so it is within experimental bounds

the reason we have this error is becasue our solutions to irrarional equations are approximate